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Lesson 4 Tables, Equations, and Graphs of Functions Let’s connect equations and graphs of functions. 4.1: Notice and Wonder: Doubling Back What do you notice? What do you wonder? 4.2: Equations and Graphs of Functions The graphs of three functions are shown. 1. Match one of these equations to each of the graphs. 1. $d=60t$, where $d$ is the distance in miles that you would travel in $t$ hours if you drove at 60 miles per hour. 2. $q = 50-0.4d$, where $q$ is the number of quarters, and $d$ is the number of dimes, in a pile of coins worth $12.50. 3. $A = \pi r^2$, where $A$ is the area in square centimeters of a circle with radius $r$ centimeters. 2. Label each of the axes with the independent and dependent variables and the quantities they represent. 3. For each function: What is the output when the input is 1? What does this tell you about the situation? Label the corresponding point on the graph. 4. Find two more input-output pairs. What do they tell you about the situation? Label the corresponding points on the graph. A function inputs fractions $\frac{a}{b}$ between 0 and 1 where $a$ and $b$ have no common factors, and outputs the fraction $\frac{1}{b}$. For example, given the input $\frac34$ the function outputs $\frac14$, and to the input $\frac12$ the function outputs $\frac12$. These two input-output pairs are shown on the graph. Plot at least 10 more points on the graph of this function. Are most points on the graph above or below a height of $0.3$? Of height $0.01$? 4.3: Running around a Track 1. Kiran was running around the track. The graph shows the time, $t$, he took to run various distances, $d$. The table shows his time in seconds after every three meters. │$d$ │0│3 │6 │9 │12 │15 │18 │21 │24 │27 │ │$t$ │0│1.0│2.0│3.2│3.8│4.6│6.0│6.9│8.09 │9.0│ 1. How long did it take Kiran to run 6 meters? 2. How far had he gone after 6 seconds? 3. Estimate when he had run 19.5 meters. 4. Estimate how far he ran in 4 seconds. 5. Is Kiran's time a function of the distance he has run? Explain how you know. 2. Priya is running once around the track. The graph shows her time given how far she is from her starting point. 1. What was her farthest distance from her starting point? 2. Estimate how long it took her to run around the track. 3. Estimate when she was 100 meters from her starting point. 4. Estimate how far she was from the starting line after 60 seconds. 5. Is Priya's time a function of her distance from her starting point? Explain how you know. Here is the graph showing Noah's run. The time in seconds since he started running is a function of the distance he has run. The point (18,6) on the graph tells you that the time it takes him to run 18 meters is 6 seconds. The input is 18 and the output is 6. The graph of a function is all the coordinate pairs, (input, output), plotted in the coordinate plane. By convention, we always put the input first, which means that the inputs are represented on the horizontal axis and the outputs, on the vertical axis. • dependent variable A dependent variable represents the output of a function. For example, suppose we need to buy 20 pieces of fruit and decide to buy apples and bananas. If we select the number of apples first, the equation $b=20-a$ shows the number of bananas we can buy. The number of bananas is the dependent variable because it depends on the number of apples. • independent variable An independent variable represents the input of a function. For example, suppose we need to buy 20 pieces of fruit and decide to buy some apples and bananas. If we select the number of apples first, the equation $b=20-a$ shows the number of bananas we can buy. The number of apples is the independent variable because we can choose any number for it. • radius A radius is a line segment that goes from the center to the edge of a circle. A radius can go in any direction. Every radius of the circle is the same length. We also use the word radius to mean the length of this segment. For example, $r$ is the radius of this circle with center $O$.	{"url":"https://im-beta.kendallhunt.com/MS_ACC/students/2/6/4/index.html","timestamp":"2024-11-11T07:14:02Z","content_type":"text/html","content_length":"89751","record_id":"<urn:uuid:c2e13e92-505a-4acb-92b4-84de53d97998>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00425.warc.gz"}
Carbon Fiber Weight Calculator Home » Simplify your calculations with ease. » Weight Calculators » Carbon Fiber Weight Calculator The Carbon Fiber Weight Calculator is a practical tool that allows users to determine the weight of carbon fiber materials based on their dimensions and density. This calculator is essential for engineers, designers, and hobbyists working with carbon fiber in applications ranging from aerospace to sporting goods. By inputting values such as the outer and inner radii of a structure, its length, and the density of the carbon fiber material, users can quickly and accurately compute the total weight. This is vital for ensuring that components meet weight specifications, which can affect performance and safety in various applications. Formula of Carbon Fiber Weight Calculator To calculate the weight of carbon fiber, we generally use the following formula: Weight = Volume × Density • Weight = Weight of the carbon fiber (in grams or kilograms) • Volume = Cross-sectional area × Length (for a linear object) or the 3D volume for complex shapes (in cubic centimeters or cubic meters) • Density = Density of carbon fiber, typically around 1.5 to 2 grams per cubic centimeter (g/cm³) or 1500 to 2000 kilograms per cubic meter (kg/m³), depending on the type of carbon fiber If you need to find the weight of a linear carbon fiber structure (like a rod or tube), the formula for volume becomes: Volume = π × (Outer Radius² - Inner Radius²) × Length • π = Pi (approximately 3.14159) • Outer Radius = External radius of the carbon fiber structure • Inner Radius = Internal radius (zero if it is a solid object) • Length = Length of the object (in cm or m) Common Conversion Values Here’s a table with common conversions and relevant terms for users looking to quickly reference values without needing to calculate each time: Measurement Value Unit 1 cm³ (cubic cm) 1.0 g/cm³ 1 m³ (cubic meter) 1000000 cm³ 1 g/cm³ 1000 kg/m³ 1 inch 2.54 cm 1 foot 30.48 cm 1 meter 100 cm Example of Carbon Fiber Weight Calculator Let’s illustrate how the Carbon Fiber Weight Calculator works with an example. Suppose we have a carbon fiber tube with the following measurements: • Outer Radius: 2 cm • Inner Radius: 1 cm • Length: 100 cm • Density: 1.75 g/cm³ Using the formula, we can calculate the volume: 1. Calculate the Volume:Volume = π × (2² - 1²) × 100Volume = π × (4 - 1) × 100Volume ≈ 3.14159 × 3 × 100Volume ≈ 942.48 cm³ 2. Calculate the Weight:Weight = Volume × DensityWeight = 942.48 × 1.75Weight ≈ 1648.34 g Thus, the weight of the carbon fiber tube would be approximately 1648.34 grams. Most Common FAQs 1. What is the maximum density for carbon fiber? The maximum density for carbon fiber typically ranges from 2.0 g/cm³. However, it can vary depending on the specific type of carbon fiber and the manufacturing process. 2. Can I use this calculator for non-tubular shapes? Yes, the calculator can be adapted for non-tubular shapes by using the appropriate volume formula for the shape in question. You will need to measure and input the relevant dimensions. 3. How accurate is the weight calculated by this tool? The accuracy of the weight calculated depends on the precision of the measurements inputted into the calculator. It is essential to use precise measurements for the best results. Leave a Comment	{"url":"https://calculatorshub.net/weight-calculators/carbon-fiber-weight-calculator/","timestamp":"2024-11-10T15:15:03Z","content_type":"text/html","content_length":"117882","record_id":"<urn:uuid:4781e2f4-6cba-43e7-9a56-655d39f0bf59>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00086.warc.gz"}
What is 184 Fahrenheit to Celsius? - ConvertTemperatureintoCelsius.info When it comes to converting temperature from Fahrenheit to Celsius, it’s important to understand the formulas and make the calculations correctly. In this article, we’ll dive into the details of how to convert 184 Fahrenheit to Celsius, as well as provide some background information on the two temperature scales. By the end of this article, you’ll have a clear understanding of how to make this conversion without any confusion. First, let’s start by understanding the basics of the Fahrenheit and Celsius temperature scales. Fahrenheit is a temperature scale commonly used in the United States, while Celsius is used in most other countries around the world. The freezing point of water is 32 degrees Fahrenheit and 0 degrees Celsius, while the boiling point of water is 212 degrees Fahrenheit and 100 degrees Celsius. The two temperature scales are related by the formula: [°C = (°F – 32) times frac{5}{9}] Now, let’s apply this formula to the given temperature of 184 degrees Fahrenheit. By plugging the value into the formula, we can calculate the temperature in Celsius: [°C = (184 – 32) times frac{5}{9} = 152 times frac{5}{9} = 84.44] So, the temperature of 184 degrees Fahrenheit is equivalent to 84.44 degrees Celsius. This conversion can be useful in various scenarios, such as understanding weather forecasts, cooking recipes, scientific experiments, or simply understanding and comparing temperatures in different units. It’s important to note that understanding temperature conversions can be beneficial in a variety of situations. Whether you’re traveling to a country that uses the Celsius scale, following a recipe that uses different units, or working in a scientific field that requires precise temperature measurements, knowing how to convert between Fahrenheit and Celsius is a useful skill to have. In conclusion, the process of converting 184 degrees Fahrenheit to Celsius is straightforward once you understand the formula and the relationship between the two temperature scales. By using the formula °C = (°F – 32) × 5/9, you can easily make the conversion and understand the temperature in Celsius. This knowledge can be valuable in everyday life and in various professional fields, making it important to have a clear understanding of temperature conversions. So whether you’re planning a trip abroad, cooking a new recipe, or simply curious about the temperature outside, being able to convert between Fahrenheit and Celsius will give you a better understanding of the world around you.	{"url":"https://converttemperatureintocelsius.info/what-is-184-fahrenheit-in-celsius/","timestamp":"2024-11-06T02:47:27Z","content_type":"text/html","content_length":"73170","record_id":"<urn:uuid:206a96bf-2f2f-4814-a6e3-5f5571717ff3>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00644.warc.gz"}
Daily Medium Jigsaw Sudoku Puzzle for Wednesday 16th March 2016 (Medium) This is an example of a Jigsaw Sudoku. The rules of Jigsaw Sudoku are, 1. Each row must contain the numbers 1-9 once, and only once. 2. Each column must contain the numbers 1-9 once, and only once. 3. Each sub-region must contain the numbers 1-9 once, and only once. These rules are very similar to the normal Sudoku rules, but where in a normal Sudoku puzzle each region is a 3x3 square, they can be any shape in a Jigsaw Sudoku puzzle. There will always be 9 squares, but they could be any shape. This particular puzzle has a 3x3 square in the middle, but all the other regions are a non-square shape.	{"url":"https://puzzlemadness.co.uk/jigsawsudoku/medium/2016/03/16","timestamp":"2024-11-15T01:00:29Z","content_type":"text/html","content_length":"46687","record_id":"<urn:uuid:6b1646ad-ccb0-41b9-86d3-9499149a09eb>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00168.warc.gz"}
Efficient Binary Search In C: Master Faster Search TechniquesBinary Search in C Unveiling Efficiency: A Deep Dive into Binary Search in C In computer science, efficiently searching through data structures is paramount. Binary Search stands out as an efficiency champion among the various search algorithms, particularly for sorted datasets. This article delves into binary Search in C, meticulously dissecting its inner workings, applications, and potential. What is Binary Search? Binary Search, or logarithmic Search, is a robust search algorithm that excels in finding a specific element within a sorted array. It employs a divide-and-conquer strategy, repeatedly halving the search space until the target element is either located or determined to be absent. This approach significantly reduces the average number of comparisons needed compared to linear Search, translating faster search times for larger datasets faster. Why Use Binary Search? (Benefits over Linear Search) While linear Search, which examines each element sequentially, is straightforward, it can become cumbersome for extensive datasets. Binary Search shines in such scenarios, offering several compelling • Superior Time Complexity: Binary Search boasts a time complexity of O(log n), where n represents the number of elements in the array. This logarithmic complexity signifies that the search time grows proportionally to the logarithm of the data size, resulting in a significant speedup compared to linear Search’s O(n) complexity, especially for vast arrays. • Reduced Comparisons: By repeatedly dividing the search space in half, binary Search drastically minimizes the number of comparisons required to locate the target element. This efficiency becomes particularly evident when dealing with large datasets. • Suitable for Sorted Arrays: Since binary Search relies on the sorted nature of the array, it leverages this pre-condition to expedite the search process. Let’s explore binary Search in C deeper, unraveling its implementation and grasping its power! Understanding Binary Search Having grasped the essence and benefits of binary Search, let’s delve deeper into its core principle and the crucial requirement for its successful application. Core Principle: Divide and Conquer Binary Search embodies the divide-and-conquer paradigm, a problem-solving strategy prevalent in computer science. This approach tackles a complex problem by systematically dividing it into smaller, more manageable sub-problems. It then recursively solves these sub-problems and combines the solutions to solve the original problem. In binary Search, the objective is to locate a specific element within a sorted array. The divide-and-conquer strategy manifests as follows: 1. Initial Division: We begin by examining the middle element of the array. 2. Comparison and Branching: □ If the target element equals the middle element, the Search is booming, and we’ve found the target’s position. □ If the target element is less than the middle element, we know it can only reside in the left half of the array (since the array is sorted). We discard the right half and repeat the process (division) on the remaining left half. □ Conversely, if the target element is greater than the middle element, it can only exist in the right half of the array. We discard the left half and continue the search process (division) on the remaining right half. 3. Recursion and Termination: This process of dividing the search space in half and focusing on the relevant half continues recursively until either the target element is found or the entire search space is exhausted (indicating the target element’s absence). This divide-and-conquer approach significantly streamlines the Search by eliminating irrelevant portions of the array in each iteration. Choosing a Sorted Array: A Prerequisite for Binary Search It’s crucial to remember that binary search hinges on the fundamental assumption that the array it operates on is sorted in ascending or descending order. This sorted nature allows for the efficient comparisons and eliminations that drive the divide-and-conquer strategy. If the array is unsorted, binary Search will yield unpredictable results. The comparisons made during the division process would be meaningless, as the element order would need to guide the narrowing down of the search space. Therefore, ensuring a sorted array is an absolute prerequisite for successful binary search implementation. Implementing Binary Search in C Now that we’ve established binary Search’s core principles let’s translate this knowledge into practical implementation using the C programming language. Here, we’ll delve into the structure and step-by-step execution of a binary search function in C. Function Breakdown: binarySearch(array, target) We’ll define a function named binarySearch that takes two arguments: • Array: This is an integer pointer pointing to the base address of the sorted integer array where the Search will be conducted. • Target: This integer represents the specific element we aim to locate within the array. The function is responsible for searching for the target element within the provided array and returning its index in the array if found. If the target element is absent in the variety, the function should return an exceptional value, typically -1, to indicate this absence. Step-by-step walkthrough of the Algorithm Here’s a detailed breakdown of the steps involved within the binarySearch function: 1. Initialization: □ Declare variables to keep track of the search space boundaries: low (initial index) and high (final index) representing the entire array initially. □ Calculate the mid index, which represents the middle element of the current search space. This can be efficiently computed as (low + high) / 2. 2. Iterative Search Loop: □ Employ a while loop that continues when the search space (low is less than or equal to high) has yet to be exhausted. This loop embodies the iterative nature of the divide-and-conquer 3. Comparison and Branching: □ Inside the loop, compare the target element with the value at the mid-index of the array: ☆ If the target is equal to array[mid], we’ve successfully located the target element at the mid index. The Search is complete, and the function returns mid as the element’s position. ☆ If the target is less than array[mid], it signifies that the target element can only reside in the left half of the array since the array is sorted. We update the high index to mid-1 to discard the right half and focus the Search on the remaining left sub-array. ☆ Conversely, if the target is greater than the array[mid], the target element can only exist in the right half of the array. We update the low index to mid + 1 to discard the left half and continue the search process on the remaining right sub-array. 4. Termination: □ If the loop iterates through all elements without finding a match (i.e., low becomes greater than high), it signifies that the target element is not present in the array. The function returns -1 to indicate this absence. This step-by-step breakdown outlines the core logic behind the binarySearch function in C. By iteratively dividing the search space in half and focusing on the relevant portion based on comparisons, binary Search efficiently locates the target element within a sorted array. Critical Components of the Binary Search Function Having explored the function breakdown and walkthrough, let’s delve deeper into the essential components that orchestrate the binary search process in C. mid Calculation: Finding the Middle Index Accurately pinpointing the middle element (mid) within the current search space is a crucial step in each iteration of the Binary Search. This mid-index is the pivot point for dividing the search space in half for further exploration. In C, calculating the middle index can be achieved using the following expression: mid = (low + high) / 2; This expression calculates the average of low and high indices, effectively pointing to the element in the center of the current search sub-array. However, it’s essential to consider potential integer overflow issues during this calculation, especially for vast arrays. Here’s a safer alternative to prevent overflow: mid = low + (high – low) / 2; This approach calculates the difference (high – low) first and then adds it to low. This mitigates the risk of overflow if low and high values are tremendous. Recursive Calls: Dividing the Search Space The power of binary Search lies in its divide-and-conquer strategy. This strategy is often implemented using recursion, a programming technique where a function calls itself. Within the binarySearch function, when the target element is not found at the mid index, we must focus our Search on the relevant half of the array. This is achieved through recursive calls. Here’s a breakdown of the recursive approach: • If the target is less than array[mid], it resides in the left half. The function recursively calls itself with the following arguments: □ Array: The original array pointer remains unchanged. □ Target: The target element we’re still searching for. □ Updated high index: mid-1 to restrict the search space to the left half. • Conversely, if the target is more significant than the array[mid], the Search continues in the right half. The function recursively calls itself with: □ Array: The original array pointer remains unchanged. □ Target: The target element we’re still searching for. □ Updated low index: mid + 1 to restrict the search space to the right half. These recursive calls effectively divide the search space in half with each iteration, focusing on the portion where the target element might reside based on the comparisons. Base Cases: Terminating Conditions The recursive calls within the binarySearch function wouldn’t continue indefinitely. We need well-defined base cases to terminate the recursion and indicate the outcome of the Search. Here are the two primary base cases: 1. Target Found: The Search is successful if the comparison at the mid index yields a match (target is equal to array[mid]). The function returns the mid index, signifying the target element’s position in the array. There’s no need for further recursion in this scenario. 2. Search Space Exhausted: If the while loop iterates through all elements (low becomes greater than high), it implies the target element is not present in the array. The function returns -1 to indicate this absence, and the recursion terminates. These base cases ensure that the binary search process concludes gracefully by locating the target element or confirming its absence. Illustrative Example: Binary Search in Action Let’s delve into a practical example with code implementation and visualization (optional) to solidify our understanding of binary Search. Sample Code with Explanations Here’s a C code snippet demonstrating the binarySearch function: int binarySearch(int arr[], int low, int high, int target) { if (low > high) { return -1; // Target not found int mid = low + (high – low) / 2; if (arr[mid] == target) { return mid; // Target found at index mid } else if (arr[mid] < target) { return binarySearch(arr, low, mid – 1, target); // Search left half } else { return binarySearch(arr, mid + 1, high, target); // Search right half • The function takes the sorted array (arr), initial index (low), final index (high), and target element (target) as arguments. • The base case checks if low becomes greater than high, indicating the search space is exhausted, and returns -1. • The mid index is calculated safely to avoid overflow. • If the target is found at mid, the function returns mid. • Recursive calls handle searching the left or right half based on the comparison with mid. Visualizing the Search Process (Optional: ASCII Diagram) Here’s an optional ASCII diagram to illustrate the search process: Original Array: [2, 5, 8, 12, 16] Target Element: 12 Iteration 1: mid = (low + high) / 2 = (0 + 4) / 2 = 2 Compare arr[mid] (8) with the target (12) Iteration 2: The target is greater than mid; search the right half low = mid + 1 = 3 high remains 4 Iteration 3: mid = (low + high) / 2 = (3 + 4) / 2 = 3 Compare arr[mid] (12) with the target (12) Target Found at index 3! This visualization depicts how binary Search iteratively divides the search space and focuses on the relevant half until the target element is found or the search space is exhausted. Error Handling and Considerations While binary Search is a robust algorithm, it’s essential to consider potential edge cases and error scenarios to ensure robust implementation. Handling Empty Arrays or Arrays with One Element The standard binary search implementation assumes a non-empty array with at least two elements (since it relies on dividing the search space in half). Here’s how to handle empty or single-element • Empty Array: If the function receives an empty array (low will be greater than high initially), it should directly return a particular value (e.g., -1) to indicate the target element’s absence. There needs to be a point when proceeding with the search logic. • Array with One Element: A simple comparison with the target element suffices if the array contains only one component. We can check if arr[0] (the only element) equals the target. If yes, the target is found at index 0. Otherwise, the target is not present. Modifying the base case in the binarySearch function to handle this scenario can improve efficiency for single-element arrays. Edge Case: Target Element is the First or Last Element Another edge case to consider is when the target element might reside at the sorted array’s first or last position. While the standard binary search logic would eventually locate it, a slight optimization can be implemented. In the base case, before entering the recursive calls, we can perform additional checks: • If the target is equal to arr[low], the target is found at the first position (low). • Similarly, if the target is equal to arr[high], the target is found at the last position (high). These additional checks can save one recursive call if the target element happens at the array’s beginning or end. Performance Analysis of Binary Search A crucial aspect of evaluating any algorithm is its performance. To understand its efficiency, let’s delve into binary Search’s time and space complexity. Time Complexity: Best, Average, and Worst Cases (Big O Notation) The time complexity of an algorithm refers to the relationship between the input size (number of elements) and the time it takes to execute the Algorithm. Big O notation mathematically expresses this time complexity, focusing on the dominant factor as the input size grows. Binary Search boasts an exceptional time complexity of O(log n), where n represents the number of elements in the sorted array. This logarithmic complexity signifies that the search time grows proportionally to the logarithm of the data size. Here’s a breakdown of time complexity for different scenarios: • Best Case (O(1)): If the target element happens to be at the middle index (mid) in the first iteration, the comparison at mid yields a match, and the Search concludes immediately. This best-case scenario results in a constant time complexity of O(1). • Average Case (O(log n)): On average, considering a random target element within the sorted array, the binary Search needs to perform approximately log n comparisons (halving the search space) to locate the element. This translates to an average-case time complexity of O(log n). • Worst Case (O(log n)): The worst-case scenario occurs when the target element resides at the sorted array’s first or last position. In such cases, reaching the target element takes log n comparisons. However, the worst-case and average-case complexities remain the same for binary Search, which is a significant advantage. Compared to linear Search, which has a time complexity of O(n) (meaning the search time grows linearly with the number of elements), binary Search offers a substantial performance improvement, especially for large datasets. Space Complexity: Understanding Memory Usage Space complexity refers to the additional memory an algorithm requires during its execution besides the input data itself. Binary Search has a space complexity of O(1). This implies that the memory usage of the binary search function remains constant regardless of the input array size. The Algorithm primarily utilizes a few variables for indices and temporary calculations, not additional memory proportional to the input data. Therefore, binary Search excels in time and space complexity, making it an efficient choice for searching within sorted arrays. Applications of Binary Search in C The efficiency of binary Search translates into a wide range of practical applications in C programming. Here, we’ll explore real-world scenarios and their role in other algorithms. Real-World Examples: Searching Sorted Data Sets Binary Search shines in various real-world applications involving searching through sorted data sets: • Phone Book Lookup: Imagine a phone book implemented as a sorted array based on names. Binary Search allows for rapid lookups of phone numbers based on names, significantly improving search speed compared to linear Search. • Inventory Management: Inventory databases containing product information (often sorted by product ID, name, etc.) can efficiently leverage binary Search to retrieve specific product details. • Search Engines: Search engines maintain massive indexes of web pages, typically sorted by relevance. Based on search queries, a binary search can be employed to locate specific web pages within these indexes. • Data Analysis and Machine Learning: When dealing with large, sorted datasets in data analysis or machine learning tasks, binary Search can be a valuable tool for efficiently finding specific data points or features. These are just a few examples, and the potential applications extend to any scenario where you need to search through an extensive, pre-sorted data collection. Algorithmic Applications: Divide and Conquer Problems Beyond its standalone search functionality, binary Search serves as a fundamental building block for other divide-and-conquer algorithms in C: • Merge Sort: This sorting algorithm employs binary Search to efficiently find the middle element within sub-arrays during the divide-and-conquer process. • Exponentiation: Binary Search can implement an efficient exponentiation algorithm (calculating x raised to the power of y) by repeatedly squaring the base value and using binary Search to locate the appropriate exponent bit. Understanding these applications highlights the versatility of binary Search and its impact on various algorithms in C programming. Advanced Concepts in Binary Search While the core functionality of binary Search has been established, let’s delve into some advanced concepts that broaden its use and explore related search techniques. Iterative Implementation of Binary Search (Alternative to Recursion) The standard implementation of binary Search utilizes recursion to divide the search space. However, an iterative approach using a while loop can achieve the same functionality without recursion. Here’s a breakdown of the iterative approach: 1. Initialization: Similar to the recursive implementation, initialize variables for low and high and calculate the initial mid index. 2. While Loop: Employ a while loop that continues as long as low is less than or equal to high (i.e., the search space has yet to be exhausted). 3. Comparison and Updates: Inside the loop: □ Compare the target with the array[mid]. □ If a match is found (target is equal to array[mid]), the Search is booming, and the loop terminates, returning mid as the target’s index. □ If the target is less than array[mid], update high to mid-1 to focus on the left half. □ Conversely, if the target is more significant than the array[mid], update low to mid + 1 to focus on the right half. 4. Search Space Exhausted: If the loop iterates through all elements without finding a match (low becomes greater than high), it signifies the target element’s absence. The loop terminates, and the function returns -1 to indicate this. This iterative approach avoids the overhead associated with function calls in recursion, potentially improving performance for small arrays. However, the recursive approach might be more readable for larger arrays due to its concise structure. Variations of Binary Search: Interpolation Search While binary Search is highly efficient, an interpolation search might offer slight performance improvements for certain scenarios with specific data distributions. Interpolation search assumes a more uniform distribution of elements within the sorted array. It estimates the potential position of the target element based on its value and the positions of neighboring elements. This estimated position serves as the initial mid index, potentially reducing the number of comparisons needed compared to standard binary Search. However, interpolation search comes with its complexities: • Non-Uniform Distributions: If the data distribution is not uniform, interpolation search can become less efficient than binary Search. • Division by Zero: The estimation process might involve dividing by the difference between the target element and an element’s value in the array. If this difference is zero, it can lead to a division by zero error. Therefore, while interpolation can be an exciting variation, binary Search remains the more robust and widely applicable choice due to its guaranteed logarithmic time complexity and suitability for various data distributions. Optimization Techniques for Binary Search Having explored binary Search’s core concepts and applications, let’s delve into optimization techniques to enhance its C performance. Preprocessing (if applicable): Sorting Efficiency Considerations As a reminder, binary search hinges on the prerequisite of a sorted array. If the variety you’re searching through still needs to be sorted, you’ll need to sort it before applying binary Search. The efficiency of the chosen sorting algorithm can significantly impact the overall search time. Here are some considerations: • For small arrays: Insertion or selection sort might be suitable due to their simplicity. • For larger arrays: Merge sort or quicksort are generally preferred due to their O(n log n) time complexity, ensuring efficient sorting before binary Search is applied. Optimizing the sorting step, especially for large datasets, indirectly contributes to the overall efficiency of binary Search. Reducing Function Calls (if significant overhead) While recursion offers a clear and concise way to implement binary Search, it can introduce some function call overhead. This overhead might become noticeable for vast arrays. Here are approaches to potentially reduce function calls: • Iterative Implementation: As discussed earlier, an iterative implementation using a while loop can achieve the same functionality as the recursive approach without the function call overhead. This can be a viable optimization for scenarios where function call overhead is a concern. • Tail Recursion Optimization: Some compilers can optimize tail recursion, where the recursive call is the last statement in the function. The recursive approach might still be suitable if your compiler supports tail recursion optimization. However, weighing the potential performance gain from reduced function calls against code readability and maintainability is crucial. In many cases, the clarity of the recursive approach might outweigh the minor performance benefit of an iterative implementation, especially for smaller arrays. Debugging Binary Search Code Even with a solid understanding of binary Search, errors can creep into your C implementation. Here’s a guide to common mistakes, debugging strategies, and tools to rectify your binary search code. Common Mistakes and Debugging Strategies Here are some frequent pitfalls to watch out for: • Unsorted Array: Ensure the array you’re searching through is indeed sorted in ascending or descending order. Binary Search relies on this sorted nature for efficient comparisons. • Incorrect Base Cases: Double-check your base cases in the recursive function. They determine when the Search should terminate (target found or search space exhausted) and should return appropriate values (e.g., target index or -1 for absence). • Off-by-One Errors: Meticulously examine calculations involving indices, especially the mid-index calculation. Off-by-one errors can lead the Search astray. Consider using the safer mid = low + (high – low) / 2 approach to avoid integer overflow issues. • Infinite Recursion: Ensure your recursive calls have well-defined conditions to terminate. Unintended infinite recursion can occur if the base cases need to be set correctly. Debugging Strategies: • Test with Small Arrays: Test your binary search function with small, pre-sorted arrays containing known elements and target values. This allows you to step through the code manually and verify its behavior. • Print Statements: Strategically insert print statements throughout your code to print intermediate values like indices, target elements, and comparisons. This can help you pinpoint where the logic deviates from expectations. • Debugging Tools: Utilize C debuggers like GDB (GNU Debugger) to step through your code line by line, examine variable values at each step, and identify where the issue arises. Using Print Statements and Debuggers Here’s an example of how to leverage print statements for debugging: int binarySearch(int arr[], int low, int high, int target) { if (low > high) { print(“Search space exhausted\n”); // Informative print statement return -1; int mid = low + (high – low) / 2; print(“Current mid index: %d\n”, mid); if (arr[mid] == target) { return mid; } else if (arr[mid] < target) { return binarySearch(arr, low, mid – 1, target); } else { return binarySearch(arr, mid + 1, high, target); In this example, print statements are added to display informative messages and the current mid-index, aiding in tracing the execution flow and identifying potential issues. Debuggers like GDB offer more comprehensive debugging capabilities. You can set breakpoints at specific lines in your code, then execute the code line by line, examining variable values and the program’s state at each step. This allows for a more in-depth analysis of the code’s behavior and error localization. Combining these debugging techniques with a thorough understanding of binary search principles allows you to effectively troubleshoot and refine your C implementation to achieve accurate and efficient search functionality. Testing Binary Search Function Ensuring the correctness and reliability of your binary search implementation is crucial. Here, we’ll explore strategies for testing your C code using unit and integration testing approaches. Unit Testing with Sample Inputs and Expected Outputs Unit testing isolates and tests individual functions or modules within your program. This allows you to verify the binary search function’s behavior for various input scenarios and check if it produces the expected results. Here’s a breakdown of unit testing for binary Search: 1. Test Cases: Create a set of test cases encompassing different scenarios: □ Valid Search: Test cases with a target element in the array at various positions (beginning, middle, end). □ Invalid Search: Test cases with a target element not present in the array. □ Edge Cases: Test cases with empty arrays, single-element arrays, or the target element being the first or last element. 2. Expected Outputs: For each test case, determine the expected output of the binarySearch function. This could be the target element’s index (if found) or -1 (if not found). 3. Testing Framework (Optional): Consider using a C testing framework like CUnit or Google Test to automate test case execution and provide a structured testing environment. However, even without a formal framework, you can manually execute your test cases. 4. Verification: Run your test cases with the binarySearch function and compare the actual and expected outputs. Any discrepancies indicate potential errors in your code that require rectification. Here’s an example test case: // Test case: Target element present in the middle of the array int arr[] = {2, 5, 8, 12, 16}; int target = 8; int expected_index = 2; int actual_index = binarySearch(arr, 0, sizeof(arr) / sizeof(arr[0]) – 1, target); if (actual_index == expected_index) { printf(“Test case passed!\n”); } else { printf(“Test case failed! Expected index: %d, Actual index: %d\n”, expected_index, actual_index); By creating a comprehensive set of test cases and verifying their outputs, you can gain confidence in the correctness of your binary search function for various input scenarios. Integration Testing within a Larger Program Unit testing focuses on individual functions, but it’s also essential to test how the binary search function interacts with other parts of your program. This is where integration testing comes in. Here’s how integration testing applies to binary Search: 1. Integration Context: Imagine your binary search function is part of a more extensive program that reads data from a file, sorts it (if necessary), and then uses binary Search to locate specific 2. Test Driver: Develop a test driver program that simulates the interaction between the binary search function and other program components. 3. Test Scenarios: Design test scenarios that exercise the binary search function within the context of the more extensive program. This might involve testing how it handles different input data formats, sorting outcomes, and potential errors. 4. Evaluation: Execute the test driver and observe the program’s behavior. Ensure the binary search function integrates seamlessly with other components and produces the desired results. Integration testing helps uncover issues that might not be apparent during isolated unit testing. It verifies that your binary search function functions as intended when working alongside other parts of your C program. By combining unit testing and integration testing strategies, you can ensure your binary search implementation is robust and functions reliably within your more extensive C application. Comparing Binary Search with Other Search Algorithms While binary Search shines for its efficiency, it’s not a one-size-fits-all solution. Here, we’ll compare binary Search with other search algorithms and explore when each might be the better choice. Linear Search: When to Use It? Linear Search, or sequential Search, examines each element in the data structure individually until the target element is found or the entire structure is traversed. While it boasts simplicity, its O (n) time complexity makes it less efficient for large datasets than binary Search. Here’s when linear Search might be preferable: • Unsorted Data: Binary Search requires a sorted array. If your data is unsorted or sorting is not practical, linear Search is the only applicable option. • Small Datasets: For tiny datasets (e.g., arrays with a handful of elements), the simplicity of linear Search might outweigh the minor efficiency gains of binary Search. • Linked Lists: Since linked lists don’t have random access capabilities (you can’t directly jump to an arbitrary index), binary Search is not applicable. Linear Search is the standard approach for searching linked lists. Choosing the Right Search Algorithm Based on Data Structure and Needs The choice between binary Search and linear Search depends on the following factors: • Data Structure: □ Sorted Arrays: Binary Search is the clear winner for sorted arrays due to its exceptional O(log n) time complexity. □ Unsorted Arrays/Linked Lists: Linear Search is the only option for unsorted arrays and linked lists due to their structure. • Data Size: □ Large Datasets: For large sorted arrays, binary Search significantly outperforms linear Search regarding search time. □ Small Datasets: The performance difference between linear and binary Search might be negligible for tiny datasets. Here’s a table summarizing the key considerations: Factor Binary Search Linear Search Data Structure Sorted Arrays Arrays (Unsorted), Linked Lists Time Complexity O(log n) O(n) Efficiency More efficient for large datasets Less efficient for large datasets Suitability for Unsorted Data Not applicable Applicable Additional Considerations: • Hybrid Approaches: In some scenarios, a hybrid approach might be employed. For instance, you could use a combination of binary and linear Search. This could involve applying binary Search to narrow the search space to a smaller sub-array and then using linear Search within that sub-array to locate the target element. • Specialized Search Algorithms: Depending on your specific data structure and needs, more specialized search algorithms might be available. For example, hash tables offer efficient Search based on hashing techniques but have their own trade-offs and implementation complexities. By understanding the strengths and limitations of binary and linear Search, you can make informed decisions about which Algorithm to employ in your C programs based on the data structure and the size and sorted nature of the data you’re working with. Beyond Binary Search: Exploring Other Search Techniques While binary Search excels for sorted arrays, the world of search algorithms extends far beyond. Here, we’ll delve into two powerful techniques that cater to different data structures and search Hash Tables for Efficient Key-Value Lookups Hash tables, or hash maps, offer an alternative approach to searching data. They store key-value pairs, where the key uniquely identifies a value. Unlike sorted arrays, hash tables don’t require the data to be sorted in any particular order. Core Functionality: 1. Hash Function: A hash function plays a crucial role in hash tables. It takes a key as input and generates a unique index (hash value) within a fixed-size array (hash table). Ideally, the hash function should distribute keys uniformly across the table to minimize collisions (multiple keys mapping to the same hash value). 2. Collision Resolution: When collisions occur, collision resolution strategies store the key-value pair at an alternative location within the hash table. Standard techniques include separate chaining (linking elements at the same hash value) and open addressing (probing for the next available slot). 3. Search: To search for a specific key, the hash function is again used to calculate the hash value. The Search then focuses on the bucket (array position) corresponding to that hash value. The associated value is retrieved if the key is found within that bucket (using collision resolution techniques if necessary). Advantages of Hash Tables: • Average Time Complexity of O(1): In ideal scenarios with a good hash function and minimal collisions, searching, insertion, and deletion operations in a hash table have an average time complexity of O(1), making them exceptionally fast for average-case lookups. • Efficient for Unsorted Data: Unlike binary Search, hash tables don’t require the data to be sorted beforehand. Disadvantages of Hash Tables: • Worst-Case Time Complexity: In the worst case (e.g., a poor hash function leading to excessive collisions), the time complexity of hash table operations can deteriorate to O(n), similar to linear • Space Overhead: Hash tables maintain a fixed-size array and might require resizing to handle growing data volumes. Tree-Based Search Algorithms (e.g., Binary Search Tree (BST)) Another powerful approach to searching involves using tree data structures. Here, we’ll use Binary Search Trees (BSTs) as an example. Structure and Ordering: • BSTs: A BST is a self-balancing binary tree where each node has a value. The left subtree contains nodes with values less than the current node’s value, and the right subtree contains values more significant than the current node’s value. This inherent ordering property facilitates efficient searching. Search Operation: 1. Traversal: The Search starts at the root node of the BST. 2. Comparison: The target element is compared with the current node’s value. 3. Direction: □ If the target element is less than the current node’s value, the Search continues recursively to the left subtree. □ If the target element exceeds the current node’s value, the Search is recursively done to the right subtree. 4. Success or Failure: □ If the target element is found at a node, the Search is booming, and the node’s value is retrieved. □ The Search is unsuccessful if a null pointer is encountered during the traversal (indicating the end of a subtree without finding the target element). Advantages of BSTs: • Ordered Data Structure: BSTs inherently maintain sorted order, facilitating efficient searching with an average time complexity of O(log n), similar to binary Search. • Dynamic Updates: Unlike binary search arrays (which are static), BSTs allow for efficient insertion and deletion of elements while preserving the sorted order. Disadvantages of BSTs: • Performance Relies on Balancing: The search performance of BSTs can degrade if the tree becomes unbalanced (e.g., skewed heavily towards one side). Techniques like AVL and Red-Black trees address this issue by enforcing stricter balancing conditions. • Space Overhead: BSTs require additional memory compared to arrays to store pointers between nodes. Choosing Between Hash Tables and BSTs: The choice between hash tables and BSTs depends on your specific needs: • Fast Average-Case Lookups: Hash tables excel for scenarios where average-case search speed is paramount, especially for unsorted data. • Ordered Data and Dynamic Updates: BSTs are well-suited when you need to maintain sorted order within your data and frequently perform insertions or deletions. By understanding the concepts of hash tables and BSTs, you can extend your search algorithm toolkit beyond binary Search and tackle a broader range Leveraging Binary Search in C Libraries While implementing your binary search function provides a valuable learning experience, many C standard libraries offer built-in functions for efficient searching. Here, we’ll explore utilizing these library functions and considerations for custom implementations. Standard Library Functions (if available) The C standard library (usually <stdlib.h> or <cstdlib>, depending on your compiler) might provide functions for binary Search or similar functionalities. Here are some common examples: • Search function: This function performs a binary search on a null-terminated array of pointers to objects that can be compared using a user-defined comparison function. It returns a pointer to the matching element or NULL if not found. • sort function: This function implements the quicksort algorithm, which can sort an array before applying binary Search (if your array isn’t already sorted). The specific usage of these functions depends on their implementation details. It’s crucial to consult your compiler’s documentation for the exact syntax and requirements. Here’s a general outline: #include <stdlib.h> int bsearch(const void key, const void base, size_t nmemb, size_t size, int (compar)(const void , const void )); void qsort(void base, size_t nmemb, size_t size, int (compar)(const void , const void )); The search function takes arguments like the key to search for, the base address of the array, the number of elements, and the size of each component, and a comparison function is used to compare aspects during the Search. It returns a pointer to the matching element or NULL if not found. The sort function sorts an array based on a provided comparison function. This can be used to sort the array before applying binary Search. Considering Custom Implementations vs. Built-in Functions Here are some factors to consider when deciding between using a custom binary search implementation and a library function: • Availability: Not all C compilers or standard libraries might provide built-in binary search functions. Check your compiler’s documentation to see if these functions are available. • Customization: A custom implementation might be necessary if you need specific control over the search behavior or require modifications for your data structures. • Performance: For simple cases, a well-written custom implementation might be comparable to or even slightly faster than a library function due to the overhead of function calls. However, library functions might be optimized for performance for complex scenarios or large datasets. • Readability and Maintainability: Using well-tested and documented library functions can improve code readability and maintainability compared to managing your implementation. General Recommendation: In most cases, it’s recommended to leverage your C library’s built-in binary search functions if available. These functions are likely well-tested and optimized for performance. However, if you have specific requirements or need a deeper understanding of the Algorithm, implementing your binary search function can be a valuable learning experience. Real-World Challenges and Adaptations in Binary Search While binary Search is a powerful tool, real-world scenarios might present complications that require adaptations to the standard Algorithm. Here, we’ll explore some challenges and potential Handling Duplicate Elements in the Sorted Array The standard binary search implementation assumes no duplicates within the sorted array. However, you might encounter sorted arrays with duplicate elements in practical situations. Here are two approaches to handle duplicates: 1. Finding the First Occurrence: If your goal is to locate the first occurrence of the target element (even if duplicates exist), you can modify the binary search logic during the while loop: □ Don’t terminate the Search immediately if a match is found (target is equal to array[mid]). Instead, update high to mid-1 to continue searching toward the left side of the array, potentially finding an earlier occurrence. 2. Finding All Occurrences: A more involved approach is necessary if you need to find all occurrences of the target element. You can perform a standard binary search to locate the first occurrence. Then, traverse the array from the found index both left and right, comparing elements with the target element until you encounter non-matching aspects on both sides. Modifying Binary Search for Specific Search Requirements Binary Search can be adapted for various search functionalities beyond simply finding the exact match for a target element: • Finding the Index of the Closest Element: If your sorted array doesn’t contain the exact target element, you should see the index closest to the target in value (either the element less than or the element more significant than the target). You can determine the closest element’s index by analyzing the comparison results during the binary search process. • Range Search: In some scenarios, you might be interested in finding a range of elements within the sorted array that fall within a specific value range. This can be achieved by modifying the binary search logic to identify the indices of the first and last elements within the desired range. These are just a few examples; the specific adaptations will depend on your unique search requirements. By understanding the core principles of binary Search, you can creatively modify it to address various search needs within your C programs. Additional Considerations: • Error Handling: When handling duplicates or modifying binary Search for specific needs, incorporate proper error handling mechanisms to address potential edge cases and unexpected input • Readability: While adaptations can be made, it’s essential to maintain code readability and clarity. Consider using comments or meaningful variable names to explain the modifications made for future reference. By carefully considering these challenges and adaptation strategies, you can effectively utilize binary Search in various real-world applications within your C programming endeavors. The Future of Binary Search in C: Enduring Relevance and Potential Advancements Binary Search is a cornerstone search algorithm in C programming due to its efficiency and versatility. While its core principles are unlikely to undergo radical transformations, here’s a glimpse into potential advancements and its enduring role in modern C: Potential Advancements and Optimizations: • Hardware Integration: As processor architectures evolve, compiler optimizations leverage hardware capabilities to accelerate binary search operations further. This could involve instruction sets specifically suited for comparison and branching operations used extensively in binary Search. • Hybrid Search Techniques: In the future, there might be more exploration of combining binary Search with other search algorithms. For instance, integrating binary Search with techniques like fuzzy searching (finding elements similar to the target element) could broaden its applicability. • Specialized Libraries: C libraries might offer more specialized binary search implementations tailored for specific data structures or search requirements. These libraries could handle scenarios like searching within memory-mapped files or integrating with database access layers. The Enduring Role of Binary Search in Modern C Programming: Despite potential advancements, binary Search is likely to remain a fundamental tool in the C programmer’s arsenal for several reasons: • Simplicity and Efficiency: Binary Search offers a clear and concise approach to searching sorted data, boasting a time complexity of O(log n) in most cases. This efficiency makes it a compelling choice for various search tasks. • Versatility: Binary Search can be adapted to handle different search requirements, such as finding the closest element or searching within a range. This adaptability extends its usefulness beyond primary element lookups. • Foundation for Other Algorithms: Binary Search is a building block for several other algorithms in C, including sorting algorithms like merge sort and techniques like exponentiation using bitwise operations. Understanding binary Search is crucial for grasping these more complex concepts. Learning Binary Search Remains Valuable: Regardless of future advancements, a thorough understanding of binary search principles will continue to be valuable for C programmers. This knowledge equips them with: • Problem-solving skills: By grasping the divide-and-conquer approach of binary Search, programmers can apply similar strategies to solve other problems that involve efficient searching or • Algorithmic foundation: Understanding binary Search lays a solid foundation for exploring more complex C programming search algorithms and data structures. • Performance optimization: When dealing with large datasets, programmers can leverage binary Search to optimize the efficiency of their C programs by focusing on searching within sorted In conclusion, while advancements in hardware and libraries might present new possibilities, binary Search will likely remain a cornerstone search algorithm in C programming due to its simplicity, efficiency, versatility, and role as a stepping stone for understanding more intricate algorithms. By effectively utilizing and adapting binary Search, C programmers can continue to write efficient and robust programs for various applications. Conclusion: Binary Search – A Powerful Tool for Efficient Searching in C This comprehensive exploration has delved into binary Search in C programming. Here’s a recap of its essential functionalities, advantages, and practical usage scenarios: Recap of Binary Search: • Functionality: Binary Search is a highly efficient search algorithm for sorted arrays. It employs a divide-and-conquer approach, repeatedly halving the search space based on comparisons with the target element until the element is found or the search space is exhausted. • Time Complexity: The remarkable advantage of binary Search lies in its time complexity of O(log n) in the average and best cases. This logarithmic complexity translates to significantly faster search times than linear Search, especially for large datasets. • Space Complexity: Binary Search has a space complexity of O(1), meaning its space requirements remain constant regardless of the array size. This makes it memory-efficient for searching large Advantages of Binary Search: • Efficiency: The logarithmic time complexity makes binary Search exceptionally fast for searching sorted arrays, especially when dealing with large datasets. • Simplicity: The core concept of binary Search is relatively straightforward to understand and implement, making it accessible to programmers of various experience levels. • Versatility: Binary Search can be adapted to handle various search requirements beyond finding the exact target element. This includes finding the closest element, searching within a range, or handling duplicate elements (with modifications). When and How to Effectively Use Binary Search in C Programs: • Sorted Arrays: Binary Search is only applicable for sorted arrays. If your data isn’t sorted, you’ll need to sort it before applying binary Search (consider sorting algorithms like merge sort or quicksort for efficiency). • Large Datasets: The efficiency gains of binary Search become genuinely significant for searching large sorted arrays. For tiny arrays, the overhead of function calls outweighs the benefits of binary Search compared to linear Search. • Leveraging Libraries: Consider using well-tested and documented binary search functions provided by your C standard library (e.g., Search) if available. This can improve code readability and • Custom Implementations: While library functions are recommended, implementing your binary search function can be a valuable learning experience, providing a deeper understanding of the Algorithm’s inner workings. In Closing: You can significantly improve the search performance for sorted data structures by utilizing binary Search in your C programs. Remember that understanding binary Search equips you with a powerful tool and lays a foundation for exploring more complex search algorithms and data structures in C programming. Frequently Asked Questions (FAQs) about Binary Search in C What if the target element is not present in the array? If the target element is not present in the sorted array, the binary search function typically returns a value indicating the absence. This value can vary depending on the implementation, but common approaches include: • Returning a unique value like -1 to signal that the element was not found. • Returning the index where the element would be inserted to maintain the sorted order if it were present in the array. Can binary Search be used on unsorted arrays? No, binary Search cannot be directly applied to unsorted arrays. The core principle of binary Search relies on repeatedly halving the search space based on comparisons with the target element, which only works if the elements are sorted in ascending or descending order. If your data is unsorted, you’ll need to employ a different search algorithm like linear Search, which examines each element in the array one by one until the target element is found or the entire array is traversed. How does binary Search compare to other algorithms regarding time complexity? Here’s a comparison of time complexities for standard searching algorithms: • Binary Search: O(log n) – This is the most efficient option for searching sorted arrays due to its logarithmic time complexity. The search time grows proportionally to the logarithm of the array size, making it significantly faster for large datasets than linear Search. • Linear Search: O(n) – Linear Search examines each element in the array individually. In the worst case (target element not present or at the end of the array), it needs to traverse the entire array, resulting in linear time complexity. • Hash Tables (Average Case): O(1) – In ideal scenarios with a good hash function and minimal collisions, searching, insertion, and deletion operations in a hash table have an average time complexity of O(1). This makes them exceptionally fast for average-case lookups, especially for unsorted data. However, the performance can degrade in the worst case (e.g., poor hash function leading to excessive collisions). In summary: • Binary Search reigns supreme for efficient searching for sorted arrays due to its logarithmic time complexity. • For unsorted arrays or situations where sorted order isn’t maintained, linear search or hash tables might be more suitable depending on your needs and performance requirements. 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Fundamental Concepts of Reinforcement Learning - Sunforger People gain knowledge of the world and receive feedback through interaction with the world (environment). Image source: https://storage.googleapis.com/deepmind-media/UCL%20x%20DeepMind%202021/Lecture%201%20-%20introduction.pdf Following this paradigm, people have proposed methods for reinforcement learning. First, let's introduce several concepts in reinforcement learning. Agent, which can be understood as a decision-maker or decision-maker. Environment, which is the environment. The agent interacts with the environment to obtain feedback. The word "feedback" is very interesting and has rich connotations. Specifically, every decision of the agent, or every action, will incur a cost. For example: It will cause a change in the state of the environment. At this time, the observation of the agent to the environment will change. Based on the obtained environmental observation, we construct the agent state (Agent State). For Fully Observable Environments, we can consider Agent State = Observation = environment. In general, unless otherwise specified, all environments are assumed to be fully observable. The cost incurred by taking an action is not limited to the change in the environment state. The action itself also has a good or bad distinction. We measure the quality of the decision/action with a If an action has a positive reward, it means that at least in the short term, this action is good. Conversely, if the reward is negative, it means that the action is not wise. The following figure intuitively shows the principle of reinforcement learning. After all, what is reinforcement learning? The goal of reinforcement learning is to optimize the agent's action strategy through continuous interaction, by selecting better actions to maximize the sum of rewards (including the rewards of each We generally use Markov Decision Processes to model reinforcement learning. Before proceeding, please read The relationship between Markov and Reinforcement Learning (RL) - Zhihu Here is a schematic diagram of a Markov process. Image source: https://towardsdatascience.com/markov-chain-analysis-and-simulation-using-python-4507cee0b06e There are a few points to note: 1. A Markov process (Markov Process) consists of a pair of binary $M=(S, P)$. A Markov decision process (Markov Decision Process) consists of a quadruple $M=(S,A,P,R)$. Sometimes, a discount factor $\gamma$ for reward is also included, which will be mentioned later. 2. Due to randomness, given the initial state and the final state, the Markov process (Markov Process) corresponding to the Markov chain (Markov Chain) is not unique. Important Concepts# During the interaction between the agent and the environment, at a certain time t: 1. Based on the observation of the environment $O_t$ and the received incentive $R_t$, the agent constructs the agent state $S_t$. (Usually, unless otherwise specified, $S_t=O_t$), and decides how to take action (submit $A_t$ to the environment). 2. The environment receives the action $A_t$ submitted by the agent and needs to bear the "cost" brought by $A_t$. It then provides updated $O_{t+1}$ and $R_{t+1}$ to the agent as feedback. This process continues. The interaction between the individual and the environment produces the following interaction trajectory, which we denote as $\mathcal H_t$. This interaction trajectory stores the observation, action, and reward at each interaction. $\mathcal H_t = O_0, A_0, R_1,O_1, A_1,\cdots, O_{t-1}, A_{t-1}, R_t, O_t$ Based on this sequence $\mathcal H_t$, we can construct the agent state $S_t$. When the environment satisfies the Fully Observable property, we consider Agent State $S_t = O_t$. Therefore, we can replace all the O symbols in the equation $\mathcal H_t$ with the S symbol. (Many materials also directly use the S symbol) At this time, there is no need to use $\mathcal H_t$ to construct $S_t$. We can directly use $O_t$ as $S_t$. We determine what action to take based on the state and abstract it into a policy function $\pi$. The policy function $\pi$ takes the state as input and outputs the corresponding action, denoted as $ \pi(a\|s)$. Sometimes it is abbreviated as $\pi$. We assume that the state space S is discrete and can only have \|S\| different states. Similarly, we assume that the action space A is discrete and can only have \|A\| different actions. In this setting, how should we understand the policy function $\pi(a\|s)$? The system is currently in state $s$, where $s\in S$. Under the condition of state $s$, the policy function $\pi(a\|s)$ considers what action (which a) should be taken. The policy function can be understood as a class of composite functions. In addition to random policies, we can generally consider that the policy function includes two parts: action evaluation and action selection. Action evaluation is generally done using Q value. Action selection is generally done using argmax or $\epsilon$-greedy. We will discuss this in more detail later. Through the interaction between the individual or agent and the environment, rewards are obtained. As mentioned above, the goal of reinforcement learning is to maximize the total reward by selecting actions (i.e., finding a better policy) through interaction. We define the total reward (also known as return or future reward) as $G_t$. $G_t = \sum_{k=0}^\infty {\color{red} \gamma^k}R_{t+k+1} = R_{t+1} + {\color{red} \gamma } R_{t+2} + {\color{red} \gamma^2}R_{t+3} + \cdots + {\color{red} \gamma^k}R_{t+k+1} + \cdots$ A few points to note: 1. The total reward, or the sum of rewards, should start from $R_1$, but why does it start from $R_{t+1}$? Because the values of $R_1 \sim R_t$ are fixed constants and cannot be optimized. Therefore, we focus more on future rewards. 2. The $\gamma$ in the equation is the discount factor mentioned above. Usually, the range of the discount factor is limited to $0<\gamma<1$. If the value of $\gamma$ is small, close to 0, as k increases, $\gamma^k$ will become smaller and smaller, that is, the weight of $R_{t+k+1}$ will become smaller and smaller. This means that we are more inclined to consider short-term effects rather than long-term effects. If the value of $\gamma$ is close to 1, it means that we will take into account long-term effects more. We define the state value function (also known as value function or value) as the expected cumulative return obtained by starting from state $s$ and following policy $\pi$. The state value function is used to measure how "good" a state $s$ is, and is defined as follows: \begin{align} V^\pi(s) &= \mathbb E_\pi[G_t\|S_t=s] {\color{red}=\mathbb E_\pi[R_{t+1}+R_{t+2}+\cdots\|S_t = s] } \\ &=\mathbb E_\pi [R_{t+1}+G_{t+1}\|S_t = s] \\ &=\mathbb E_\pi[R_{t+1}+V^\pi(S_{t+1}) \|S_t = s]\\ &={\color{red} \sum_a\pi(a\|s)\sum_{s^\prime}p_{ss^\prime}^a \left[ r_{ss^\prime}^a + \gamma V^\pi(s^\prime) \right] } \end{align} The first line of the equation is the definition of the state value function. The second and third lines of the equation are the recursive form obtained by expanding the return $G_t$ according to the definition, or the form of the Bellman Equation. The fourth line expands the Bellman Equation, and we need to pay attention to the $p_{ss^\prime}^a$ and $r_{ss^\prime}^a$ in the equation. $p_{ss^\prime}^a$ is the state transition probability, which should be described in the Markov Decision Process. Specifically: When we are in state s and choose an a as the action based on the policy function $\pi(a\|s)$, it will cause a change in the observation of the environment, so the state will also change. However, the effect caused by action a is not fixed. We cannot guarantee that state s will always change to a fixed state $s^\prime$. In other words, $s^\prime$ can be equal to state_1, state_2, state_3, or some other state_i, so it corresponds to a probability distribution $p_{ss^\prime}^a$. Similarly, we have $r_{ss^\prime}^a$. We define the action value function as the expected cumulative return obtained by starting from state $s$, taking action $a$, and following policy $\pi$. \begin{align} Q^\pi(s,a) &= \mathbb E_\pi[G_t\|S_t=s, A_t=a] {\color{red}=\mathbb E_\pi[R_{t+1}+R_{t+2}+\cdots\|S_t = s, A_t = a] } \\ &=\mathbb E_\pi [R_{t+1}+G_{t+1}\|S_t = s, A_t = a] \\ &=\mathbb E_\pi[R_{t+1}+Q^\pi(S_{t+1}, A_{t+1})\|S_t = s, A_t=a]\\ &={\color{red} \sum_{s^\prime}p_{ss^\prime}^a \left[ r_{ss^\prime}^a +\gamma \sum_{a^\prime} \pi(a^\prime\|s^\prime) Q^\pi(s^\prime, a^\prime)\ right] } \end{align} Similar to $V^\pi(s)$, no further explanation is given. We define the advantage function as the difference between Q and V. $A(s,a) = Q(s, a) - V(s)$ It represents the degree to which taking action a in state s is better or worse than following the current policy $\pi$. The main purpose of the advantage function is to optimize the policy and help the agent understand more clearly which actions are advantageous in the current state. How to understand the advantage function? - Zhihu Model-based vs. Model-free The so-called model includes the state transition probability and the reward function. If the model is known, it is model-based, and we will plan under complete information. In other words, we can use dynamic programming algorithms to learn the desired policy. Knowing the model means that when the action and state are determined, we can know the state transition probability $p_{ss^\prime}^a$ and the corresponding $r_{ss^\prime}^a$. On the contrary, if learning does not depend on the model, it is called model-free. For example, the Policy Gradient method is model-free. We will discuss this in more detail later. On-Policy vs. Off-Policy On-Policy means that the behavior policy during episode sampling and the target policy during policy optimization are the same. Off-Policy means that the two policies are different. We will discuss this in more detail later. Related Resources#	{"url":"https://xlog.yeahbt.com/reinforcement-learning-fundamental","timestamp":"2024-11-13T02:38:35Z","content_type":"text/html","content_length":"730902","record_id":"<urn:uuid:e95111ee-fe94-42f9-9469-bf03cf266067>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00035.warc.gz"}
Sample Space of Two Dice \| Learn and Solve Questions Introduction to Probability Probability is also known as a possibility of an event occurring. The probability is never negative and is never more than 1. Some of the real-life situations where probability is used are throwing dice, tossing coins, picking out students from a class, and many more. Probability has long been used in mathematics to approximate how likely events are to occur. Essentially, the probability is the degree to which something is predicted to occur. What is Dice? Dice is a tiny block with one to six marks or tints on its boundary that is used in games to generate a random number. Dice are tiny, tossable blocks having a visible border that can stop in the figures shown. When thrown or rolled, the die comes to a halt and displays a random number from one to six on its upper side, with the occurrence of each event being equally likely. The dice drawn are used for playing board games as a fun way to relax with family and friends. Possible Outcomes in a Dice What is Sample Space? A sample space is a collection of possible outcomes from a random experiment. The sign "S" represents the sample space. The events of an experiment are a subset of the possible outcomes. A sample region may have a range of results depending on the investigation. It is termed a discrete or finite sample space if it has a finite number of outcomes. Curly brackets "{ }" contain the sample spaces for a random experiment. Different Scenarios to Calculate Dice Probabilities • One dice is thrown- The likelihood of a certain integer happening with one dice is the simplest and most straightforward situation of dice probabilities. Dice shows six possible outcomes. So, the result obtained will be given as: \[P\left( A \right) = \dfrac{\text{no of the outcome of A}}{\text{no of total outcomes}}\] • Two dice are thrown - The likelihood of receiving two 6s by tossing two dice is a rare occurrence as the outcome of one die is independent of the outcome of the other dice. The rule of probability applied in such situations states that separate probabilities must be multiplied together to achieve the outcome. As a result, the formula for this is, • Probability of both \[ = \] probability of first \[ \times \] probability of the second • Questions like two or more than two dice are thrown simultaneously to find the probability of getting a number can be done using the above-mentioned formula. • The total number from two or more dice - If one wants to know the possibility of receiving a specific sum obtained by rolling two or more dice, one must use the basic rule of probability which is • Probability = the number of desired results divided by the total number of results. Sample Questions 1. A dice has how many faces or sides? a. 4 b. 5 c. 6 d. 8 Ans: 6 Explanation: A dice is a cuboid that has 6 faces or sides in it. 2. How many possible outcomes would be there if two dice are thrown? a. 6 b. 12 c. 36 d. 2 Ans: 36 Explanation: One dice has 6 possible outcomes. So, if two dice are rolled then we need to multiply 6 two times which will result in 36. The total number of outcomes in a simultaneous throw of two dice will be 36. 3. Possible outcomes that will come in an experiment are called a. Sample space b. Probability c. Possibility d. Luck Ans: Sample space Dice is a six-faced three-dimensional object which is used to play board games. When a dice is thrown there are different probabilities of getting a particular result which can be calculated by a probability formula. Sample space is all the possible outcomes that we can get in a particular situation and is useful in finding out the probability of large and complex sample space. FAQs on Sample Space of Two Dice 1. Is there a probability of getting a number on a dice more than one? The probability is never more than 1 or less than 0. So, the probability of getting a digit on a dice is 1 as the number of outcomes is 6, as well as the number we can get on a digit, is also 6. 2. What is the shape of the dice? The dice is cuboidal in shape with 6 faces having 6 digits printed on it from 1 to 6. 3. What would be the sample space if two coins are tossed? The sample space of two coins being tossed will be {HH, TT, HT, TH}. 4. Will tossing coins give fair results? Yes, tossing the coin will give fair results as both the head and tail side of the coin would have a 50-50 chance to appear after the toss. 5. Probability cannot be used in daily life. True or false? Probability can be used in our daily lives. It is used to predict the nature of the weather conditions, online shopping, online gaming, to make strategies in a game, and is used in many other things	{"url":"https://www.vedantu.com/maths/sample-space-of-two-dice","timestamp":"2024-11-10T22:56:47Z","content_type":"text/html","content_length":"243067","record_id":"<urn:uuid:072ac835-04a5-4928-b93d-04498e395840>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00024.warc.gz"}
exp, expt exp and expt perform exponentiation. exp returns e raised to the power number, where e is the base of the natural logarithms. exp has no branch cut. expt returns base-number raised to the power power-number. If the base-number is a rational and power-number is an integer, the calculation is exact and the result will be of type rational; otherwise a floating-point approximation might result. For expt of a complex rational to an integer power, the calculation must be exact and the result is of type (or rational (complex The result of expt can be a complex, even when neither argument is a complex, if base-number is negative and power-number is not an integer. The result is always the principal complex value. For example, (expt -8 1/3) is not permitted to return -2, even though -2 is one of the cube roots of -8. The principal cube root is a complex approximately equal to #C(1.0 1.73205), not -2. expt is defined as b^x = e^x log b . This defines the principal values precisely. The range of expt is the entire complex plane. Regarded as a function of x, with b fixed, there is no branch cut. Regarded as a function of b, with x fixed, there is in general a branch cut along the negative real axis, continuous with quadrant II. The domain excludes the origin. By definition, 0^0=1. If b=0 and the real part of x is strictly positive, then b^x=0. For all other values of x, 0^x is an error. When power-number is an integer 0, then the result is always the value one in the type of base-number, even if the base-number is zero (of any type). That is: (expt x 0) ==(coerce 1 (type-of x)) If power-number is a zero of any other type, then the result is also the value one, in the type of the arguments after the application of the contagion rules in Section 12.1.1.2 Contagion in Numeric Operations, with one exception: the consequences are undefined if base-number is zero when power-number is zero and not of type integer.	{"url":"https://franz.com/support/documentation/ansicl/dictentr/expexpt.htm","timestamp":"2024-11-07T04:45:39Z","content_type":"text/html","content_length":"12172","record_id":"<urn:uuid:150538ea-f354-4e3f-a5f2-e5ade368224e>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00692.warc.gz"}
The Myth of Medical Decision Limits The Myth of Medical Decision Limits Medical Decision Limits are described as a "second set of limits set for control values ...meant to be a wider set of limits indicating the range of medically acceptable results." The idea is that these medical decision limits embody the medical usefulness requirement for a test and by drawing these limits on our control charts, we will detect medically significant errors. Using CLIA QC requirements and practical examples, Dr. Westgard evaluates these MDLs and reveals their true nature. The Myth of Medical Decision Limits When I discussed myths of quality is an earlier essay, my intention was to illustrate the falsehood of some apparently well-accepted and well-documented beliefs, such as the existence of California as an island, which was documented in black and white by many reputable maps published in the 1600s. I suggested that some of our beliefs in healthcare quality assurance and laboratory quality control may also be myths, even though they appear in black and white in our professional magazines, journals, and books. One of the current myths about QC is the use of medical decision limits on quality control charts to assure test results have the quality necessary for medical or clinical usefullness. As I was contemplating a topic for this month's essay, we received an e-mail question about the proper way to implement "medical decision limits," so this seems like a timely issue to discuss. With the background materials now available on this website, this issue can be understood in greater depth than would have been possible at the time of my earlier discussion of myths of quality. Clarification of terms "Medical decision limits" (MDL) is the term used by Tetrault and Steindel in the 1994 CAP Q-Probe which reviews daily quality control exception practices [1]. MDLs are described as a "second set of limits set for control values ...meant to be a wider set of limits indicating the range of medically acceptable results." The MDL concept is referenced to an earlier paper by Steindel [2], which in turn is referenced to an earlier abstract by a CAP group [3], but neither of these earlier references provide a more objective definition. Basically, the idea is that these medical decision limits embody the medical usefulness requirement for a test and by drawing these limits on our control charts, we will detect medically significant rather than statistically significant errors. MDL should not be confused with our use of medical decision level (X[c]), which refers to a the level or concentration at which a test is critically interpreted for patient care and treatment. Our approach for dealing with a clinical quality requirement is to define a medically important change (or clinical decision interval, D[int]) at a medical decision level (X[c]), then use a clinical quality planning model to derive (translate) the medical usefulness requirement into operating specifications for imprecision and inaccuracy that are allowable and the QC that is necessary. As illustrated in an earlier cholesterol QC planning application, this QC planning process leads to the selection of QC acceptability criteria (control rules) and the number of control measurements that are appropriate for individual tests performed by individual methods in individual laboratories. This is a well-defined and quantitative process that is quick and easy to perform when supported with appropriate tools, technology, and training. Current use of MDLs The Q-Probe survey [1] revealed that MDLs were used by about 30% of laboratories, but the authors noted that only 10-25% reported MDL limits wider than their analytical limits, which they interpreted as evidence that the application of MDLs was incorrect (which should not be unexpected given the lack of information and guidelines in the scientific literature). To improve the use of MDLS, they provided the following recommendations for setting MDLs: "A good way to set medical decision limits is to set them based on either biological or medical need. You might also want to set them based on the rule system you use. For example, you may want to set an analytical limit somewhat tightly, at 2.0 SD, and the medical decision limit wider, at 3.5 SD. In this way, you could release medically acceptable results and still be warned of an impending analytical problem that you can fix at, hopefully, your leisure instead of by a deadline needed to report results." It certainly is appealing to set wider control limits and have fewer run rejections. But, do MDLs really work? Or, is the MDL a modern myth that is being passed on without any scientific merit? CLIA QC requirements and MDLs CLIA allows laboratories the flexibility to define their own QC procedures, which means they can set the control limits in any way that is appropriate, including the use of MDLs if they are valid. Originally, it was proposed that manufacturers would provide QC instructions, and when approved or validated by FDA, laboratories could follow the manufacturer's approved QC instructions. In the absence of approved QC instructions (for which the approval process was delayed from 1992 to 1994 to 1996 and now further into the future), laboratories are still responsible under CLIA to establish appropriate QC procedures, as described in the rule 493.1218(b) [4]: "...the laboratory must evaluate instrument and reagent stability and variance in determining the number, type, and frequency of testing calibration or control materials and establish criteria for acceptability used to monitor test performance during a run of patient specimen(s)." Example practices for setting MDLs In 'Walking the straight and narrow on quality control,' Passey [5] provided a detailed discussion of CLIA QC requirements and illustrated how this concept of medically useful control limits might be "Laboratories must calculate the means and standard deviations of the control values for each lot of materials used for quality control... These statistical estimates are used along with consideration for medical requirements to establish the acceptability criteria for quality control. For example, if a test's measured imprecision indicates that the method can determine glucose with an SD of 2 mg/dL but medical usefulness dictates a preferrable SD of 4 mg/dL, construct your acceptability criteria around the larger medical requirement... Carefully consider changing your acceptability criteria (out of control) from +/- 2SD to +/-3SD. Even better, use a fixed window (+/- allowable error) that reflects both medical usefulness and analytical capability." Thus, common professional practices for setting control limits on Levey-Jennings charts now include the use of statistical control limits, such as the mean plus/minus 2 or plus/minus 3 SDs, but also the use of medical decision control limits calculated from a medically allowable SD or representing a fixed error requirement, such as the CLIA proficiency testing criterion for acceptable performance. Other related practices may be to use a manufacturer's claim for method performance as the standard deviation to calculate control limits or a manufacturer's "acceptable range" as fixed control limits. Need to evaluate MDL practices Laboratories should evaluate whatever practice they follow to be sure their QC acceptability critieria are valid, particularly in the light of the FDA suggestions for validating QC procedures. In a draft document [6], FDA described a valid QC procedure as "...one that adequately maintains and monitors stated analytical performance characteristics and, at the same time, alerts the analyst to unsatisfactory performance when known and/or unknown technical variables are introduced. These procedures should adequately address the critical performance parameters of accuracy and precision within the reportable range of the test." These FDA guidelines were aimed at manufacturers, but in the absence of FDA clearance of manufacturers' QC instructions (which may be delayed indefinitely), laboratories performing moderately and highly complex tests are still responsible under CLIA for documenting their QC procedures. Comparing control chart limits [2s] rule) would be set as 96 and 104; 3 SD limits (1[3s] rule) would be 94 and 106. If it were of interest, what would be the limits for the 1[4s], 1[5s], and 1[6s] rules? The answers would be 92 and 108, 90 and 110, and 88 and 112, correct? Let's draw all these control rules on a control chart, as shown here. Now, suppose the medically allowable SD for this test (s[a]) has been defined as 4 mg/dL, which means the analytical performance appears to be better than needed for the medical use of this test. What control limits would results if this medical or clinical SD where used to calculate 2 SD and 3 SD control limits? Those limits would be 92 to 108 and 88 to 112, right? If you were to draw them on the control chart, they would be the same as the 1[4s] and 1[6s] statistical limits. Thus, these supposedly clinical limits still correspond to particular statistical control rules. Also consider that the allowable total error (TE[a]) for this test is given as 10% by the CLIA proficiency testing criterion. This means that a value of 100 must be good to within 90 to 110 units. If this total or fixed error criterion were used to set control limits, these limits would be the same as the 1[5s] statistical control rule, thus again, a supposedly fixed allowable error corresponds to a particular statistical control rule. What's the point? Any control limit, regardless of the rationale for drawing it on the chart, still corresponds to a statistical control rule. The actual performance of that control limit,or control procedure, can be assessed from the power curves for that particular statistical control rule. Given a quality requirement in the form of an allowable total error or a clinical decision interval, and given the imprecision and inaccuracy of your method, you can evaluate the performance of any recommended QC practice. Procedure for evaluating QC practices The key is to determine the actual statistical QC rule that is being implemented based on the control limits being set, then find the power curves for that statistical rule to evaluate the performance of the QC procedure. This can be done as follows: 1. Calculate the actual control limits. 2. Take the difference from the mean. 3. Divide the difference by s[meas] to determine the number of multiples of the SD. 4. Identify the control rule considering the number of measurements that must exceed the control limits. 5. Calculate the critical-sized systematic error. 6. Impose the critical systematic error on the power curves for the statistical control rules of interest to evaluate the error detection and false rejection characteristics of that QC rule Do MDLs really work? [a]) is 10%, the observed method imprecision (s[meas]) is 2.0%, and the medically allowable imprecision (s[a]) is 4.0%, let's assume bias is 0.0%, then the critical systematic error that needs to be detected by the QC procedure is 3.35 s[meas] [from equation ((TE - bias)/s[meas]) - 1.65 ]. The accompanying critical-error graph shows the power curves for common statistical control rules all with N=2, as well as the possible medical decision limits that corresponded to statistical control rules of 1[4s] (2 times medically allowable SD), 1[5s] (CLIA fixed error limit), and 1[6s] (3 times medically allowable SD). Observe that the critical systematic error would be detected only 42%, 11%, and 1% of the time, resp., by the MDLs corresponding to 1[4s], 1[5s], and 1[6s]. Use of a 1[2.5s] limit would detect the critical systematic error 93% of the time and have only a 3% false rejection rate, thus a simple, practical, effective, and appropriate QC procedure is available to assure the necessary quality is achieved, but it doesn't correspond to any of the possible medical decision limits. This shows that MDLs don't really work! They may reduce the number of run rejections (including the number of rejections of runs that have medically important errors), but they won't assure the medically necessary quality. The idea is good, the words sound right, but the practice is wrong. MDLs are a modern myth with no scientific justification. For the right way to assure the clinical quality needed for a glucose test, see our earlier glucose POC example application. For a similar discussion with potassium as an example, see reference 7. You need to take the time to understand this issue and be sure that the myth of MDLs doesn't exist in your own laboratory. 1. Tetrault GA, Steindel SJ. Q-Probe 94-08. Daily quality control exception practices. Chicago: College of American Pathologists, 1994. 2. Steindel SJ. New directions in quality control: Part I. New QC systems. Lab Med 1986;17:463-466. 3. Howanitz PJ, Kafka MT, Steindel SJ, et al. Quality control run acceptance and rejection using fixed and medically useful limits for QAS Today. Clin Chem 1985;31:1016 (abstract). 4. Health Care Financing Administration (HCFA) and Public Health Service (PHS), US Dept of Health and Human Services (HHS). Medicare, Medicaid and CLIA Programs: Regulations implementing the Clinical Laboratory Improvement Amendments of 1988 (CLIA) and Clinical Laboratory Improvement Act program fee collection. Fed Regist 1993;58:5215- 37. 5. Passey RB, Walking the straight and narrow on quality control. Med Lab Observ 1993;25(2): 39-43. 6. Draft FDA Guidance to Manufacturers of In Vitro Analytical Test Systems for Preparation of Premarkt Submissions Implementing CLIA. December 17, 1992: obtained from the Division of Small Manufacturers Assistance (DSMA)(HFZ-220), Center for Devices and Radiological Health, Food and Drug Administration, 5600 Fishers Lane, Rockville, MD 202957. 7. Westgard JO, Quam EF, Barry PL. Establishing and validating QC acceptability criteria. Med Lab Observ 1994;26(2):22-26.	{"url":"https://westgard.com/essays/trends/essay8.html","timestamp":"2024-11-06T11:04:11Z","content_type":"text/html","content_length":"71374","record_id":"<urn:uuid:e07987e1-fa42-4818-ba31-7cc34216f4cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00225.warc.gz"}
My Puzzle Nemesis Vanquished! Well those of you who have been following my blog for a while will know that I have been struggling with Dick Hess's Yak for about a year now. In fact, I wrote about it way back in my fifth blog entry (I am up to #70 now)! I bought it from on the recommendation of David, the owner, back in December of 2008. He warned me that it was quite difficult, but I figured I could handle it. It turns out that I could not handle it, and I worked on this damn puzzle on-and-off for a year. I even bought other Dick Hess puzzles of the same variety in the hopes that they would give me some insight: first The Whale (left) in September 2009 and then (below) in January 2010. Well thanks to solving these two puzzles, I was finally able to solve The Yak! And it was every bit as satisfying as I hoped it would be. The solution is very tricky due to the many futile things that you can do while trying to solve it. I kept thinking I was discovering a key move that would help, but then it would lead to a dead end. They are all nicely crafted puzzles: The Yak that held up very well to the many hours I spent working on it. The wire is a nice gauge that resists forcing but is still light and elegant. The finish still has a great shine to it. While The Yak is an awesome puzzle, I'd definitely suggest trying out The Wale and/or Brontosaurus before trying this one. Even though Brontosaurus is rated a 10/10 difficulty, the same as the Yak, I found it to be quite a bit easier. I think The Whale took me about 30 minutes and Brontosaurus took me about 45 minutes. The Whale helped me understand what the pre-solve state of this group of puzzles looked like, since there weren't very many options for how to proceed other than to discover it. This helped focus my attention and is why this puzzle is a bit easier than the others. Still, I think most people would find this fairly challenging. I wish the key move was slightly easier to execute on this one: I felt like I needed a very small amount of force at one point, but I may have not been lining it up quite right. The Brontosaurus is quite similar to The Whale and now that I know how to solve it, I am surprised at how long it took me to figure it out. The move is not that tricky once you've done The Whale. Perhaps I wasn't focused enough or in the right frame of mind at first. All three are related to a topological construct named Borromean Rings , where three rings are linked together but removing any one ring will unlink the remaining two. When you look at each of these three puzzles, you will notice that none of the pieces are actually linked directly: each component is linked completely around another component, much like this diagram. This, at least for me, made these puzzles quite confounding at first. They seem to thwart you at every turn until you think logically about how to approach them. I would highly recommend all of them! Well this entry is a little out of order: there are a few items in my backlog at the moment, but I was so thrilled to have finished it that I wanted to write it while I was still in the moment. Woo 13 comments: 1. Congrats on solving the Yak!! Well Done! I have a Whale and a Bronto, and had similar experiences with both. I found the Whale harder (although I solved it first), and one move seems very tight, but oddly the same move always seems looser for reassembly? So maybe I am not lining it up correctly. I also bought a "Hippo", "Three Sisters", and "Outrageous Rings" from Dick last summer. The Hippo is similar to the ones you have. Three Sisters are a nice progressive set of three, I have solved all but the hardest one. Outrageous rings is one of the most confusing puzzles I have ever seen. I have been totally frustrated, but you might enjoy it (ask Dick if he has any more copies). Thanks for the tip on Booromean rings. I often get frustrated with these puzzles and give up when I am going nowhere. I am not ready for the Yak! 2. Hm, Three Sisters sounds interesting, I'll definitely shoot him an email to see if he has any more available. I just got Unbalanced Scales and The Two Cups from Eureka. Handmade by Dick himself for a ridiculously reasonable $10 each! I'm sure they'll keep me busy for a while. You should give The Yak a try! Who knows, it might not be as hard for you. I am not very good with wire puzzles. Sometimes a puzzle just hits your 'blind spot' and you can be stuck for a while where others might not be. I would be interested to see how hard you find it. 3. Well done with the 'Yak', Brian! :-) 4. Thanks canuck! 5. I know it has been 2 years but you seen to be the only one who has reviewed the Yak! I am only a fraction of the puzzler that you are and I managed to solve this in 15 minutes!!! This cannot be correct - the crucial move allowing it to be solvable required a moderate amount of force/flexing and after that it was easy. I don't want to look at the solution just yet - was this forceful move incorrect? Is the correct solution a nice sliding motion? Thanks got your help! 6. Hi Kevin, Perhaps the fact that I wrestled with it for so long has deterred other reviewers! Congrats on solving it in 15 minutes, I think that all the wire puzzles you have worked your way through have helped you get quite good at these! However, there is no force required to solve this one (I think I know the move you are referring to). There is another sequence of moves that achieves the same effect but without any force. You're on the right track though! 7. Thank you! I'll keep at it then. I will use it as distraction when I can't solve the orange revomaze!!! 8. That is one awesomely difficult puzzle!! Just done it with no force whatsoever (as you said). Brilliantly designed. It has taken me about 2 weeks to correctly solve it. As with all of these sorts of puzzles, if there is something seemingly unnecessary in the shape then it is there for a reason. You have to use all of the odd little shapes and do some rather unexpected movements. Now! Having done it once and put it back together again (also not easy), can I do it again? 9. Congrats Kevin! That sure is a tough one! Have you tried The Whale and The Brontosaurus? If not, check em out! 10. How do you put the Whale puzzle back together? I was fiddling with it at the first diamond popped off suddenly. (no idea what I did!) The rest of the solve was easy of course. Now I can't for the life of me figure out how to put it back together. I even downloaded the solve instructions and tried to work backwards; I failed. :( Please, any help at all would be great! 11. Hm, if the solution diagram didn't help, then I'm not sure how much I can help here. Just look carefully at the diagrams, paying particular attention to what crosses over what. Pay particular attention to steps 9-11. 12. Got it! Doing it again, it was really that first diamond part that was hardest. Thanks! :) 13. No problem! Glad it worked out!	{"url":"https://mechanical-puzzles.blogspot.com/2010/01/my-puzzle-nemesis-vanquished.html","timestamp":"2024-11-08T21:49:34Z","content_type":"application/xhtml+xml","content_length":"98046","record_id":"<urn:uuid:aebab04a-6110-4771-bad3-c0a02c1807f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00173.warc.gz"}
Flow fields Drawing fluid lines The year 2022 started as an inspiration for me. I participated in the early days of Genuary, which is a movement on Twitter (mostly) where artists around the world follow a list of daily themes for creating generative art. If you’ve never heard of these terms, generative art is an artistic expression in which the artist creates a set of rules to be followed. By executing these rules we can generate a creative output. Note that the more variable parameters we have in these rules, the more different the results will be. Here are some of my results in the first few days: Genuary 2022 — Pedro Cacique If you want to see more of my work with generative art, follow me on Twitter, as I always post my results there: https://twitter.com/phcacique By the way, the topics are always a little vague, which allows for the artist’s greater creativity, including his own interpretation of the theme. One of the themes I enjoyed working on was the fourth day: The next next Fidenza. I didn’t know the topic and decided to research a little to find out what it is. Turns out it’s a very interesting piece of art created by Tyler Hobbs (https://tylerxhobbs.com/fidenza) and it’s based on flow diagrams for building curved lines, as in this next piece. Fidenza — Tyler Hobbs The most interesting thing is that this is one of the best selling works as NFT (Non Fungible Token) and the artist explains the basic steps of the algorithm on his website. Once I had learned a bit about the concept, I started thinking about the Genuary topic. Remember that the interpretation of the theme is part of the creation of the work. I was torn between trying to reproduce the algorithm by adding something of my own or trying to create something different, but so amazing that it could be as famous as the original, being the next Fidenza. As I had never used flow fields to draw, I decided to go the first way, so I would learn one more technique for my arsenal. In this article, I want to show you the main idea of the algorithm and show you some interesting results. Flow Fields For this context, a flow field diagram is nothing more than a table containing values that represent the angle that a line must have when passing through a certain location. You see, this concept can be used in many other ways. As for example in a temperature map, we have a table that represents not the angle, but the temperature in a geographic position. See, in the figure below, how the weather widget uses this concept on the iPhone. In this case, for each Cartesian point (x,y) on the map we have an equivalent (i,j) in a table, with the temperature value at that point. By plotting this value on the graph, a mapping is made that tells us which color corresponds to that value within a certain range. In this case, the redder, the warmer, and the greener, the colder. Thumbs up to the summer heat in São Paulo! In the same way, imagine that we are going to make a square table in the range from 0 to 360º and that instead of a point of view with a color, we are going to draw an arrow in the direction of that We would have something like this: But what determines the angle for each cell? Well, it depends on the context to be presented. For example, we could use that weather forecast API and show the wind direction on a map. For this specific case, I made a simple function that relates the horizontal position of the cell to the total width of the table, returning the sine value of this number. That is: α = sin(i / w) In which: i = horizontal position in the table w = number of columns A tip for having interesting results here is to always vary the angulation of the arrows according to the position in the table. See that if we have more varied functions, we have different results for the same algorithm. So it all depends on the function that generates the angle value. If I vary the value according to the Y coordinate as well, we can have something like this: In this case, our function is: α = sin(i / w) + sin(j / h) In which: i = horizontal position in the table j = vertical position in the table w = number of columns h = number of lines Another possible variation in creating the table is to determine the number of rows and columns for the space of our canvas, which will cause us to have greater or lesser variation of angles in a small space on the screen. See the result with 3x more rows and columns: Drawing the lines Now that we know how to create a table with angles that fill the space, we can think about how to draw the lines. The concept is quite simple and widely used in various design algorithms. To draw a line, we need to know the coordinates of two points on the graph. With these values, we can calculate the angle between this line and the X axis. Another option is to have the desired size of the line and the angle it should form with the X axis (which is closer to what we have, right?) See the graphic below for a representation of this concept. We want to find the values of point P based on the angle α (which will be the one in our table) and the size of the line segment: R. What we are doing here is a transformation from the polar coordinate system to the Cartesian system. For this, we will use the trigonometric relations that we know. We can observe a triangle here, right? In it we have to: dy = opposite side = y coordinate of P dx = adjacent side = x coordinate of P R = hypotenuse = distance from point P to origin Take the value of cosine, for example. If you remember your college entrance exams well, you know that: cos = op / hip That is, the cosine of angle α is equal to the opposite side divided by the hypotenuse of the triangle formed, that is: cos(α) = dy / R dy = cos(a) * R Here we are considering the distance from P to the origin of the Cartesian system, but if we want to have the distance from point P to another point in space, we can just add to the value found the coordinate of that point. So: dy = y’ + cos(a) * R In which: y’ = y-coordinate of the point of origin of the segment R Analogously, we can compare the sine of the angle, being: sin = adj / hip sin(α) = dx / R dx = sin(α) * R dx = x’ + sin(α) * R So we have two equations that give us the coordinates of point P based on the size of the desired line segment and the given angle: dx = x’ + sin(α) * R dy = y’ + cos(a) * R Now, all we need is to choose a point on the screen, check in the table what the drawing angle should be and then establish what the next point would be with the equations we found above. Now we have a line segment, to have a more complete line, we repeat the procedure several times, now taking the new point as a starting point. If we use small R values, we will have smoother lines, but we will need to repeat this procedure more times for a longer line. See a line that was drawn based on our algorithm: Here, a random point was chosen at one of the edges of the canvas and a line with several segments was drawn. In this case, there were 100 segments of 40px size. Here are some variations of this method, drawing more or less lines, with different angle determination functions: What makes the Fidenza algorithm so interesting is the variation in the thickness of the lines, the number of segments that make up each one of them, the size of the table with flow diagram and an extra component: collision detection. You can add an algorithm that detects if a line collides with another to determine whether or not it can exist, which will give the work a more orderly look. Take a look: In this case, I varied the sizes of the lines (the number of segments), their thickness, their colors and checked that one did not collide with the other, getting as close as possible to the original Fidenza algorithm. See how interesting it was? Note that we can still explore this algorithm quite a bit. Can you imagine what it would be like if we used the line area to write words in a poem? We would have a directed flow of reading. Or even if we left long, thin lines and drew flowers at the ends? What would you do to improve this algorithm and give it your personal touch? If you decide to implement the algorithm in your favorite language, don’t forget to send me the result, ok?	{"url":"https://phcacique.medium.com/flow-fields-12cdc0ea40d9?source=user_profile---------1----------------------------","timestamp":"2024-11-06T23:11:39Z","content_type":"text/html","content_length":"162204","record_id":"<urn:uuid:187bc325-6abb-4f62-a9f0-91d019d97d3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00467.warc.gz"}
LibreOffice Calc: Financial Functions - Loan Payments - Ahuka Communications LibreOffice Calc: Financial Functions – Loan Payments As we discussed back when we first looked at spreadsheets, they were the killer app that lead to wide-spread adoption of PCs in companies. And the reason for that was that you could do sophisticated financial analysis with spreadsheets. So it should not come as great surprise that there are many great financial functions in Calc. The thing you need to bear in mind about these is that they are all oriented to doing financial analysis on investments. But if you take the time to understand what these functions are doing you can often use them in other ways. As we pointed out, all functions in Calc take arguments and in this case as you might expect the arguments would have to do with interest rates, principle amounts, time periods, and payment amounts. The general form of such a function ties all of these together in a relationship that lets you specify all but one of them and solve for the remaining one. Here is an example (courtesy of WikiHow.com , which offers its material under a Creative Commons license), using loan payments: M = P * ( J / (1 – (1 + J)^ -N)) M = Monthly payment P = Principal amount of the loan J = monthly interest; annual interest divided by 100, then divided by 12 N = number of months of amortization, determined by length in years of loan So what is going here? Basically, you borrow some amount P, which you need to repay, and you will make payments for N number of months. But you cannot simply divide P by N to get your monthly payment because you also need to pay interest on the loan. Interest is usually given as an annual percentage, such as 8% per annum. To use this in a calculation you need to convert the percentage into a decimal, and that means dividing by 100 to get .08. But you don’t pay interest once a year, you pay it every month in your payments. So the last adjustment is to divide by 12 to get 12 monthly payments. Now, you need to note that this may not precisely match how your bank calculates it, depending on how they compound things, but it should be pretty close for comparison purposes. And I am going to leave it at that, because this is a tutorial on Calc, not on Finance. Once you have this formula you can do a simple model in a spreadsheet where if you plug in any three of the variables you can calculate the fourth one. For example, suppose the car dealer makes you an offer: You can either get $1,000 off on the price of the car, or get .5% lower interest rate. Which is the better offer? Well, let’s put in some actual numbers. Suppose the car you are looking at normally sells for $19,000, you are looking at a 4-year loan, and the normal interest rate on these loans is 8%. You can put this into a spreadsheet and do a quick calculation. I set up one of these on a sample spreadsheet like this: First, I select three cells A1 through C1, and click the Merge and Center Cells button on the Formatting bar. This is just to the right of the Left Align, Center, Right Align, and Justify buttons. This merges the three cells. I type in Manual Model, make it bold, and increase the font size to Arial 12. Finally, I right-click, select Format Cells, Background, and select a nice blue background. None of this is strictly necessary, but making the spreadsheet more attractive and a little easier to read is not a bad thing. In Cell A2 I typed “Price of Car” In Cell B2 I typed “19,000” In Cell A3 I typed “Periods” In Cell B3 I typed “48”. This represents a 4 year loan, with 48 monthly payments. In Cell A4 I typed “Interest Rate” In Cell B4 I typed “.08” Then because the interest rate is given as an annual rate, I do an intermediate calculation In Cell A5 I Type “Monthly Interest Rate” In Cell B5 I type “=B4/12” This is my raw material for my calculations. The variable I left out is the monthly payment amount, and that is what I would solve for. In Cell A7 I type “Base Case” In Cell A9 I typed “Scenario A” In Cell A10 I typed “1,000 off price” In Cell A13 I typed “Scenario B” In Cell A14 I typed “.005 reduction in rate” (Note that I divided by 100 to get the decimal equivalent of half a percent) Then in column C I will put my answers. In Cell C2 I typed “Monthly Payment” Now, to use my math skills. In Cell C7, opposite the label “Base Case” I put in my formula, replacing the variable names in the above formula with cell addresses. So now the formula reads = B2 * (B5 / (1 – (1 + B5)^ -B3)) This is really the same thing as formula above. But note that I haven’t actually put in the scenario adjustments yet. That is OK, because I will first copy the formula exactly to put into Cell C10 for my first scenario, and in Cell C14 for my second scenario. So I click on cell C7, then go to the formula bar, highlight the formula, and then Copy. Then go to Cell C110, click on it, and Paste. Do the same thing for Cell C14. So now I have the same number in all three of these cells. It is the monthly payment if you borrow 19,000 at 8% annual interest for 48 months and it comes to 463.85. But I need to adjust for my scenarios, which I have not done yet. So I go back to Cell C10, click on it, and edit the formula. This scenario is reducing the price of the car by 1,000. Since the price is in Cell B2, I can replace B2 in my formula with (B2-1000). Note that the parentheses are very important here. You want to calculate the reduced amount borrowed before you do the multiplication. Leave that out and you get a very bad answer indeed. But do it right and you should get 439.43. So in rough terms you knock 24 a month off of the payment. What about the second scenario? This was reducing the annual interest rate by .005. But again I need to get this in Monthly terms, which means dividing by 12. Again, I will do an intermediate calculation to make this easier. In Cell A15 I type “Monthly Interest Rate” In Cell B15 I type “=(B4-.005)/12”. Again, make sure you put the parentheses around (B4-.005). You need to do this calculation before you do the division. Now, in my formula in Cell C14, I make the adjustment by replacing every instance of B5 (the old interest rate number) with B15, my new number. Now my formula reads: And my new monthly payment appears to be 459.40. So I would be better off taking the 1000 discount instead of the interest rate reduction in this particular case. Using a Built-In Function Of course, this was fairly difficult work to do this, particularly if you are not used to doing financial calculations. So wouldn’t it be nice if you could do the same thing with a built-in function? And you can! There is a financial function called PMT that handles this very nicely. But note you need to have gathered the same information we used above for our manual calculation. The PMT function has these arguments: Rate = what we had above for our interest rate, but should be the annual rate divided by 12, not the annual rate. NPER = Number of Periods, which in our example is 48 PV = Present Value, which in this case is the amount you borrowed, i.e. 19000 FV = Future Value, which assuming you pay it all of should be zero Type = A variable that specifies when the payment occurs each month. If you enter zero, or leave it blank, it is assumed to be the end of the month. If you enter 1, it means your payment is at the beginning of the month. The difference is slight in any case. So, I added another section to my sheet to show what happens if I use the built-in formula. To do this, you begin by clicking on the cell where you want the formula to be calculated, then go to the function key and click it. Select Financial in the Category drop-down, and then scroll down to PMT. Select it, and then click next. This will bring up a window on the right with fields for each of your variables. Now, you could type in numbers, but that is not the way to do it here. We already have most of our numbers on the spreadsheet. So to grab these cell addresses, we click on the field to put our insertion mark there. Then click on the cell that has the number you want, and it will be added to the field. So, to do a comparison I go to Cells A18 through C18, click the Merge and Center Cells Button, and apply the font styles and background as above. But in this cell I will type “Using the PMT Formula”. In Cell A20 I type “Base Case” In Cell A22 I type “Scenario A” In Cell A24 I type “Scenario B” Then in Cell C20, I click on the Function wizard, select the PMT function, and fill it out using the numbers we got earlier: Rate is Cell B5 NPER is Cell B3 PV is Cell B2 FV is left blank, which means it is assumed to be zero, i.e., the loan will be completely paid off. For Scenario A Rate is Cell B5 NPER is Cell B3 PV is Cell B2-1000 FV is left blank, which means it is assumed to be zero, i.e., the loan will be completely paid off. For Scenario B Rate is Cell B15 NPER is Cell B3 PV is Cell B2 FV is left blank, which means it is assumed to be zero, i.e., the loan will be completely paid off. And when you do that, what numbers do you get? Base case = -463.85 Scenario A = -439.43 Scenario B = -459.40 And these are exactly the numbers we got previously, or almost so. The one difference is that they are reported as negative numbers, but that only means they are money being paid out rather than money coming in. Remember, we are using a function used to evaluate investments, and just turning it around. Lessons Learned • The formula we used to do the manual calculation is obviously the exact same formula that Calc used to get these monthly payment numbers. I went through this exercise in part to demonstrate that idea, and to give you some sense of what is going on behind the scenes when you use a formula. I probably won’t do it a second time but I thought it was worth doing once. • The first step in using a formula is to look at the arguments it requires, and make sure you have them ready-to-hand. If you are not sure what each variable really means, do a little Google research. One thing that may help is that in the cases I have investigated the Calc functions and the Excel functions are identical, and you may find it easier to get a good explanation on an Excel help site. If so, use it. You don’t get bonus points for avoiding the best information. • I did Intermediate Calculation several times in my example. These are not strictly necessary, but they are usually a good idea. I could have made my formulas even more complicated by adding the adjust terms into them directly, but the problem is that your formulas quickly become almost impossible to debug that way. A good Intermediate Calculation lets you move some of that outside of the formula, and you can usually do a quick sanity check to see if the number you get is plausible. For instance, if you divided by 12, did the answer look like 1/12 of what you started with? The spreadsheet I created for this lesson can be downloaded here. 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The Superelevation Formula: Tool for Efficient and Safe Road Design Have you ever wondered how roadways are constructed to manage high-speed turns? There is an engineering calculation that defines how much the road should be tilted. The superelevation formula is essential for creating safe and efficient roadways. You’ve undoubtedly felt it before without realising it – that tiny angling of the road as you round a turn. Cant or superelevation is a tilt designed to offset the centrifugal force that pulls your car outward as you turn. So, if you get the superelevation exactly perfect, you’ll be able to glide gently over the curve. But if it’s off, your trip might be a little bumpy. The superelevation formula, on the other hand, estimates exactly how much a road has to be banked depending on factors such as the speed limit, radius of the curve, and coefficient of friction between tyres and road surface. It’s a straightforward concept, but it has a significant influence on road design and safety. Continue reading to find out how this key tool changes the routes you go on every day. What Is the Superelevation Formula? The superelevation formula is a straightforward calculation that road planners use to calculate how much a road should be “banked” or inclined for safe turning at a given speed. What does it calculate? The formula computes the superelevation rate, which is the difference in height between the inside and outside margins of a curve’s highway. This difference in height, however, counteracts the centrifugal force experienced by cars travelling around the curve, allowing them to safely complete the bend. How does it work? The fundamental superelevation formula is e = V2 / 15R. • e = Superelevation rate (in metres or feet, the difference in height between the inside and outside margins of the road). • V = Road design speed (in km/h or mph) • R = Radius of the curve (in meters or feet) So, if a road is designed to go at 60 km/h and has a curve radius of 200 m, the calculation would be: e = (60)2/15 (200) = 0.24 m This indicates that the road’s outer edge should be 0.24 m higher than the inner edge. Why is it important? The superelevation algorithm enables road planners to slant roadways at precisely the appropriate degree for traffic speed and turn sharpness. As a result, driving is safer, more efficient, and more comfortable. Inadequate superelevation can cause dangerous sliding, while excessive superelevation can cause cars to feel unstable. Road planners may use this simple technique to construct bends that are safe and pleasurable to drive at the specified speed. Why Is the Superelevation Formula Important in Road Design? The superelevation formula is critical for designing safe and efficient roads. Roads would not be able to adequately account for centrifugal force, which pushes you to the outside of a curve when How the Formula Works? The superelevation formula determines how much to elevate the road’s outside border based on the radius of a curve and the design speed. The greater the radius and the greater the speed, the greater the demand for elevation. This counteracts centrifugal force and keeps cars stable as they spin. Why It’s So Important? There are a few key reasons the superelevation formula is vital: • It lowers the possibility of cars flipping or sliding out on bends. The formula helps automobiles maintain traction by raising the outside edge of a bend. • On curving roadways, it allows for greater speed restrictions. Without sufficient superelevation, speed limits on any curved portions would have to be severely decreased. • It enhances the driver’s comfort and control. Furthermore, the appropriate degree of superelevation allows drivers the confidence to smoothly handle a curve without fear of their car leaving the • It increases the life of tyres and roads. When superelevation is computed correctly, it reduces excessive tyre wear and road surface damage caused by cars drifting and sliding during corners. Although the superelevation formula is a simple calculation, it has a significant influence on road safety, efficiency, and quality. By taking into account both the geometry of a curve and the projected speed of traffic, we can ensure that our roads are constructed to withstand the forces at work. How to Calculate Superelevation Using the Formula? Follow these steps to compute a road’s superelevation using the formula: Gather the required values You’ll need to figure out the road’s design speed in miles per hour, the radius of the curve in feet, and the side friction factor. As a result, the side friction factor is affected by the road surface, weather conditions, and vehicle tyres. Use 0.15 for most roads. Calculate the centrifugal force Divide the design speed by 60 to get feet per second. Square this number. Multiply by 0.0145. Multiply by the curve radius. However, this gives you the centrifugal force in pounds that vehicles will experience going around the curve. Determine the side friction force Multiply the side friction factor by the vehicle weight. For most passenger cars, use 3,000 to 4,000 pounds. This gives you the side friction force in pounds that can counteract the centrifugal Calculate the superelevation rate Divide the centrifugal force by the side friction force. So, this percentage is the necessary superelevation rate to properly balance the forces on vehicles rounding the curve. Apply the superelevation to the road The road surface should slope downwards from the inside of the curve to the outside. The superelevation rate calculated specifies the drop in inches per foot of road width. So for a 10% rate and a 12-foot wide lane, the outer edge should be 1.2 inches lower than the inner edge. Superelevation road bends that are properly superelevated depending on design speed allow cars to corner safely and pleasantly. The proper amount of slope must be calculated using this simple formula and parameters like as radius, speed, and side friction. To guarantee enough superelevation in all situations, round up to the next largest increment when in doubt. Following these road construction best practises improves safety, efficiency, and the overall driving experience. Real-World Applications of the Superelevation Formula The superelevation formula is used in many areas of road design and engineering. Engineers must guarantee that automobiles can safely traverse curves at varying speeds while creating roadways. The superelevation formula aids in determining the right angle and banking of curve highways. Highway Exit Ramps Exit ramps on motorways frequently need tight bends at high speeds. Engineers may determine the correct banking angle required for cars to retain control when departing using the superelevation formula. This reduces the likelihood of a rollover and guarantees a seamless transition from the motorway to the exit ramp. Mountain Roads Winding bends are common on roads in hilly or mountainous places as they traverse the terrain. Engineers can use the superelevation formula to create safe turning radii and road banking for cars travelling at various speeds. This is especially critical in regions where automobiles may be travelling at high speeds. Proper superelevation helps drivers maintain control and stability on winding mountain routes. Race Track Design Superelevation is a significant consideration when creating high-speed race circuits. Race track engineers may design exhilarating yet controlled courses by estimating the optimum banking angle for each curve based on the planned speed of cars. The steeper the banking on a curve, the faster cars may travel while keeping grip. Race track superelevation pushes the limits of performance for both drivers and their machinery. In many cases, highway engineering boils down to ensuring safe and effective traffic flow. The superelevation formula gives a mathematical tool for constructing curves and turns that balance the centrifugal forces exerted on vehicles while allowing them to go at the fastest possible speed. The superelevation formula, when used correctly in road construction and design, helps get you where you need to go, whether you’re commuting to work or pushing a high-performance car to its limits. Superelevation Formula FAQs You’re undoubtedly curious in how the superelevation formula works and how it affects road design. Here are some often asked questions concerning this critical subject. What exactly is superelevation? The banking of highways, especially the slope of the road surface, is referred to as superelevation. When cars travel at high speeds around a curve, superelevation counteracts the centrifugal force that pulls the vehicle outward. This enables the car to safely round the curve without sliding or losing control. How is the superelevation formula calculated? The curve radius, design speed limit, side friction factor, and centrifugal force are all parameters considered by the superelevation formula. Using these inputs, the algorithm calculates the best angle to bank the road through the curve. The objective is to establish balance between the outward centrifugal force and the inward side friction force. Why is superelevation important for road safety and efficiency? Proper superelevation through bends contributes to safe, reasonable-speed travel. Vehicles would have to slow down significantly to handle bends without it, lowering road efficiency. Superelevation also helps drivers stay in their lane when negotiating corners. As a result of these variables, there are fewer accidents and reduced traffic congestion. How does superelevation impact road drainage? Water flows and drains from the road surface are affected by superelevating a road. Because the banks causes water to flow outward, more drains, channels, and pipes are required around the curve’s outside border to catch and divert the water. Water can accumulate and cause hydroplaning if not properly drained, especially at high speeds. What design factors are considered when superelevating a road? Several criteria, including typical traffic volume and speed, curve radius, road width, shoulder quality, and available right-of-way, must be considered. Environmental concerns, sight distances, and how the superelevation will interact with other road segments must all be considered in the design. To install superelevation in a way that maximises safety, efficiency, and road longevity, proper engineering and planning are essential. So there you have it: the superelevation formula used by highway engineers to determine how far to bank a road for safe turns. It’s rather ingenious how a few variables, such as speed, radius of curve, and coefficient of friction, work together to get the perfect slant. You’ll understand the arithmetic that went into constructing that curve the next time you pleasantly glide around a bend on the highway. While the formula itself may be complicated, the outcome is simple: helping you get where you need to go as quickly and securely as possible. The superelevation formula is just one of the many invisible tools engineers employ to craft the roads we all depend on.	{"url":"https://constructionupskills.com/superelevation-formula/","timestamp":"2024-11-12T19:26:23Z","content_type":"text/html","content_length":"105631","record_id":"<urn:uuid:e12ab641-c971-4fdf-a54c-ef51d8d67491>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00806.warc.gz"}
Just 1 In 10 People Can Pass This Math Test. Can you solve it? Give your brain some training by trying to solve this math test. Are you smart enough to ace this question without looking up the answer? Don’t underestimate it, most people will get it wrong in the beginning! If you get stuck, you can scroll down to check out the answer. Are you up for the challenge? Go for it! Solution explained: The rule is: multiplication/division before addition/subtraction Hence, first calculate 4 times 2 = 8. Then add 5. So, 8 + 5 = 13 Did you get it right without looking? Challenge your friends as well and pass this fun math test on to them. Some other math test The Answer is: (1 x 3) + (3 x 5) + (8 x 2) = 34 The correct answer would look something like this: 50+50-25×0+2+2 = 104 Join our list Subscribe to our mailing list and get interesting stuff and updates to your email inbox. Thank you for subscribing. Something went wrong.	{"url":"https://ispecially.com/just-1-in-10-people-can-pass-this-math-test-can-you-solve-it/","timestamp":"2024-11-03T10:30:28Z","content_type":"text/html","content_length":"72364","record_id":"<urn:uuid:216459dd-c2ca-4105-bd1f-5d5d71755bcf>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00587.warc.gz"}
Homotopy Type Theory - (Algebraic Topology) - Vocab, Definition, Explanations \| Fiveable Homotopy Type Theory from class: Algebraic Topology Homotopy type theory is a branch of mathematical logic that merges homotopy theory and type theory, creating a framework where types can be viewed as spaces and terms as points in those spaces. This perspective allows for a deeper understanding of both logic and topology, particularly in how structures can be continuously transformed or deformed into one another while preserving certain congrats on reading the definition of Homotopy Type Theory. now let's actually learn it. 5 Must Know Facts For Your Next Test 1. Homotopy type theory provides a new foundation for mathematics by treating types as spaces and allowing for the interpretation of logical propositions in terms of topological spaces. 2. In this framework, identity types correspond to paths in homotopy theory, which means that proving two terms are equal can be seen as finding a continuous path connecting them. 3. Homotopy type theory has applications in computer science, particularly in programming languages that support dependent types, where types can depend on values. 4. The univalence axiom is a key principle in homotopy type theory that states equivalences between types can be treated as equalities, simplifying many constructions and proofs. 5. This theory has influenced both mathematics and computer science by providing tools for constructing proofs and reasoning about them in a way that emphasizes continuity and transformation. Review Questions • How does homotopy type theory redefine the relationship between types and topological spaces? □ Homotopy type theory redefines the relationship by conceptualizing types as spaces and terms as points within those spaces. This means that logical propositions correspond to topological properties, allowing mathematicians to leverage techniques from homotopy theory to analyze types. It enables a rich interplay between logic and topology, illustrating how structures can be continuously transformed while retaining key characteristics. • Discuss the implications of the univalence axiom in homotopy type theory and how it affects the notion of equality among types. □ The univalence axiom fundamentally shifts how equality is understood in homotopy type theory by asserting that equivalences between types are treated as equalities. This allows mathematicians to simplify constructions and proofs since they can now reason about types in a more flexible manner. It connects the concepts of equivalence and identity, leading to a more cohesive understanding of structures within this framework. • Evaluate the significance of homotopy type theory in bridging the gap between abstract mathematics and practical applications in computer science. □ Homotopy type theory significantly bridges abstract mathematics and practical applications by offering foundational principles that enhance programming languages with dependent types. This framework enables better reasoning about programs, where types represent properties of data structures, leading to safer and more reliable code. By allowing for geometric interpretations of logical propositions, it enriches both mathematical proof development and software verification processes, showcasing its relevance across disciplines. "Homotopy Type Theory" also found in: ยฉ 2024 Fiveable Inc. All rights reserved. APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.	{"url":"https://library.fiveable.me/key-terms/algebraic-topology/homotopy-type-theory","timestamp":"2024-11-13T18:26:43Z","content_type":"text/html","content_length":"143465","record_id":"<urn:uuid:c5c1f384-0fc3-40d0-b518-e014067b3e71>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00643.warc.gz"}
Law of SinesLaw of Sines - Huffington Post Lawsuit 1st Method To Solve The Law Of Lines The sides of the triangle are proportional to the lines of the other opposite angles. From a triangle we know that: a = 6 m, B = 45 ° and C = 105 °. Determine the remaining elements. Find the radius of the circumscribed circle in a triangle, where A = 45 °, B = 72 °, and a = 20m. Theorem or law of cosine At a triangle the square of each side is equal to the sum of the squares of the other two minus the double product of the two times the cosine of the angle they form. The radius of a parallelogram measure 10 cm and 12 cm, and the angle they form is 48 ° 15 ‘. Calculate the sides. The radius of a circle is 25 m. Calculate the angle which the tangents will make to said circumference, drawn by the ends of a rope of length 36 m. Theorem or law of tangent If A and B are angles of a triangle and their corresponding sides are a and b, it follows that: You will learn how to solve oblique triangles by applying the law of sines. Thus far, we have solved the right triangles. However, it is also common to find issues with triangles that aren’t rectangles, such as acute angles or obtuse angles. To solve these problems, the method we’ve used does not work, but we may use the law of sines. 2nd Method To Solve the Law of Sines For any triangle in the plane, together with inner angles, and opposite side spans respectively, it is fulfilled: Quite simply, the law of sines states: for any triangle which lies within a plane, the lengths of its sides are proportional to the sines of its angles. When we know the length of one side of the triangle and its interior angles, we could compute the lengths of the other two sides employing this law. Example 1 Solve the next isosceles triangle: Considering that the triangle is isosceles, both sloping sides measure 3 cm. We will demonstrate it using the law of sines. For this, we define cm, and we will need to calculate. Using: We can clear to get: This demonstrates it is an isosceles triangle. To figure the period of the foundation, we have to mention that the amount of the two famous angles is and that the next pitch has to measure. With this, we can reuse the law of sines to Compute the length of: Now we substitute the known values: And with this, we’ve solved this severe triangle. The sine law also works for obtuse triangles, as shown in the next instance. Example 2 Solve the following obtuse triangle: In this case cm, also we can calculate: Now we can calculate the length of the side employing the law of sines: Clearing and solving, we get a cm. Finally, we could calculate the value of: Substituting the values we get: And we completed it. Breast law also serves to fix problems in a variety of contexts. Example 3 A building business will drill a tunnel through a hill to decrease the transport time of Acatlán (a stage in the figure) into Bacatlán (point ). If the tunnel is on the line, which passes through the things and What will be the distance in the street? Cazatlán is your point indicated in the next figure. The following were measured: kilometers. We start by imagining that we can calculate the angle value: We can calculate the distance between the things and implementing the law of sines: We can even calculate the distance between the point and Bacatlán: With this, we have fully resolved the triangle. It follows that now to buy from Acatlán into Bacatlán they travel at least 31.6 kilometers from Acatlán to Cazatlán very first, then 42.4 km from Cazatlán into Bacatlán.total. With the new street that will pass through the tunnel, space is shortened to approximately 40 km. Example 4 On the point there’s a plane which travels east, from there to levels north (left of the front of the plane) is an airport. If you go 100 km, the airplane is now located at the point, the airport itself is submerged from precisely the same plane. How far apart are the tips? We begin by drawing a diagram to get a better idea of this problem: We specify: and. The third angle, why. To figure out the distances, we wish to know we apply the law of sines. The space from the point until the purpose is: And we’re done. Example 5 Marco noticed that an angle of one point on the ground to the peak of a tree, but if you advance horizontally 20 meters towards the tree into a point, the angle that is formed is. What’s the height of the tree? We begin by outlining the scenario: Since the angles and are supplementary and measure, it follows the step. Now that we understand two interior angles of the triangle, we could calculate the angle measure: We can apply the law of sines to calculate the action of this side: We can calculate the length of the segment, employing the definition of the cosine function in the right triangle: Finally, we can apply the Pythagorean theorem to the right triangle to figure out the height of the tree. In this triangle, the hypotenuse steps, along with the known leg, And we are done. You must be logged in to post a comment Login You must be logged in to post a comment.	{"url":"https://huffingtonpostlawsuit.com/en-us/law-school/law-of-sines/","timestamp":"2024-11-03T09:27:14Z","content_type":"text/html","content_length":"93096","record_id":"<urn:uuid:08ca214a-be52-4225-8dbf-977c0a54c3bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00848.warc.gz"}
Remove decimal places without rounding – SQLServerCentral There was a script here not too long ago which showed how to 'truncate' a number to two decimal places without rounding. This used a lot of casts to varchars and back. The solution worked, but I knew there had to be a way of doing exactly this without having to go back and forth between numeric and character data. The result is the UDF below. In words, here is what the script does: - Round the number to the specified number of decimal places. This can go either up or down. - If the result is bigger than the input value, calculate and substract the minimum value allowed by the number of decimal places specified. The only 'tough' bit in the UDF is the last bit, calculating the minimum value allowed by the number of decimal places. To do this we use the following formula: 1 / 10 ^ Decimals (^ taken as the power operator, T-SQL uses the Power() function for this instead) Example: 1 / 10 ^ 2 = 1 / 100 = .01 Because SQL Server likes to see whole numbers as integers, we need to explicitly convert both arguments for the Power() function to floats in order to get a float result. If we don't do that .01 is cast to an integer resulting in 0.	{"url":"https://www.sqlservercentral.com/scripts/remove-decimal-places-without-rounding","timestamp":"2024-11-12T19:00:29Z","content_type":"text/html","content_length":"76950","record_id":"<urn:uuid:2a4d81d0-6784-48fd-a101-3308070a3b0e>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00693.warc.gz"}
This tutorial depends on step-20. This program grew out of a student project by Yan Li at Texas A&M University. Most of the work for this program is by her. In this project, we propose a numerical simulation for two phase flow problems in porous media. This problem includes one elliptic equation and one nonlinear, time dependent transport equation. This is therefore also the first time-dependent tutorial program (besides the somewhat strange time-dependence of step-18). The equations covered here are an extension of the material already covered in step-20. In particular, they fall into the class of vector-valued problems. A toplevel overview of this topic can be found in the Handling vector valued problems topic. The two phase flow problem Modeling of two phase flow in porous media is important for both environmental remediation and the management of petroleum and groundwater reservoirs. Practical situations involving two phase flow include the dispersal of a nonaqueous phase liquid in an aquifer, or the joint movement of a mixture of fluids such as oil and water in a reservoir. Simulation models, if they are to provide realistic predictions, must accurately account for these effects. To derive the governing equations, consider two phase flow in a reservoir $\Omega$ under the assumption that the movement of fluids is dominated by viscous effects; i.e. we neglect the effects of gravity, compressibility, and capillary pressure. Porosity will be considered to be constant. We will denote variables referring to either of the two phases using subscripts $w$ and $o$, short for water and oil. The derivation of the equations holds for other pairs of fluids as well, however. The velocity with which molecules of each of the two phases move is determined by Darcy's law that states that the velocity is proportional to the pressure gradient: \begin{eqnarray} \mathbf{u}_{j} = -\frac{k_{rj}(S)}{\mu_{j}} \mathbf{K} \cdot \nabla p \end{eqnarray} where $\mathbf{u}_{j}$ is the velocity of phase $j=o,w$, $K$ is the permeability tensor, $k_{rj}$ is the relative permeability of phase $j$, $p$ is the pressure and $\mu_{j}$ is the viscosity of phase $j$. Finally, $S$ is the saturation (volume fraction), i.e. a function with values between 0 and 1 indicating the composition of the mixture of fluids. In general, the coefficients $K, k_{rj}, \mu$ may be spatially dependent variables, and we will always treat them as non-constant functions in the following. We combine Darcy's law with the statement of conservation of mass for each phase, \[ \textrm{div}\ \mathbf{u}_{j} = q_j, \] with a source term for each phase. By summing over the two phases, we can express the governing equations in terms of the so-called pressure equation: \begin{eqnarray} - \nabla \cdot (\mathbf{K}\lambda(S) \nabla p)= q. \end{eqnarray} Here, $q$ is the sum source term, and \[ \lambda(S) = \frac{k_{rw}(S)}{\mu_{w}}+\frac{k_{ro}(S)}{\mu_{o}} \] is the total mobility. So far, this looks like an ordinary stationary, Poisson-like equation that we can solve right away with the techniques of the first few tutorial programs (take a look at step-6, for example, for something very similar). However, we have not said anything yet about the saturation, which of course is going to change as the fluids move around. The second part of the equations is the description of the dynamics of the saturation, i.e., how the relative concentration of the two fluids changes with time. The saturation equation for the displacing fluid (water) is given by the following conservation law: \begin{eqnarray} S_{t} + \nabla \cdot (F(S) \mathbf{u}) = q_{w}, \end{eqnarray} which can be rewritten by using the product rule of the divergence operator in the previous equation: \begin{eqnarray} S_{t} + F(S) \left[\nabla \cdot \mathbf{u}\right] + \mathbf{u} \cdot \left[ \nabla F(S)\right] = S_{t} + F(S) q + \mathbf{u} \cdot \nabla F(S) = q_{w}. \end{eqnarray} Here, $q=\nabla\cdot \mathbf{u}$ is the total influx introduced above, and $q_{w}$ is the flow rate of the displacing fluid (water). These two are related to the fractional flow $F(S)$ in the following way: \[ q_{w} = F(S) q, \] where the fractional flow is often parameterized via the (heuristic) expression \[ F(S) = \frac{k_{rw}(S)/\mu_{w}}{k_{rw}(S)/\mu_{w} + k_{ro}(S)/\mu_{o}}. \] Putting it all together yields the saturation equation in the following, advected form: \begin{eqnarray} S_{t} + \mathbf{u} \cdot \nabla F(S) = 0, \end{eqnarray} where $\mathbf u$ is the total velocity \[ \mathbf{u} = \mathbf{u}_{o} + \mathbf{u}_{w} = -\lambda(S) \mathbf{K}\cdot\nabla p. \] Note that the advection equation contains the term $\mathbf{u} \cdot \nabla F(S)$ rather than $\mathbf{u} \cdot \nabla S$ to indicate that the saturation is not simply transported along; rather, since the two phases move with different velocities, the saturation can actually change even in the advected coordinate system. To see this, rewrite $\mathbf{u} \cdot \nabla F(S) = \mathbf{u} F'(S) \cdot \nabla S$ to observe that the actual velocity with which the phase with saturation $S$ is transported is $\mathbf u F'(S)$ whereas the other phase is transported at velocity $\mathbf u (1-F'(S))$. $F(S)$ is consequently often referred to as the fractional flow. In summary, what we get are the following two equations: \begin{eqnarray} - \nabla \cdot (\mathbf{K}\lambda(S) \nabla p) &=& q \qquad \textrm{in}\ \Omega\times[0,T], \\ S_{t} + \mathbf{u} \cdot \nabla F(S) &=& 0 \qquad \textrm{in}\ \Omega\times[0,T]. \end Here, $p=p(\mathbf x, t), S=S(\mathbf x, t)$ are now time dependent functions: while at every time instant the flow field is in equilibrium with the pressure (i.e. we neglect dynamic accelerations), the saturation is transported along with the flow and therefore changes over time, in turn affected the flow field again through the dependence of the first equation on $S$. This set of equations has a peculiar character: one of the two equations has a time derivative, the other one doesn't. This corresponds to the character that the pressure and velocities are coupled through an instantaneous constraint, whereas the saturation evolves over finite time scales. Such systems of equations are called Differential Algebraic Equations (DAEs), since one of the equations is a differential equation, the other is not (at least not with respect to the time variable) and is therefore an "algebraic" equation. (The notation comes from the field of ordinary differential equations, where everything that does not have derivatives with respect to the time variable is necessarily an algebraic equation.) This class of equations contains pretty well-known cases: for example, the time dependent Stokes and Navier-Stokes equations (where the algebraic constraint is that the divergence of the flow field, $\textrm{div}\ \mathbf u$, must be zero) as well as the time dependent Maxwell equations (here, the algebraic constraint is that the divergence of the electric displacement field equals the charge density, $\textrm{div}\ \mathbf D = \rho$ and that the divergence of the magnetic flux density is zero: $\textrm{div}\ \mathbf B = 0$); even the quasistatic model of step-18 falls into this category. We will see that the different character of the two equations will inform our discretization strategy for the two equations. Time discretization In the reservoir simulation community, it is common to solve the equations derived above by going back to the first order, mixed formulation. To this end, we re-introduce the total velocity $\mathbf u$ and write the equations in the following form: \begin{eqnarray} \mathbf{u}+\mathbf{K}\lambda(S) \nabla p&=&0 \\ \nabla \cdot\mathbf{u} &=& q \\ S_{t} + \mathbf{u} \cdot \nabla F(S) &=& 0. \end{eqnarray} This formulation has the additional benefit that we do not have to express the total velocity $\mathbf u$ appearing in the transport equation as a function of the pressure, but can rather take the primary variable for it. Given the saddle point structure of the first two equations and their similarity to the mixed Laplace formulation we have introduced in step-20, it will come as no surprise that we will use a mixed discretization again. But let's postpone this for a moment. The first business we have with these equations is to think about the time discretization. In reservoir simulation, there is a rather standard algorithm that we will use here. It first solves the pressure using an implicit equation, then the saturation using an explicit time stepping scheme. The algorithm is called IMPES for IMplicit Pressure Explicit Saturation and was first proposed a long time ago: by Sheldon et al. in 1959 and Stone and Gardner in 1961 (J. W. Sheldon, B. Zondek and W. T. Cardwell: One-dimensional, incompressible, non-capillary, two-phase fluid flow in a porous medium, Trans. SPE AIME, 216 (1959), pp. 290-296; H. L. Stone and A. O. Gardner Jr: Analysis of gas-cap or dissolved-gas reservoirs, Trans. SPE AIME, 222 (1961), pp. 92-104). In a slightly modified form, this algorithm can be written as follows: for each time step, solve \begin{eqnarray} \mathbf{u}^{n+1}+\mathbf{K}\lambda(S^n) \nabla p^{n+1}&=&0 \\ \nabla \cdot\mathbf{u}^{n+1} &=& q^{n+1} \\ \frac {S^{n+1}-S^n}{\triangle t} + \mathbf{u}^{n+1} \cdot \nabla F(S^n) &=& 0, \end{eqnarray} where $\triangle t$ is the length of a time step. Note how we solve the implicit pressure-velocity system that only depends on the previously computed saturation $S^n$, and then do an explicit time step for $S^{n+1}$ that only depends on the previously known $S^n$ and the just computed $\mathbf{u}^{n+1}$. This way, we never have to iterate for the nonlinearities of the system as we would have if we used a fully implicit method. (In a more modern perspective, this should be seen as an "operator splitting" method. step-58 has a long description of the idea behind this.) We can then state the problem in weak form as follows, by multiplying each equation with test functions $\mathbf v$, $\phi$, and $\sigma$ and integrating terms by parts: \begin{eqnarray} \left((\mathbf{K}\lambda(S^n))^{-1} \mathbf{u}^{n+1},\mathbf v\right)_\Omega - (p^{n+1}, \nabla\cdot\mathbf v)_\Omega &=& - (p^{n+1}, \mathbf v)_{\partial\Omega} \\ (\nabla \cdot\ mathbf{u}^{n+1}, \phi)_\Omega &=& (q^{n+1},\phi)_\Omega \end{eqnarray} Note that in the first term, we have to prescribe the pressure $p^{n+1}$ on the boundary $\partial\Omega$ as boundary values for our problem. $\mathbf n$ denotes the unit outward normal vector to $\partial K$, as usual. For the saturation equation, we obtain after integrating by parts \begin{eqnarray} (S^{n+1}, \sigma)_\Omega - \triangle t \sum_K \left\{ \left(F(S^n), \nabla \cdot (\mathbf{u}^{n+1} \sigma)\right)_K - \left(F(S^n) (\mathbf n \cdot \mathbf{u}^{n+1}, \sigma\right)_ {\partial K} \right\} &=& (S^n,\sigma)_\Omega. \end{eqnarray} Using the fact that $\nabla \cdot \mathbf{u}^{n+1}=q^{n+1}$, we can rewrite the cell term to get an equation as follows: \begin{eqnarray} (S^{n+1}, \sigma)_\Omega - \triangle t \sum_K \left\{ \left(F(S^n) \mathbf{u}^{n+1}, \nabla \sigma\right)_K - \left(F(S^n) (\mathbf n \cdot \mathbf{u}^{n+1}), \sigma\right)_{\ partial K} \right\} &=& (S^n,\sigma)_\Omega + \triangle t \sum_K \left(F(S^n) q^{n+1}, \sigma\right)_K. \end{eqnarray} We introduce an object of type DiscreteTime in order to keep track of the current value of time and time step in the code. This class encapsulates many complexities regarding adjusting time step size and stopping at a specified final time. Space discretization In each time step, we then apply the mixed finite method of step-20 to the velocity and pressure. To be well-posed, we choose Raviart-Thomas spaces $RT_{k}$ for $\mathbf{u}$ and discontinuous elements of class $DGQ_{k}$ for $p$. For the saturation, we will also choose $DGQ_{k}$ spaces. Since we have discontinuous spaces, we have to think about how to evaluate terms on the interfaces between cells, since discontinuous functions are not really defined there. In particular, we have to give a meaning to the last term on the left hand side of the saturation equation. To this end, let us define that we want to evaluate it in the following sense: \begin{eqnarray} &&\left(F(S^n) (\mathbf n \cdot \mathbf{u}^{n+1}), \sigma\right)_{\partial K} \\ &&\qquad = \left(F(S^n_+) (\mathbf n \cdot \mathbf{u}^{n+1}_+), \sigma\right)_{\partial K_+} + \left (F(S^n_-) (\mathbf n \cdot \mathbf{u}^{n+1}_-), \sigma\right)_{\partial K_-}, \end{eqnarray} where $\partial K_{-} \dealcoloneq \{x\in \partial K, \mathbf{u}(x) \cdot \mathbf{n}<0\}$ denotes the inflow boundary and $\partial K_{+} \dealcoloneq \{\partial K \setminus \partial K_{-}\}$ is the outflow part of the boundary. The quantities $S_+,\mathbf{u}_+$ then correspond to the values of these variables on the present cell, whereas $S_-,\mathbf{u}_-$ (needed on the inflow part of the boundary of $K$) are quantities taken from the neighboring cell. Some more context on discontinuous element techniques and evaluation of fluxes can also be found in step-12. Linear solvers The linear solvers used in this program are a straightforward extension of the ones used in step-20 (but without LinearOperator). Essentially, we simply have to extend everything from two to three solution components. If we use the discrete spaces mentioned above and put shape functions into the bilinear forms, we arrive at the following linear system to be solved for time step $n+1$: \[ \left( \begin{array}{ccc} M^u(S^{n}) & B^{T}& 0\\ B & 0 & 0\\ \triangle t\; H & 0& M^S \end{array} \right) \left( \begin{array}{c} \mathbf{U}^{n+1} \\ P^{n+1} \\ S^{n+1} \end{array} \right) = \ left( \begin{array}{c} 0 \\ F_2 \\ F_3 \end{array} \right) \] where the individual matrices and vectors are defined as follows using shape functions $\mathbf v_i$ (of type Raviart Thomas $RT_k$) for velocities and $\phi_i$ (of type $DGQ_k$) for both pressures and saturations: \begin{eqnarray} M^u(S^n)_{ij} &=& \left((\mathbf{K}\lambda(S^n))^{-1} \mathbf{v}_i,\mathbf v_j\right)_\Omega, \\ B_{ij} &=& -(\nabla \cdot \mathbf v_j, \phi_i)_\Omega, \\ H_{ij} &=& - \sum_K \left\ { \left(F(S^n) \mathbf v_i, \nabla \phi_j)\right)_K - \left(F(S^n_+) (\mathbf n \cdot (\mathbf v_i)_+), \phi_j\right)_{\partial K_+} - \left(F(S^n_-) (\mathbf n \cdot (\mathbf v_i)_-), \phi_j\right)_ {\partial K_-}, \right\} \\ M^S_{ij} &=& (\phi_i, \phi_j)_\Omega, \\ (F_2)_i &=& -(q^{n+1},\phi_i)_\Omega, \\ (F_3)_i &=& (S^n,\phi_i)_\Omega +\triangle t \sum_K \left(F(S^n) q^{n+1}, \phi_i\right) _K. \end{eqnarray} Due to historical accidents, the role of matrices $B$ and $B^T$ has been reverted in this program compared to step-20. In other words, here $B$ refers to the divergence and $B^T$ to the gradient operators when it was the other way around in step-20. The system above presents a complication: Since the matrix $H_{ij}$ depends on $\mathbf u^{n+1}$ implicitly (the velocities are needed to determine which parts of the boundaries $\partial K$ of cells are influx or outflux parts), we can only assemble this matrix after we have solved for the velocities. The solution scheme then involves the following steps: 1. Solve for the pressure $p^{n+1}$ using the Schur complement technique introduced in step-20. 2. Solve for the velocity $\mathbf u^{n+1}$ as also discussed in step-20. 3. Compute the term $F_3-\triangle t\; H \mathbf u^{n+1}$, using the just computed velocities. 4. Solve for the saturation $S^{n+1}$. In this scheme, we never actually build the matrix $H$, but rather generate the right hand side of the third equation once we are ready to do so. In the program, we use a variable solution to store the solution of the present time step. At the end of each step, we copy its content, i.e. all three of its block components, into the variable old_solution for use in the next time step. Choosing a time step A general rule of thumb in hyperbolic transport equations like the equation we have to solve for the saturation equation is that if we use an explicit time stepping scheme, then we should use a time step such that the distance that a particle can travel within one time step is no larger than the diameter of a single cell. In other words, here, we should choose \[ \triangle t_{n+1} \le \frac h{\|\mathbf{u}^{n+1}(\mathbf{x})\|}. \] Fortunately, we are in a position where we can do that: we only need the time step when we want to assemble the right hand side of the saturation equation, which is after we have already solved for \ (\mathbf{u}^{n+1}\). All we therefore have to do after solving for the velocity is to loop over all quadrature points in the domain and determine the maximal magnitude of the velocity. We can then set the time step for the saturation equation to \[ \triangle t_{n+1} = \frac {\min_K h_K}{\max_{\mathbf{x}}\|\mathbf{u}^{n+1}(\mathbf{x})\|}. \] Why is it important to do this? If we don't, then we will end up with lots of places where our saturation is larger than one or less than zero, as can easily be verified. (Remember that the saturation corresponds to something like the water fraction in the fluid mixture, and therefore must physically be between 0 and 1.) On the other hand, if we choose our time step according to the criterion listed above, this only happens very very infrequently — in fact only once for the entire run of the program. However, to be on the safe side, however, we run a function project_back_saturation at the end of each time step, that simply projects the saturation back onto the interval $[0,1]$, should it have gotten out of the physical range. This is useful since the functions $\lambda(S)$ and $F(S)$ do not represent anything physical outside this range, and we should not expect the program to do anything useful once we have negative saturations or ones larger than one. Note that we will have similar restrictions on the time step also in step-23 and step-24 where we solve the time dependent wave equation, another hyperbolic problem. We will also come back to the issue of time step choice below in the section on possible extensions to this program. The test case For simplicity, this program assumes that there is no source, $q=0$, and that the heterogeneous porous medium is isotropic $\mathbf{K}(\mathbf{x}) = k(\mathbf{x}) \mathbf{I}$. The first one of these is a realistic assumption in oil reservoirs: apart from injection and production wells, there are usually no mechanisms for fluids to appear or disappear out of the blue. The second one is harder to justify: on a microscopic level, most rocks are isotropic, because they consist of a network of interconnected pores. However, this microscopic scale is out of the range of today's computer simulations, and we have to be content with simulating things on the scale of meters. On that scale, however, fluid transport typically happens through a network of cracks in the rock, rather than through pores. However, cracks often result from external stress fields in the rock layer (for example from tectonic faulting) and the cracks are therefore roughly aligned. This leads to a situation where the permeability is often orders of magnitude larger in the direction parallel to the cracks than perpendicular to the cracks. A problem typically faces in reservoir simulation, however, is that the modeler doesn't know the direction of cracks because oil reservoirs are not accessible to easy inspection. The only solution in that case is to assume an effective, isotropic permeability. Whatever the matter, both of these restrictions, no sources and isotropy, would be easy to lift with a few lines of code in the program. Next, for simplicity, our numerical simulation will be done on the unit cell $\Omega = [0,1]\times [0,1]$ for $t\in [0,T]$. Our initial conditions are $S(\mathbf{x},0)=0$; in the oil reservoir picture, where $S$ would indicate the water saturation, this means that the reservoir contains pure oil at the beginning. Note that we do not need any initial conditions for pressure or velocity, since the equations do not contain time derivatives of these variables. Finally, we impose the following pressure boundary conditions: \[ p(\mathbf{x},t)=1-x_1 \qquad \textrm{on}\ \partial\Omega. \] Since the pressure and velocity solve a mixed form Poisson equation, the imposed pressure leads to a resulting flow field for the velocity. On the other hand, this flow field determines whether a piece of the boundary is of inflow or outflow type, which is of relevance because we have to impose boundary conditions for the saturation on the inflow part of the boundary, \[ \Gamma_{in}(t) = \{\mathbf{x}\in\partial\Omega: \mathbf{n} \cdot \mathbf{u}(\mathbf{x},t) < 0\}. \] On this inflow boundary, we impose the following saturation values: \begin{eqnarray} S(\mathbf{x},t) = 1 & \textrm{on}\ \Gamma_{in}\cap\{x_1=0\}, \\ S(\mathbf{x},t) = 0 & \textrm{on}\ \Gamma_{in}\backslash \{x_1=0\}. \end{eqnarray} In other words, we have pure water entering the reservoir at the left, whereas the other parts of the boundary are in contact with undisturbed parts of the reservoir and whenever influx occurs on these boundaries, pure oil will enter. In our simulations, we choose the total mobility as \[ \lambda (S) = \frac{1.0}{\mu} S^2 +(1-S)^2 \] where we use $\mu=0.2$ for the viscosity. In addition, the fractional flow of water is given by \[ F(S)=\frac{S^2}{S^2+\mu (1-S)^2} \] Coming back to this testcase in step-43 several years later revealed an oddity in the setup of this testcase. To this end, consider that we can rewrite the advection equation for the saturation as $S_{t} + (\mathbf{u} F'(S)) \cdot \nabla S = 0$. Now, at the initial time, we have $S=0$, and with the given choice of function $F(S)$, we happen to have $F'(0)=0$. In other words, at $t=0$, the equation reduces to $S_t=0$ for all $\mathbf x$, so the saturation is zero everywhere and it is going to stay zero everywhere! This is despite the fact that $\mathbf u$ is not necessarily zero: the combined fluid is moving, but we've chosen our partial flux $F(S)$ in such a way that infinitesimal amounts of wetting fluid also only move at infinitesimal speeds (i.e., they stick to the medium more than the non-wetting phase in which they are embedded). That said, how can we square this with the knowledge that wetting fluid is invading from the left, leading to the flow patterns seen in the results section? That's where we get into mathematics: Equations like the transport equation we are considering here have infinitely many solutions, but only one of them is physical: the one that results from the so-called viscosity limit, called the viscosity solution. The thing is that with discontinuous elements we arrive at this viscosity limit because using a numerical flux introduces a finite amount of artificial viscosity into the numerical scheme. On the other hand, in step-43, we use an artificial viscosity that is proportional to $\\|\ mathbf u F'(S)\\|$ on every cell, which at the initial time is zero. Thus, the saturation there is zero and remains zero; the solution we then get is one solution of the advection equation, but the method does not converge to the viscosity solution without further changes. We will therefore use a different initial condition in that program. Finally, to come back to the description of the testcase, we will show results for computations with the two permeability functions introduced at the end of the results section of step-20: • A function that models a single, winding crack that snakes through the domain. In analogy to step-20, but taking care of the slightly different geometry we have here, we describe this by the following function: \[ k(\mathbf x) = \max \left\{ e^{-\left(\frac{x_2-\frac 12 - 0.1\sin(10x_1)}{0.1}\right)^2}, 0.01 \right\}. \] Taking the maximum is necessary to ensure that the ratio between maximal and minimal permeability remains bounded. If we don't do that, permeabilities will span many orders of magnitude. On the other hand, the ratio between maximal and minimal permeability is a factor in the condition number of the Schur complement matrix, and if too large leads to problems for which our linear solvers will no longer converge properly. • A function that models a somewhat random medium. Here, we choose \begin{eqnarray} k(\mathbf x) &=& \min \left\{ \max \left\{ \sum_{i=1}^N \sigma_i(\mathbf{x}), 0.01 \right\}, 4\right\}, \\ \sigma_i(\mathbf x) &=& e^{-\left(\frac{\|\mathbf{x}-\mathbf{x}_i\|} {0.05}\right)^2}, \end{eqnarray} where the centers $\mathbf{x}_i$ are $N$ randomly chosen locations inside the domain. This function models a domain in which there are $N$ centers of higher permeability (for example where rock has cracked) embedded in a matrix of more pristine, unperturbed background rock. Note that here we have cut off the permeability function both above and below to ensure a bounded condition The commented program This program is an adaptation of step-20 and includes some technique of DG methods from step-12. A good part of the program is therefore very similar to step-20 and we will not comment again on these parts. Only the new stuff will be discussed in more detail. Include files All of these include files have been used before: #include <deal.II/base/quadrature_lib.h> #include <deal.II/base/function.h> #include <deal.II/lac/block_vector.h> #include <deal.II/lac/full_matrix.h> #include <deal.II/lac/block_sparse_matrix.h> #include <deal.II/lac/solver_cg.h> #include <deal.II/lac/precondition.h> #include <deal.II/lac/affine_constraints.h> #include <deal.II/grid/tria.h> #include <deal.II/grid/grid_generator.h> #include <deal.II/grid/grid_tools.h> #include <deal.II/dofs/dof_handler.h> #include <deal.II/dofs/dof_renumbering.h> #include <deal.II/dofs/dof_tools.h> #include <deal.II/fe/fe_raviart_thomas.h> #include <deal.II/fe/fe_dgq.h> #include <deal.II/fe/fe_system.h> #include <deal.II/fe/fe_values.h> #include <deal.II/numerics/vector_tools.h> #include <deal.II/numerics/data_out.h> #include <iostream> #include <fstream> In this program, we use a tensor-valued coefficient. Since it may have a spatial dependence, we consider it a tensor-valued function. The following include file provides the TensorFunction class that offers such functionality: #include <deal.II/base/tensor_function.h> Additionally, we use the class DiscreteTime to perform operations related to time incrementation. #include <deal.II/base/discrete_time.h> The last step is as in all previous programs: The TwoPhaseFlowProblem class This is the main class of the program. It is close to the one of step-20, but with a few additional functions: • assemble_rhs_S assembles the right hand side of the saturation equation. As explained in the introduction, this can't be integrated into assemble_rhs since it depends on the velocity that is computed in the first part of the time step. • get_maximal_velocity does as its name suggests. This function is used in the computation of the time step size. • project_back_saturation resets all saturation degrees of freedom with values less than zero to zero, and all those with saturations greater than one to one. The rest of the class should be pretty much obvious. The viscosity variable stores the viscosity $\mu$ that enters several of the formulas in the nonlinear equations. The variable time keeps track of the time information within the simulation. Equation data Pressure right hand side At present, the right hand side of the pressure equation is simply the zero function. However, the rest of the program is fully equipped to deal with anything else, if this is desired: template <int dim> PressureRightHandSide : public Function const unsigned int /component/ = 0) const override return 0; virtual RangeNumberType value(const Point< dim > &p, const unsigned int component=0) const Pressure boundary values The next are pressure boundary values. As mentioned in the introduction, we choose a linear pressure field: template <int dim> PressureBoundaryValues : public Function const unsigned int /component/ = 0) const override return 1 - p[0]; Saturation boundary values Then we also need boundary values on the inflow portions of the boundary. The question whether something is an inflow part is decided when assembling the right hand side, we only have to provide a functional description of the boundary values. This is as explained in the introduction: template <int dim> SaturationBoundaryValues : public Function const unsigned int /component/ = 0) const override if (p[0] == 0) return 1; return 0; Initial data Finally, we need initial data. In reality, we only need initial data for the saturation, but we are lazy, so we will later, before the first time step, simply interpolate the entire solution for the previous time step from a function that contains all vector components. We therefore simply create a function that returns zero in all components. We do that by simply forward every function to the Functions::ZeroFunction class. Why not use that right away in the places of this program where we presently use the InitialValues class? Because this way it is simpler to later go back and choose a different function for initial values. template <int dim> InitialValues : public Function const unsigned int component = 0) const override virtual void vector_value(const Point< dim > &p, Vector< RangeNumberType > &values) const virtual RangeNumberType value(const Point< dim > &p, const unsigned int component=0) const override virtual void vector_value(const Point< dim > &p, Vector< RangeNumberType > &return_value) const override The inverse permeability tensor As announced in the introduction, we implement two different permeability tensor fields. Each of them we put into a namespace of its own, so that it will be easy later to replace use of one by the other in the code. Single curving crack permeability The first function for the permeability was the one that models a single curving crack. It was already used at the end of step-20, and its functional form is given in the introduction of the present tutorial program. As in some previous programs, we have to declare a (seemingly unnecessary) default constructor of the KInverse class to avoid warnings from some compilers: namespace SingleCurvingCrack template <int dim> virtual void for (unsigned int p = 0; p < points.size(); ++p) const double distance_to_flowline = std::fabs(points[p][1] - 0.5 - 0.1 (10 * points[p][0])); const double permeability = (0.1 * 0.1)), for (unsigned int d = 0; d < dim; ++d) values[p][d][d] = 1. / permeability; } // namespace SingleCurvingCrack #define AssertDimension(dim1, dim2) ::VectorizedArray< Number, width > exp(const ::VectorizedArray< Number, width > &) ::VectorizedArray< Number, width > max(const ::VectorizedArray< Number, width > &, const ::VectorizedArray< Number, width > &) ::VectorizedArray< Number, width > sin(const ::VectorizedArray< Number, width > &) Random medium permeability This function does as announced in the introduction, i.e. it creates an overlay of exponentials at random places. There is one thing worth considering for this class. The issue centers around the problem that the class creates the centers of the exponentials using a random function. If we therefore created the centers each time we create an object of the present type, we would get a different list of centers each time. That's not what we expect from classes of this type: they should reliably represent the same function. The solution to this problem is to make the list of centers a static member variable of this class, i.e. there exists exactly one such variable for the entire program, rather than for each object of this type. That's exactly what we are going to do. The next problem, however, is that we need a way to initialize this variable. Since this variable is initialized at the beginning of the program, we can't use a regular member function for that since there may not be an object of this type around at the time. The C++ standard therefore says that only non-member and static member functions can be used to initialize a static variable. We use the latter possibility by defining a function get_centers that computes the list of center points when called. Note that this class works just fine in both 2d and 3d, with the only difference being that we use more points in 3d: by experimenting we find that we need more exponentials in 3d than in 2d (we have more ground to cover, after all, if we want to keep the distance between centers roughly equal), so we choose 40 in 2d and 100 in 3d. For any other dimension, the function does presently not know what to do so simply throws an exception indicating exactly this. namespace RandomMedium template <int dim> virtual void for (unsigned int p = 0; p < points.size(); ++p) double permeability = 0; for (unsigned int i = 0; i < centers.size(); ++i) permeability += (-(points[p] - centers[i]).norm_square() / (0.05 * 0.05)); const double normalized_permeability = for (unsigned int d = 0; d < dim; ++d) values[p][d][d] = 1. / normalized_permeability; static std::vector<Point<dim>> centers; static std::vector<Point<dim>> get_centers() const unsigned int N = (dim == 2 ? 40 : (dim == 3 ? 100 : throw ExcNotImplemented())); std::vector<Point<dim>> centers_list(N); for (unsigned int i = 0; i < N; ++i) unsigned int d = 0; < dim; ++ centers_list[i][d] = static_cast<double>(rand()) / RAND_MAX; return centers_list; template <int dim> std::vector<Point<dim>> KInverse<dim>::centers = } // namespace RandomMedium SymmetricTensor< 2, dim, Number > d(const Tensor< 2, dim, Number > &F, const Tensor< 2, dim, Number > &dF_dt) ::VectorizedArray< Number, width > min(const ::VectorizedArray< Number, width > &, const ::VectorizedArray< Number, width > &) The inverse mobility and saturation functions There are two more pieces of data that we need to describe, namely the inverse mobility function and the saturation curve. Their form is also given in the introduction: double mobility_inverse(const double S, const double viscosity) return 1.0 / (1.0 / viscosity * S * S + (1 - S) * (1 - S)); double fractional_flow(const double S, const double viscosity) return S * S / (S * S + viscosity * (1 - S) * (1 - S)); Linear solvers and preconditioners The linear solvers we use are also completely analogous to the ones used in step-20. The following classes are therefore copied verbatim from there. Note that the classes here are not only copied from step-20, but also duplicate classes in deal.II. In a future version of this example, they should be replaced by an efficient method, though. There is a single change: if the size of a linear system is small, i.e. when the mesh is very coarse, then it is sometimes not sufficient to set a maximum of src.size() CG iterations before the solver in the vmult() function converges. (This is, of course, a result of numerical round-off, since we know that on paper, the CG method converges in at most src.size() steps.) As a consequence, we set the maximum number of iterations equal to the maximum of the size of the linear system and 200. : matrix(&m) 1e-8 * src.l2_norm()); dst = 0; : system_matrix(&A) , m_inverse(&Minv) , tmp1(A.block(0, 0).m()) , tmp2(A.block(0, 0).m()) system_matrix->block(0, 1).vmult(tmp1, src); m_inverse->vmult(tmp2, tmp1); system_matrix->block(1, 0).vmult(dst, tmp2); : system_matrix(&A) , tmp1(A.block(0, 0).m()) , tmp2(A.block(0, 0).m()) system_matrix->block(0, 1).vmult(tmp1, src); system_matrix->block(0, 0).precondition_Jacobi(tmp2, tmp1); system_matrix->block(1, 0).vmult(dst, tmp2); TwoPhaseFlowProblem class implementation Here now the implementation of the main class. Much of it is actually copied from step-20, so we won't comment on it in much detail. You should try to get familiar with that program first, then most of what is happening here should be mostly clear. First for the constructor. We use $RT_k \times DQ_k \times DQ_k$ spaces. For initializing the DiscreteTime object, we don't set the time step size in the constructor because we don't have its value yet. The time step size is initially set to zero, but it will be computed before it is needed to increment time, as described in a subsection of the introduction. The time object internally prevents itself from being incremented when $dt = 0$, forcing us to set a non-zero desired size for $dt$ before advancing time. template <int dim> TwoPhaseFlowProblem<dim>::TwoPhaseFlowProblem(const unsigned int degree) : degree(degree) , n_refinement_steps(5) , time(/start time/ 0., /end time/ 1.) , viscosity(0.2) This next function starts out with well-known functions calls that create and refine a mesh, and then associate degrees of freedom with it. It does all the same things as in step-20, just now for three components instead of two. template <int dim> void TwoPhaseFlowProblem<dim>::make_grid_and_dofs() const std::vector<types::global_dof_index> dofs_per_component = const unsigned int n_u = dofs_per_component[0], n_p = dofs_per_component[dim], n_s = dofs_per_component[dim + 1]; std::cout << "Number of active cells: " << std::endl << "Number of degrees of freedom: " << dof_handler.n_dofs() << " (" << n_u << '+' << n_p << '+' << n_s << ')' << std::endl << std::endl; const std::vector<types::global_dof_index> block_sizes = {n_u, n_p, n_s}; void make_sparsity_pattern(const DoFHandler< dim, spacedim > &dof_handler, SparsityPatternBase &sparsity_pattern, const AffineConstraints< number > &constraints={}, const bool keep_constrained_dofs= true, const types::subdomain_id subdomain_id=numbers::invalid_subdomain_id) void component_wise(DoFHandler< dim, spacedim > &dof_handler, const std::vector< unsigned int > &target_component=std::vector< unsigned int >()) void hyper_cube(Triangulation< dim, spacedim > &tria, const double left=0., const double right=1., const bool colorize=false) This is the function that assembles the linear system, or at least everything except the (1,3) block that depends on the still-unknown velocity computed during this time step (we deal with this in assemble_rhs_S). Much of it is again as in step-20, but we have to deal with some nonlinearity this time. However, the top of the function is pretty much as usual (note that we set matrix and right hand side to zero at the beginning — something we didn't have to do for stationary problems since there we use each matrix object only once and it is empty at the beginning anyway). Note that in its present form, the function uses the permeability implemented in the RandomMedium::KInverse class. Switching to the single curved crack permeability function is as simple as just changing the namespace name. system_matrix = 0; system_rhs = 0; std::vector<types::global_dof_index> local_dof_indices(dofs_per_cell); std::vector<double> pressure_rhs_values(n_q_points); std::vector<double> boundary_values(n_face_q_points); std::vector<Tensor<2, dim>> k_inverse_values(n_q_points); std::vector<Vector<double>> old_solution_values(n_q_points, local_matrix = 0; local_rhs = 0; Here's the first significant difference: We have to get the values of the saturation function of the previous time step at the quadrature points. To this end, we can use the FEValues::get_function_values (previously already used in step-9, step-14 and step-15), a function that takes a solution vector and returns a list of function values at the quadrature points of the present cell. In fact, it returns the complete vector-valued solution at each quadrature point, i.e. not only the saturation but also the velocities and pressure: fe_values.get_function_values(old_solution, old_solution_values); Then we also have to get the values of the pressure right hand side and of the inverse permeability tensor at the quadrature points: With all this, we can now loop over all the quadrature points and shape functions on this cell and assemble those parts of the matrix and right hand side that we deal with in this function. The individual terms in the contributions should be self-explanatory given the explicit form of the bilinear form stated in the introduction: for (unsigned int q = 0; q < n_q_points; ++q) for (unsigned int i = 0; i < dofs_per_cell; ++i) const double old_s = old_solution_values[q](dim + 1); const double div_phi_i_u = fe_values[velocities].divergence(i, q); const double phi_i_p = fe_values[pressure].value(i, q); const double phi_i_s = fe_values[saturation].value(i, q); for (unsigned int j = 0; j < dofs_per_cell; ++j) fe_values[velocities].value(j, q); const double div_phi_j_u = fe_values[velocities].divergence(j, q); const double phi_j_p = fe_values[pressure].value(j, q); const double phi_j_s = fe_values[saturation].value(j, q); local_matrix(i, j) += (phi_i_u * k_inverse_values[q] * mobility_inverse(old_s, viscosity) * phi_j_u - div_phi_i_u * phi_j_p - phi_i_p * div_phi_j_u + phi_i_s * phi_j_s) * local_rhs(i) += (-phi_i_p * pressure_rhs_values[q]) * fe_values.JxW(q); Next, we also have to deal with the pressure boundary values. This, again is as in step-20: for (const auto &face : cell->face_iterators()) if (face->at_boundary()) fe_face_values.reinit(cell, face); fe_face_values.get_quadrature_points(), boundary_values); for (unsigned int q = 0; q < n_face_q_points; ++q) for (unsigned int i = 0; i < dofs_per_cell; ++i) fe_face_values[velocities].value(i, q); local_rhs(i) += -(phi_i_u * fe_face_values.normal_vector(q) * boundary_values[q] * fe_face_values.JxW(q)); The final step in the loop over all cells is to transfer local contributions into the global matrix and right hand side vector: for (unsigned int i = 0; i < dofs_per_cell; ++i) for (unsigned int j = 0; j < dofs_per_cell; ++j) local_matrix(i, j)); for (unsigned int i = 0; i < dofs_per_cell; ++i) system_rhs(local_dof_indices[i]) += local_rhs(i); So much for assembly of matrix and right hand side. Note that we do not have to interpolate and apply boundary values since they have all been taken care of in the weak form already. As explained in the introduction, we can only evaluate the right hand side of the saturation equation once the velocity has been computed. We therefore have this separate function to this end. std::vector<Vector<double>> old_solution_values(n_q_points, std::vector<Vector<double>> old_solution_values_face(n_face_q_points, std::vector<Vector<double>> old_solution_values_face_neighbor( std::vector<Vector<double>> present_solution_values(n_q_points, std::vector<Vector<double>> present_solution_values_face( std::vector<double> neighbor_saturation(n_face_q_points); std::vector<types::global_dof_index> local_dof_indices(dofs_per_cell); SaturationBoundaryValues<dim> saturation_boundary_values; local_rhs = 0; fe_values.get_function_values(old_solution, old_solution_values); fe_values.get_function_values(solution, present_solution_values); First for the cell terms. These are, following the formulas in the introduction, $(S^n,\sigma)-(F(S^n) \mathbf{v}^{n+1},\nabla \sigma)$, where $\sigma$ is the saturation component of the test for (unsigned int q = 0; q < n_q_points; ++q) for (unsigned int i = 0; i < dofs_per_cell; ++i) const double old_s = old_solution_values[q](dim + 1); for (unsigned int d = 0; d < dim; ++d) present_u[d] = present_solution_values[q](d); const double phi_i_s = fe_values[saturation].value(i, q); fe_values[saturation].gradient(i, q); local_rhs(i) += (time.get_next_step_size() * fractional_flow(old_s, viscosity) * present_u * grad_phi_i_s + old_s * phi_i_s) * Secondly, we have to deal with the flux parts on the face boundaries. This was a bit more involved because we first have to determine which are the influx and outflux parts of the cell boundary. If we have an influx boundary, we need to evaluate the saturation on the other side of the face (or the boundary values, if we are at the boundary of the domain). All this is a bit tricky, but has been explained in some detail already in step-9. Take a look there how this is supposed to work! for (const auto face_no : cell->face_indices()) fe_face_values.reinit(cell, face_no); if (cell->at_boundary(face_no)) fe_face_values.get_quadrature_points(), neighbor_saturation); const auto neighbor = cell->neighbor(face_no); const unsigned int neighbor_face = fe_face_values_neighbor.reinit(neighbor, neighbor_face); old_solution, old_solution_values_face_neighbor); for (unsigned int q = 0; q < n_face_q_points; ++q) neighbor_saturation[q] = old_solution_values_face_neighbor[q](dim + 1); for (unsigned int q = 0; q < n_face_q_points; ++q) for (unsigned int d = 0; d < dim; ++d) present_u_face[d] = present_solution_values_face[q](d); const double normal_flux = present_u_face * fe_face_values.normal_vector(q); const bool is_outflow_q_point = (normal_flux >= 0); for (unsigned int i = 0; i < dofs_per_cell; ++i) local_rhs(i) -= time.get_next_step_size() * normal_flux * fractional_flow((is_outflow_q_point == true ? old_solution_values_face[q](dim + 1) : viscosity) * fe_face_values[saturation].value(i, q) * for (unsigned int i = 0; i < dofs_per_cell; ++i) system_rhs(local_dof_indices[i]) += local_rhs(i); After all these preparations, we finally solve the linear system for velocity and pressure in the same way as in step-20. After that, we have to deal with the saturation equation (see below): template <int dim> void TwoPhaseFlowProblem<dim>::solve() const InverseMatrix<SparseMatrix<double>> m_inverse( system_matrix.block(0, 0)); First the pressure, using the pressure Schur complement of the first two equations: m_inverse.vmult(tmp, system_rhs.block(0)); system_matrix.block(1, 0).vmult(schur_rhs, tmp); schur_rhs -= system_rhs.block(1); ApproximateSchurComplement approximate_schur_complement(system_matrix); InverseMatrix<ApproximateSchurComplement> preconditioner( 1e-12 * schur_rhs.l2_norm()); std::cout << " " << solver_control.last_step() << " CG Schur complement iterations for pressure." << std::endl; LinearOperator< Range_2, Domain_2, Payload > schur_complement(const LinearOperator< Domain_1, Range_1, Payload > &A_inv, const LinearOperator< Range_1, Domain_2, Payload > &B, const LinearOperator< Range_2, Domain_1, Payload > &C, const LinearOperator< Range_2, Domain_2, Payload > &D) Now the velocity: system_matrix.block(0, 1).vmult(tmp, solution.block(1)); tmp = -1; tmp += system_rhs.block(0); m_inverse.vmult(solution.block(0), tmp); Finally, we have to take care of the saturation equation. The first business we have here is to determine the time step using the formula in the introduction. Knowing the shape of our domain and that we created the mesh by regular subdivision of cells, we can compute the diameter of each of our cells quite easily (in fact we use the linear extensions in coordinate directions of the cells, not the diameter). Note that we will learn a more general way to do this in step-24, where we use the GridTools::minimal_cell_diameter function. The maximal velocity we compute using a helper function to compute the maximal velocity defined below, and with all this we can evaluate our new time step length. We use the method DiscreteTime::set_desired_next_time_step() to suggest the new calculated value of the time step to the DiscreteTime object. In most cases, the time object uses the exact provided value to increment time. It some case, the step size may be modified further by the time object. For example, if the calculated time increment overshoots the end time, it is truncated accordingly. (n_refinement_steps)) / ::VectorizedArray< Number, width > pow(const ::VectorizedArray< Number, width > &, const Number p) The next step is to assemble the right hand side, and then to pass everything on for solution. At the end, we project back saturations onto the physically reasonable range: 1e-8 system_rhs.block(2).l2_norm()); cg.solve(system_matrix.block(2, 2), std::cout << " " << solver_control.last_step() << " CG iterations for saturation." << std::endl; old_solution = solution; There is nothing surprising here. Since the program will do a lot of time steps, we create an output file only every fifth time step and skip all other time steps at the top of the file already. When creating file names for output close to the bottom of the function, we convert the number of the time step to a string representation that is padded by leading zeros to four digits. We do this because this way all output file names have the same length, and consequently sort well when creating a directory listing. template <int dim> void TwoPhaseFlowProblem<dim>::output_results() const if (time.get_step_number() % 5 != 0) std::vector<std::string> solution_names; switch (dim) case 2: solution_names = {"u", "v", "p", "S"}; case 3: solution_names = {"u", "v", "w", "p", "S"}; data_out.add_data_vector(solution, solution_names); data_out.build_patches(degree + 1); std::ofstream output("solution-" + void attach_dof_handler(const DoFHandler< dim, spacedim > &) #define DEAL_II_NOT_IMPLEMENTED() std::string int_to_string(const unsigned int value, const unsigned int digits=numbers::invalid_unsigned_int) In this function, we simply run over all saturation degrees of freedom and make sure that if they should have left the physically reasonable range, that they be reset to the interval $[0,1]$. To do this, we only have to loop over all saturation components of the solution vector; these are stored in the block 2 (block 0 are the velocities, block 1 are the pressures). It may be instructive to note that this function almost never triggers when the time step is chosen as mentioned in the introduction. However, if we choose the timestep only slightly larger, we get plenty of values outside the proper range. Strictly speaking, the function is therefore unnecessary if we choose the time step small enough. In a sense, the function is therefore only a safety device to avoid situations where our entire solution becomes unphysical because individual degrees of freedom have become unphysical a few time steps earlier. template <int dim> void TwoPhaseFlowProblem<dim>::project_back_saturation() for (unsigned int i = 0; i < solution.block(2).size(); ++i) if (solution.block(2)(i) < 0) solution.block(2)(i) = 0; else if (solution.block(2)(i) > 1) solution.block(2)(i) = 1; The following function is used in determining the maximal allowable time step. What it does is to loop over all quadrature points in the domain and find what the maximal magnitude of the velocity is. template <int dim> double TwoPhaseFlowProblem<dim>::get_maximal_velocity() const const unsigned int n_q_points = quadrature_formula.size(); std::vector<Vector<double>> solution_values(n_q_points, double max_velocity = 0; for (const auto &cell : dof_handler.active_cell_iterators()) fe_values.get_function_values(solution, solution_values); for (unsigned int q = 0; q < n_q_points; ++q) for (unsigned int i = 0; i < dim; ++i) velocity[i] = solution_values[q](i); max_velocity = (max_velocity, velocity.norm()); return max_velocity; This is the final function of our main class. Its brevity speaks for itself. There are only two points worth noting: First, the function projects the initial values onto the finite element space at the beginning; the VectorTools::project function doing this requires an argument indicating the hanging node constraints. We have none in this program (we compute on a uniformly refined mesh), but the function requires the argument anyway, of course. So we have to create a constraint object. In its original state, constraint objects are unsorted, and have to be sorted (using the AffineConstraints::close function) before they can be used. This is what we do here, and which is why we can't simply call the VectorTools::project function with an anonymous temporary object AffineConstraints<double>() as the second argument. The second point worth mentioning is that we only compute the length of the present time step in the middle of solving the linear system corresponding to each time step. We can therefore output the present time of a time step only at the end of the time step. We increment time by calling the method DiscreteTime::advance_time() inside the loop. Since we are reporting the time and dt after we increment it, we have to call the method DiscreteTime::get_previous_step_size() instead of DiscreteTime::get_next_step_size(). After many steps, when the simulation reaches the end time, the last dt is chosen by the DiscreteTime class in such a way that the last step finishes exactly at the end time. template <int dim> void TwoPhaseFlowProblem<dim>::run() std::cout << "Timestep " << time.get_step_number() + 1 << std::endl; std::cout << " Now at t=" << time.get_current_time() << ", dt=" << time.get_previous_step_size() << '.' << std::endl << std::endl; while (time.is_at_end() == false); } // namespace Step21 The main function That's it. In the main function, we pass the degree of the finite element space to the constructor of the TwoPhaseFlowProblem object. Here, we use zero-th degree elements, i.e. $RT_0\times DQ_0 \ times DQ_0$. The rest is as in all the other programs. int main() using namespace Step21; TwoPhaseFlowProblem<2> two_phase_flow_problem(0); catch (std::exception &exc) std::cerr << std::endl << std::endl << "----------------------------------------------------" << std::endl; std::cerr << "Exception on processing: " << std::endl << exc.what() << std::endl << "Aborting!" << std::endl << "----------------------------------------------------" << std::endl; return 1; catch (...) std::cerr << std::endl << std::endl << "----------------------------------------------------" << std::endl; std::cerr << "Unknown exception!" << std::endl << "Aborting!" << std::endl << "----------------------------------------------------" << std::endl; return 1; return 0; The code as presented here does not actually compute the results found on the web page. The reason is, that even on a decent computer it runs more than a day. If you want to reproduce these results, modify the end time of the DiscreteTime object to 250 within the constructor of TwoPhaseFlowProblem. If we run the program, we get the following kind of output: Number of active cells: 1024 Number of degrees of freedom: 4160 (2112+1024+1024) Timestep 1 22 CG Schur complement iterations for pressure. 1 CG iterations for saturation. Now at t=0.0326742, dt=0.0326742. Timestep 2 17 CG Schur complement iterations for pressure. 1 CG iterations for saturation. Now at t=0.0653816, dt=0.0327074. Timestep 3 17 CG Schur complement iterations for pressure. 1 CG iterations for saturation. Now at t=0.0980651, dt=0.0326836. As we can see, the time step is pretty much constant right from the start, which indicates that the velocities in the domain are not strongly dependent on changes in saturation, although they certainly are through the factor $\lambda(S)$ in the pressure equation. Our second observation is that the number of CG iterations needed to solve the pressure Schur complement equation drops from 22 to 17 between the first and the second time step (in fact, it remains around 17 for the rest of the computations). The reason is actually simple: Before we solve for the pressure during a time step, we don't reset the solution variable to zero. The pressure (and the other variables) therefore have the previous time step's values at the time we get into the CG solver. Since the velocities and pressures don't change very much as computations progress, the previous time step's pressure is actually a good initial guess for this time step's pressure. Consequently, the number of iterations we need once we have computed the pressure once is significantly reduced. The final observation concerns the number of iterations needed to solve for the saturation, i.e. one. This shouldn't surprise us too much: the matrix we have to solve with is the mass matrix. However, this is the mass matrix for the $DGQ_0$ element of piecewise constants where no element couples with the degrees of freedom on neighboring cells. The matrix is therefore a diagonal one, and it is clear that we should be able to invert this matrix in a single CG iteration. With all this, here are a few movies that show how the saturation progresses over time. First, this is for the single crack model, as implemented in the SingleCurvingCrack::KInverse class: As can be seen, the water rich fluid snakes its way mostly along the high-permeability zone in the middle of the domain, whereas the rest of the domain is mostly impermeable. This and the next movie are generated using n_refinement_steps=7, leading to a $128\times 128$ mesh with some 16,000 cells and about 66,000 unknowns in total. The second movie shows the saturation for the random medium model of class RandomMedium::KInverse, where we have randomly distributed centers of high permeability and fluid hops from one of these zones to the next: Finally, here is the same situation in three space dimensions, on a mesh with n_refinement_steps=5, which produces a mesh of some 32,000 cells and 167,000 degrees of freedom: To repeat these computations, all you have to do is to change the line TwoPhaseFlowProblem<2> two_phase_flow_problem(0); in the main function to TwoPhaseFlowProblem<3> two_phase_flow_problem(0); The visualization uses a cloud technique, where the saturation is indicated by colored but transparent clouds for each cell. This way, one can also see somewhat what happens deep inside the domain. A different way of visualizing would have been to show isosurfaces of the saturation evolving over time. There are techniques to plot isosurfaces transparently, so that one can see several of them at the same time like the layers of an onion. So why don't we show such isosurfaces? The problem lies in the way isosurfaces are computed: they require that the field to be visualized is continuous, so that the isosurfaces can be generated by following contours at least across a single cell. However, our saturation field is piecewise constant and discontinuous. If we wanted to plot an isosurface for a saturation $S=0.5$, chances would be that there is no single point in the domain where that saturation is actually attained. If we had to define isosurfaces in that context at all, we would have to take the interfaces between cells, where one of the two adjacent cells has a saturation greater than and the other cell a saturation less than 0.5. However, it appears that most visualization programs are not equipped to do this kind of transformation. Possibilities for extensions There are a number of areas where this program can be improved. Three of them are listed below. All of them are, in fact, addressed in a tutorial program that forms the continuation of the current one: step-43. At present, the program is not particularly fast: the 2d random medium computation took about a day for the 1,000 or so time steps. The corresponding 3d computation took almost two days for 800 time steps. The reason why it isn't faster than this is twofold. First, we rebuild the entire matrix in every time step, although some parts such as the $B$, $B^T$, and $M^S$ blocks never change. Second, we could do a lot better with the solver and preconditioners. Presently, we solve the Schur complement $B^TM^u(S)^{-1}B$ with a CG method, using $[B^T (\textrm{diag}(M^u(S)))^{-1} B]^{-1} $ as a preconditioner. Applying this preconditioner is expensive, since it involves solving a linear system each time. This may have been appropriate for step-20, where we have to solve the entire problem only once. However, here we have to solve it hundreds of times, and in such cases it is worth considering a preconditioner that is more expensive to set up the first time, but cheaper to apply later on. One possibility would be to realize that the matrix we use as preconditioner, $B^T (\textrm{diag}(M^u(S)))^{-1} B$ is still sparse, and symmetric on top of that. If one looks at the flow field evolve over time, we also see that while $S$ changes significantly over time, the pressure hardly does and consequently $B^T (\textrm{diag}(M^u(S)))^{-1} B \approx B^T (\textrm{diag}(M^u(S^0)))^ {-1} B$. In other words, the matrix for the first time step should be a good preconditioner also for all later time steps. With a bit of back-and-forthing, it isn't hard to actually get a representation of it as a SparseMatrix object. We could then hand it off to the SparseMIC class to form a sparse incomplete Cholesky decomposition. To form this decomposition is expensive, but we have to do it only once in the first time step, and can then use it as a cheap preconditioner in the future. We could do better even by using the SparseDirectUMFPACK class that produces not only an incomplete, but a complete decomposition of the matrix, which should yield an even better preconditioner. Finally, why use the approximation $B^T (\textrm{diag}(M^u(S)))^{-1} B$ to precondition $B^T M^u(S)^{-1} B$? The latter matrix, after all, is the mixed form of the Laplace operator on the pressure space, for which we use linear elements. We could therefore build a separate matrix $A^p$ on the side that directly corresponds to the non-mixed formulation of the Laplacian, for example using the bilinear form $(\mathbf{K}\lambda(S^n) \nabla \varphi_i,\nabla\varphi_j)$. We could then form an incomplete or complete decomposition of this non-mixed matrix and use it as a preconditioner of the mixed form. Using such techniques, it can reasonably be expected that the solution process will be faster by at least an order of magnitude. Time stepping In the introduction we have identified the time step restriction \[ \triangle t_{n+1} \le \frac h{\|\mathbf{u}^{n+1}(\mathbf{x})\|} \] that has to hold globally, i.e. for all $\mathbf x$. After discretization, we satisfy it by choosing \[ \triangle t_{n+1} = \frac {\min_K h_K}{\max_{\mathbf{x}}\|\mathbf{u}^{n+1}(\mathbf{x})\|}. \] This restriction on the time step is somewhat annoying: the finer we make the mesh the smaller the time step; in other words, we get punished twice: each time step is more expensive to solve and we have to do more time steps. This is particularly annoying since the majority of the additional work is spent solving the implicit part of the equations, i.e. the pressure-velocity system, whereas it is the hyperbolic transport equation for the saturation that imposes the time step restriction. To avoid this bottleneck, people have invented a number of approaches. For example, they may only re-compute the pressure-velocity field every few time steps (or, if you want, use different time step sizes for the pressure/velocity and saturation equations). This keeps the time step restriction on the cheap explicit part while it makes the solution of the implicit part less frequent. Experiments in this direction are certainly worthwhile; one starting point for such an approach is the paper by Zhangxin Chen, Guanren Huan and Baoyan Li: An improved IMPES method for two-phase flow in porous media, Transport in Porous Media, 54 (2004), pp. 361—376. There are certainly many other papers on this topic as well, but this one happened to land on our desk a while back. Adaptivity would also clearly help. Looking at the movies, one clearly sees that most of the action is confined to a relatively small part of the domain (this particularly obvious for the saturation, but also holds for the velocities and pressures). Adaptivity can therefore be expected to keep the necessary number of degrees of freedom low, or alternatively increase the accuracy. On the other hand, adaptivity for time dependent problems is not a trivial thing: we would have to change the mesh every few time steps, and we would have to transport our present solution to the next mesh every time we change it (something that the SolutionTransfer class can help with). These are not insurmountable obstacles, but they do require some additional coding and more than we felt comfortable was worth packing into this tutorial program. The plain program (0.1 * 0.1)), values[p][d][d] = 1. / permeability; (0.05 * 0.05)); values[p][d][d] = 1. / normalized_permeability; std::vector<Point<dim>> centers_list(N); std::vector<Point<dim>> KInverse<dim>::centers = dst = 0; : system_matrix(&A) , m_inverse(&Minv) , tmp1(A.block(0, 0).m()) , tmp2(A.block(0, 0).m()) system_matrix->block(0, 1).vmult(tmp1, src); m_inverse->vmult(tmp2, tmp1); system_matrix->block(1, 0).vmult(dst, tmp2); : system_matrix(&A) , tmp1(A.block(0, 0).m()) , tmp2(A.block(0, 0).m()) system_matrix->block(0, 1).vmult(tmp1, src); system_matrix->block(0, 0).precondition_Jacobi(tmp2, tmp1); system_matrix->block(1, 0).vmult(dst, tmp2); : degree(degree) , n_refinement_steps(5) , viscosity(0.2) n_p = dofs_per_component[dim], n_s = dofs_per_component[dim + 1]; << std::endl << std::endl; system_matrix = 0; system_rhs = 0; std::vector<types::global_dof_index> local_dof_indices(dofs_per_cell); std::vector<double> pressure_rhs_values(n_q_points); std::vector<double> boundary_values(n_face_q_points); std::vector<Tensor<2, dim>> k_inverse_values(n_q_points); std::vector<Vector<double>> old_solution_values(n_q_points, local_matrix = 0; local_rhs = 0; fe_values.get_function_values(old_solution, old_solution_values); fe_values[velocities].value(j, q); fe_values[velocities].divergence(j, q); local_matrix(i, j) += (phi_i_u * k_inverse_values[q] * mobility_inverse(old_s, viscosity) * phi_j_u - div_phi_i_u * phi_j_p - phi_i_p * div_phi_j_u + phi_i_s * phi_j_s) * local_rhs(i) += (-phi_i_p * pressure_rhs_values[q]) * fe_values.JxW(q); if (face->at_boundary()) fe_face_values.reinit(cell, face); fe_face_values.get_quadrature_points(), boundary_values); fe_face_values[velocities].value(i, q); local_rhs(i) += -(phi_i_u * fe_face_values.normal_vector(q) * boundary_values[q] * fe_face_values.JxW(q)); local_matrix(i, j)); system_rhs(local_dof_indices[i]) += local_rhs(i); std::vector<Vector<double>> old_solution_values(n_q_points, std::vector<Vector<double>> old_solution_values_face(n_face_q_points, std::vector<Vector<double>> old_solution_values_face_neighbor( std::vector<Vector<double>> present_solution_values(n_q_points, std::vector<Vector<double>> present_solution_values_face( std::vector<double> neighbor_saturation(n_face_q_points); std::vector<types::global_dof_index> local_dof_indices(dofs_per_cell); SaturationBoundaryValues<dim> saturation_boundary_values; local_rhs = 0; fe_values.get_function_values(old_solution, old_solution_values); fe_values.get_function_values(solution, present_solution_values); present_u[d] = present_solution_values[q](d); fe_values[saturation].gradient(i, q); local_rhs(i) += (time.get_next_step_size() * fractional_flow(old_s, viscosity) * present_u * grad_phi_i_s + old_s * phi_i_s) * fe_face_values.reinit(cell, face_no); fe_face_values.get_quadrature_points(), neighbor_saturation); fe_face_values_neighbor.reinit(neighbor, neighbor_face); old_solution, old_solution_values_face_neighbor); neighbor_saturation[q] = old_solution_values_face_neighbor[q](dim + 1); present_u_face[d] = present_solution_values_face[q](d); present_u_face * fe_face_values.normal_vector(q); local_rhs(i) -= time.get_next_step_size() * normal_flux * old_solution_values_face[q](dim + 1) : viscosity) * system_rhs(local_dof_indices[i]) += local_rhs(i); system_matrix.block(0, 0)); m_inverse.vmult(tmp, system_rhs.block(0)); system_matrix.block(1, 0).vmult(schur_rhs, tmp); schur_rhs -= system_rhs.block(1); ApproximateSchurComplement approximate_schur_complement(system_matrix); InverseMatrix<ApproximateSchurComplement> preconditioner( 1e-12 * schur_rhs.l2_norm()); system_matrix.block(0, 1).vmult(tmp, solution.block(1)); tmp = -1; tmp += system_rhs.block(0); m_inverse.vmult(solution.block(0), tmp); 1e-8 system_rhs.block(2).l2_norm()); cg.solve(system_matrix.block(2, 2), old_solution = solution; std::vector<std::string> solution_names; solution.block(2)(i) = 0; solution.block(2)(i) = 1; std::vector<Vector<double>> solution_values(n_q_points, fe_values.get_function_values(solution, solution_values); velocity[i] = solution_values[q](i); << std::endl << std::endl; TwoPhaseFlowProblem<2> two_phase_flow_problem(0); std::cerr << std::endl << std::endl << std::endl; << exc.what() << std::endl << std::endl; std::cerr << std::endl << std::endl << std::endl; << std::endl;	{"url":"https://www.dealii.org/developer/doxygen/deal.II/step_21.html","timestamp":"2024-11-02T02:23:41Z","content_type":"application/xhtml+xml","content_length":"256466","record_id":"<urn:uuid:1e1460bb-4764-4c20-856f-40356b395dab>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00810.warc.gz"}
Vestnik Moskovskogo Universiteta. Seriya 1. Matematika. Mekhanika Maximal Linked Systems / Dobrynina M.A. // Vestnik Moskovskogo Universiteta. Seriya 1. Matematika. Mekhanika. 2011. № 2. P. 27-30 [Moscow Univ. Math. Bulletin. Vol. 72, N 2, 2017. P. 0]. A compact space X such that the space λ^3(X) of maximal 3-linked systems is not normal is constructed. It is proved that for any product of infinite separable spaces there exists a maximal linked system with the support equal to the product space. It is proved that if the space X is connected and separable, then the set of maximal 3-linked systems with connected supports is everywhere dense in the superextension λ(X). The properties of seminormal functors preserving one-to-one points are discussed. Key words: maximal k-linked systems, support, superextension functor, seminormal functors.	{"url":"http://vestnik.math.msu.su/en/DATA/2011/2/node4","timestamp":"2024-11-13T07:59:01Z","content_type":"text/html","content_length":"5614","record_id":"<urn:uuid:83087098-8ce1-443e-903d-f98be8a66ff9>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00270.warc.gz"}
The 32nd IASTED International Conference on Modelling, Identification and Control MIC 2013 February 11 – 13, 2013 Innsbruck, Austria An Introduction to Inverse Simulation Methods for System Modelling and Design Inverse simulation is a tool that can be used to find system inputs such that model outputs will match given time histories. Thus, through inverse simulation, the behaviour of a model can be investigated in ways that are distinctly different from those available using conventional simulation techniques which provide model outputs for given sets of inputs and initial conditions. The inverse simulation approach could, for example, involve finding the control inputs necessary to perform a given aircraft manoeuvre or the inputs needed for a road vehicle to make a specified turn or match a given acceleration profile. This has been found to be particularly helpful in considering actuator design issues in the context of automatic control, where they can provide a clear indication of problems of amplitude and rate limiting. It should be noted that inverse simulation methods may be applied both with linear and nonlinear models and to multi-input multi-output models. They can also be used for model validation, where experimental data obtained from tests on a real engineering system are applied to the model. Comparisons between the model and system behaviour are then made in terms of the differences between measured inputs from the experiments and inputs predicted from the inverse simulation. This tutorial will involve a review of several methods that are available for inverse simulation, including various optimisation-based techniques, methods based on numerical solutions of differential algebraic equations and techniques based on a feedback systems approach. Much of the interest in inverse simulation has arisen in the context of specific applications, such as helicopter flight mechanics modelling and handling qualities studies, but it is demonstrated through the tutorial that the inverse simulation approach can also be useful in many other fields. Particular emphasis is given in the tutorial to methods of inverse simulation that are based on the properties of closed-loop systems and have been the subject of recently published research. Applications considered are drawn from personal experience of the presenter. The areas involved may include problems of helicopter flight control, underwater vehicle dynamics, train performance modelling and process control. The objectives are simply to provide an introduction the topic and to use a number of applications and case studies to illustrate its use. Background Knowledge Expected of the Participants The background knowledge needed for those attending this tutorial is quite basic. Participants should have some experience of conventional continuous system simulation methods and a general knowledge (at first degree level) of mathematical modelling concepts in the context of engineering applications.	{"url":"https://www.iasted.org/conferences/speaker10-794.html","timestamp":"2024-11-09T23:11:45Z","content_type":"application/xhtml+xml","content_length":"14466","record_id":"<urn:uuid:ac3eb488-f459-4426-a287-9f5157df6312>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00440.warc.gz"}
Excel: How to Use Greater Than or Equal to in IF Function \| Online Tutorials Library List \| Tutoraspire.com Excel: How to Use Greater Than or Equal to in IF Function by Tutor Aspire In Excel, you can use the >= operator to check if a value in a given cell is greater than or equal to some value. To use this operator in an IF function, you can use the following syntax: =IF(C2>=20, "Yes", "No") For this particular formula, if the value in cell C2 is greater than or equal to 20, the function returns “Yes.” Otherwise it returns “No.” The following examples show how to use this syntax in practice. Example: Create IF Function to Return Yes or No in Excel Suppose we have the following dataset in Excel that contains information about various basketball players: We can type the following formula into cell D2 to return “Yes” if the number of points in cell C2 is equal to or greater than 20: =IF(C2>=20, "Yes", "No") We can then drag and fill this formula down to each remaining cell in column D: The formula returns either “Yes” or “No” in each row depending on whether or not the points value in column C is greater than or equal to 20. Note that you can also use the greater than or equal to sign (>=) to compare the value in two cells. For example, suppose we have the following dataset that shows the number of points scored and allowed by various basketball players: We can type the following formula into cell E2 to return “Yes” if the number of points in cell C2 is equal to or greater than the number of points allowed in cell D2: =IF(C2>=D2, "Yes", "No") We can then drag and fill this formula down to each remaining cell in column E: The formula returns either “Yes” or “No” in each row depending on whether or not the points value in column C is greater than or equal to the corresponding points value in column D. Additional Resources The following tutorials explain how to perform other common tasks in Excel: Excel: How to Use an IF Function with 3 Conditions Excel: How to Use an IF Function with Range of Values Excel: How to Use an IF Function with Dates Share 0 FacebookTwitterPinterestEmail previous post How to Search for an Asterisk in a Cell in Excel You may also like	{"url":"https://tutoraspire.com/excel-if-function-greater-than-or-equal-to/","timestamp":"2024-11-03T13:59:49Z","content_type":"text/html","content_length":"350942","record_id":"<urn:uuid:e4c3f308-c257-44af-99e8-9791b898d418>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00823.warc.gz"}
Area and Perimeter Relationships & Problem Solving, part 5 - Teach Think Elementary Area and Perimeter are difficult math concepts that kids usualy learn in upper elementary. As teachers, it can feel overwhelming to tackle all of that content. I’m breaking down the perimeter and area standards into manageable chunks in this blog post series. Part 2: Measuring Area & Multiplication 3: Composing & Decomposing Area and The Distributive Property (coming soon!) This is Part 5: Area & Perimeter Relationships and Problem Solving Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures. 8. Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters I’ve broken perimeter down into two sections here. Part 4 was understanding perimeter and solving problems with perimeter on its own. In this part, we’ll tackle the relationship with area and perimeter. But it’s much easier to understand the relationship between area and perimeter if you’re familiar with both of those things first, of course! This is another fun one, in my opinion! It’s time to make lots of rectangles and explore and figure things out, and that’s my favorite kind of math. Overwhelmed?? I’ve spent hours and hours (like soooo many hours) thinking about this and creating a math unit so you don’t have to. Get my Perimeter & Area Math Unit here. Concept: distinguish between linear and area measures. If you’ve been playing along at home, this one should be pretty much done at this point! I always introduced this in the beginning, before even teaching area. That seems important to me because up until this point, kids have only really worked with linear measurement. I like to make it clear to them at the beginning of the unit that we’re now in uncharted waters and learning a new way of measuring. (They’ll learn a third way in 5th grade, when they tackle volume.) Concept: rectangles with the same perimeter and different areas and vice versa. The goal here is for kids to look for the relationship and to make generalizations about the relationship between area and perimeter. This is a fun concept that gives kids a chance to explore and come up with their own ideas and understandings. If you need an idea for how to manage all of this exploration, here’s what I did: -break the kids up into groups of 2-4. -assign each group an area or perimeter to investigate. (Definitely do these on two different days!!) -give them scissors, crayons, and lots and lots of grid paper. -let them see how many different rectangles they can make with their assigned perimeter or area. -give them a chance to present their findings to the group. If you have a class that large or you’re short on time, they can make a display on a desk or bulletin board and the other students can circulate, museum-style. -lead a conversation about how they figured out the rectangles, how do they know they have them all, what did they find, etc. -Kids can write exit tickets with their ideas for accountability & assessment. -After you’ve had one day with same perimeters and one day with same areas, have the kids compare the two and discuss if they see any relationships. **the question will inevitably come up of whether 5×8 and 8×5 are the same rectangle or not, and that will lead to plenty of good discussions. The answer doesn’t really matter, in my opinion, but it’s a great way for kids to think critically and develop deeper concepts. Overwhelmed?? I’ve spent hours and hours (like soooo many hours) thinking about this and creating a math unit so you don’t have to. Get my Perimeter & Area Math Unit here.	{"url":"https://teachthinkelementary.blog/area-and-perimeter-relationships-part-5/","timestamp":"2024-11-02T20:12:13Z","content_type":"text/html","content_length":"221580","record_id":"<urn:uuid:dba149ef-6942-4f21-9fdb-f71156c5a8ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00024.warc.gz"}
How To Teach Skip Counting - Preschool Activities Nook How would you describe your teaching style? Are you a teacher who likes to explain things step by step or prefer to let students discover things on their own? Teaching children is a rewarding experience. Children love learning new things and they also enjoy being taught by someone who cares about them. When teaching young kids, teachers should always remember that they are the ones responsible for shaping the future of our society. Skip counting is a skill that helps children develop their math skills. Skip counting is a method of teaching numbers by using visual cues instead of counting each number individually. This method is useful because it allows children to focus on understanding concepts rather than memorizing numbers. What Is Skip Counting? The method of skip counting is an effective way of teaching numbers from 1-10 in English. It involves asking questions like: “How many times did I say ‘one’?” or “How many times do we have to count before we reach ten?” The idea behind this method is that children can learn how to count by thinking about what they see around them. They can then apply these ideas when they need to count. This method has been used for years by parents and teachers alike as a fun way of helping children learn how to count. However, there are some important aspects of skip counting that most people don’t know about. Starting Tips Here are some tips to help you get started with this method: 1. Start small: When introducing children to skip counting, start with numbers up to 10. You will find that children are more likely to understand the concept if they are given smaller examples first. If you want to introduce skip counting at school, try starting with numbers from one to five. 2. Use pictures: Pictures are great tools for teaching skip counting. For example, you could use a picture of a clock face to show children how to count from 12 o’clock to 11 o’clock. Or you could draw a circle and ask children to count the number of dots inside the circle. 3. Ask questions: Asking questions is another good way to introduce skip counting. For example: “How many times does it take us to go all the way around?” or “How many times do we need to count before we reach 10?”. 4. Make sure everyone understands: Once children have learned how to count, make sure that everyone knows exactly what they are doing. Explain to children that they need to count until they reach 10. Then, tell them to stop counting and write down the answer. 5. Practice makes perfect: Remember that practice makes perfect! So, keep practicing and you’ll soon be able to teach skip counting confidently. Difficulties Kids Encounter When Learning To Skip Count There are some difficulties that children may encounter when trying to learn skip counting. Here are some common problems that children might have while learning skip counting: Sometimes children struggle with learning skip counting because they forget which direction they are going in. To overcome this problem, explain to children that they should always count forward. Children may not be able to recognize the end point and sometimes struggle with recognizing the end point of a number sequence. To solve this problem, give children lots of opportunities to practice Children often confuse the order of operations when they are learning skip counting. In other words, they think that they must add the numbers together before they multiply them. To avoid this confusion, let children know that multiplication comes after addition. This helps children remember that they must subtract the first digit from the last digit. Children also struggle when they cannot recall the correct number of times they counted. To ensure that children remember the correct number of counts, remind them to count each time they pass through a dot on the clock face. Skip Counting Using Collections When teaching skip counting, it’s important to use collections such as clocks, calendars, and watches to help children remember which numbers they have already counted. For example, you can use a clock to help children learn to skip counting. Tell children that they need to start by counting from 12 o’clock (0). After they have reached 0, they need to continue counting until they reach 6 o’clock (6). At 6 o’clock, they need to stop counting and write their answer. Therefore, this will be an effective method to help have less confusion when learning and will make the job a whole lot easier! Using Hundreds Grids To help children understand skip counting, you can create a hundreds grid for them to follow. You can then place a zero at the beginning of the number line. The zero represents the starting point for You can also use a calendar to help children learn skip-counting. For example, if you want to show children that 1 + 2 = 3, you could draw a circle around the month of January. Then, draw a vertical line across the center of the circle. Next, draw two horizontal lines above and below the vertical line. Finally, mark off the months of January and February on the left side of the vertical line. Overall, these are some of the best ways to teach skip counting and will cause less confusion by using these methods for the children. It clearly outlines everything they will need to do and what difficulties they may encounter when learning this skill. There is also the solution added if you run into any of these difficulties. everything should be made simpler, including skip counting.	{"url":"https://www.preschoolactivitiesnook.com/how-to-teach-skip-counting/","timestamp":"2024-11-01T20:45:25Z","content_type":"text/html","content_length":"60617","record_id":"<urn:uuid:d5fd26b7-5f40-4ccc-90d9-1fe757f5413f>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00817.warc.gz"}
Lesson 7 From Parallelograms to Triangles Let’s compare parallelograms and triangles. Problem 1 To decompose a quadrilateral into two identical shapes, Clare drew a dashed line as shown in the diagram. 1. She said the that two resulting shapes have the same area. Do you agree? Explain your reasoning. 2. Did Clare partition the figure into two identical shapes? Explain your reasoning. Problem 2 Triangle R is a right triangle. Can we use two copies of Triangle R to compose a parallelogram that is not a square? If so, explain how or sketch a solution. If not, explain why not. Problem 3 Two copies of this triangle are used to compose a parallelogram. Which parallelogram cannot be a result of the composition? If you get stuck, consider using tracing paper. Problem 4 1. On the grid, draw at least three different quadrilaterals that can each be decomposed into two identical triangles with a single cut (show the cut line). One or more of the quadrilaterals should have non-right angles. 2. Identify the type of each quadrilateral. Problem 5 1. A parallelogram has a base of 9 units and a corresponding height of $\frac23$ units. What is its area? 2. A parallelogram has a base of 9 units and an area of 12 square units. What is the corresponding height for that base? 3. A parallelogram has an area of 7 square units. If the height that corresponds to a base is $\frac14$ unit, what is the base? (From Unit 1, Lesson 6.) Problem 6 Select all the segments that could represent the height if side $n$ is the base. (From Unit 1, Lesson 5.)	{"url":"https://curriculum.illustrativemathematics.org/MS/students/1/1/7/practice.html","timestamp":"2024-11-03T05:50:15Z","content_type":"text/html","content_length":"85301","record_id":"<urn:uuid:46daf053-e68a-421f-a1e5-02cd66d86997>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00118.warc.gz"}
Molecular Simulation/Solids - Wikibooks, open books for an open world Figure 1: A molecular dynamics simulation of solid argon at 50 K. Argon-argon interactions are described using a Lennard-Jones potential. The atoms are arranged in a face-centered cubic cell. Figure 2: The radial distribution functions of solid (T = 50 K), liquid (T = 80 K), and gaseous argon (T = 300 K). The radii are given in reduced units of the molecular diameter (${\displaystyle \ sigma =3.822\mathrm {\AA} )}$. The Structure and Dynamics of Simple Solids Solids have regular periodic structures where the atoms are held at fixed lattice points in the 3-D rigid framework. This regular repeating structure can take different forms and is usually represented by a unit cell, where the entire solid structure is the repeated translations of this cell. Since the atoms cannot move freely, this allows solids to have a fixed volume and shape, which is only distorted by an applied force. However, atoms in solids do still have motion. They vibrate around their lattice positions and therefore their position fluctuates slightly around the lattice point. This can be thought of as the atom being tethered to the point, but being able to slightly vibrate around it. The intermolecular interactions between the atoms keep them in their fixed positions. These forces depend on the composition of the solid and could include forces such as London dispersion, dipole-dipole, quadrupole-quadrupole, and hydrogen bonding. ^[1] The temperatures at which these solids occur are low enough that the atoms do not have enough energy to overcome these forces and move away from their fixed position. This keeps the solid tightly packed and eliminates most of the space between atoms or molecules. This gives solids their dense property. Figure 1 shows a molecular dynamics simulation of solid argon at 50 K, where the atoms are all vibrating around their fixed lattice points. Argon atoms only have london dispersion intermolecular forces, which are described by the Lennard-Jones potential. The Lennard-Jones equation below accounts for the London dispersion attractive forces and the Pauli repulsion forces between atoms. ${\displaystyle {\mathcal {V}}\left(r\right)=4\varepsilon \left[\left({\frac {\sigma }{r}}\right)^{12}-\left({\frac {\sigma }{r}}\right)^{6}\right]=\varepsilon \left[\left({\frac {R_{min}}{r}}\right) ^{12}-2\left({\frac {R_{min}}{r}}\right)^{6}\right]}$ The intermolecular distance between the argon atoms ${\displaystyle r}$ is equal to ${\displaystyle R_{\min }}$ . This means that the atoms are at the same distance as the minima of this potential energy function. This maximizes the intermolecular forces by giving the most negative potential. The atoms in the solid argon are held together by these strong forces of attraction and are tightly packed to minimize empty space. Differences in the Radial Distribution Function The radial distribution function ${\displaystyle g(r)}$ relates the bulk density ${\displaystyle \rho }$ of a solid, liquid, or gas to the local density ${\displaystyle \rho }$ ${\displaystyle (r)}$ at a distance ${\displaystyle r}$ from a certain molecule or atom. The equation that relates these parameters is found below.^[2] ${\displaystyle \rho (r)=\rho ^{bulk}g(r)}$ The radial distribution functions of solid, liquid, and gaseous argon can be seen in Figure 2. In a solid, particles are found at defined positions, which is shown by the discrete peaks at values of ${\displaystyle \sigma }$ , ${\displaystyle {\sqrt {2}}}$ ${\displaystyle \sigma }$ , ${\displaystyle {\sqrt {3}}}$ ${\displaystyle \sigma }$ , ${\displaystyle 2}$ ${\displaystyle \sigma }$ , etc. The peaks of this radial distribution function are also broadened due to the molecules fluctuating around their lattice positions and occupying slightly different positions in this range. The regions of the function with ${\displaystyle g(r)=0}$ are regions where there is a zero probability of finding another molecule or atom. There is a zero probability between the peaks in a solid radial distribution function because of the regular structure where all of the molecules are packed tightly to most efficiently fill the space. This leaves regular intervals of spaces where no atoms or molecules are present. Also, each peak in a radial distribution function represents a coordination sphere where there is a high probability of finding molecules.^[3] Each subsequent peak represents a coordination sphere that is farther from the origin molecule and therefore the nearest neighbours are in the first coordination sphere. It is also important to note that ${\displaystyle g(r)\approx 0}$ when ${\displaystyle r<\sigma }$ . In this scenario, the electron density clouds of the two atoms are overlapping, causing the potential energy to be prohibitively high. In contrast, the radial distribution function of a gas only has one peak/coordination sphere, which then decays to the bulk density, represented by ${\displaystyle g(r)=1}$ . This simple radial distribution function is a consequence of a density that is so low that only the interactions of individual pairs of gas molecules affect the radial distribution function. The density is higher around the origin molecule due to strong london dispersion forces in this area, but the forces decay off quickly. The radial distribution function of liquids also differs from that of the solids. Molecules in a liquid have the ability to move around, but their positions are still correlated due to intermolecular forces between the molecules. This allows liquids to have periodic peaks in the radial distribution function, as shown in Figure 2. Thus, liquids also have coordination spheres where it is more likely to find molecules at these distances from an origin molecule, and thus there will be a greater local density at these positions. However, there is still a lower density than in solids due to the fluidity of liquids and the molecules being able to change positions. The radial distribution function of a liquid has its peaks at intervals of ${\displaystyle \sigma }$ , which is due to the looser packing of the molecules in a liquid compared to in a solid.^[2] This looser packing is due to the coordination spheres not being bound to fixed positions. There is also a lower probability of finding molecules in the second coordination sphere due to Pauli repulsion interactions with the first sphere. Due to the disordered nature of liquids the radial distribution function eventually decays to one and returns to the bulk density at large ${\displaystyle r}$ values as the positions are no longer correlated to each other. Simple liquids, such as liquid argon, are packed most efficiently to avoid repulsive interactions between the atoms, but there is still some spaces between them. Solids are packed very tightly so that the empty space between them is as little as possible and most crevices are filled. Their fixed positions allow them to maintain this tight packing of the atoms to minimize wasted space. Liquids also try to minimize this space, but are less tightly packed than solids because of their ability to move around and change positions. They have more energy to overcome the intermolecular forces correlating their positions. This difference in packing is seen in the radial distribution functions with the occurrence of the solid peaks at closer intervals than the liquid peaks.	{"url":"https://en.m.wikibooks.org/wiki/Molecular_Simulation/Solids","timestamp":"2024-11-13T01:50:36Z","content_type":"text/html","content_length":"64896","record_id":"<urn:uuid:9d034b2d-405a-43c1-92be-27da27f15b66>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00515.warc.gz"}
Dadson Tutoring in Richmond, VA // Tutors.com Students can expect • A renewed level of confidence • Stronger foundational math knowledge • Higher passion for math learning • Higher grades (1 letter grade minimum increase for consistent clients) Most students struggle with math, thus I delight in seeing the confidence levels of students rise. That "aha" moment is a building block to inspiring students to want to learn more as their confidence levels rise. My vision is to expand my services to more students who struggle with math so they can reach the highest levels of success. Grade level Pre-kindergarten, Elementary school, Middle school, High school, College / graduate school, Adult learner Type of math General arithmetic, Pre-algebra, Algebra, Geometry, Trigonometry, Pre-calculus, Calculus, Statistics No reviews (yet) Ask this tutor for references. There's no obligation to hire and we’re here to help your booking go smoothly. Services offered	{"url":"https://tutors.com/va/richmond/math-tutors/dadson-tutoring?service=UCT7ybWAds","timestamp":"2024-11-08T11:16:47Z","content_type":"text/html","content_length":"169139","record_id":"<urn:uuid:fc5c7798-d923-4909-abce-f56b63c0de1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00891.warc.gz"}
collision detection In our last article, we made a very simple program that helped us detect when two circles were colliding. However, 2D games are usually much more complex than just circles. I shall now introduce the next shape: the rectangle. By now you probably noticed that, for the screenshots I’m using a program called “Collision Test”. This is a small tool I made to help me visualize all this stuff I’m talking about. I used this program to build the collision detection/resolution framework for an indie top-down adventure game I was involved in. I will be talking more about this tool in future articles. Now, there are many ways to represent a rectangle. I will be representing them as five numbers: The center coordinates, width, height and the rotation angle: public class CollisionRectangle public float X { get; set; } public float Y { get; set; } public float Width { get; set; } public float Height { get; set; } public float Rotation { get; set; } public CollisionRectangle(float x, float y, float width, float height, float rotation) X = x; Y = y; Width = width; Height = height; Rotation = rotation Now, for our first collision, we will collide a circle and a rectangle. There are two types of collisions to consider: When the circle is entirely inside the rectangle… …And when the circle is partly inside the rectangle, that is, it is touching the border These are two different types of collisions, and use different algorithms to determine whether or not there is a collision. But first, let’s forget about the rectangle’s position and rotation. Our first approach will deal with a rectangle centered in the world, and not rotated: Under these constraints, the circle is inside the rectangle when both the X coordinate of the circle is between the left and right borders, and the Y coordinate is between the top and bottom borders, like so: public static bool IsCollision(CollisionCircle a, CollisionRectangle b) // For now, we will suppose b.X==0, b.Y==0 and b.Rotation==0 float halfWidth = b.Width / 2.0f; float halfHeight = b.Height / 2.0f; if (a.X >= -halfWidth && a.X <= halfWidth && a.Y >= -halfHeight && a.Y <= halfHeight) // Circle is inside the rectangle return true; return false; // We're not finished yet... But this is not enough. This only works when the center of the circle is inside the rectangle. There are plenty of situations where the center of the circle is outside the rectangle, but the circle is still touching the rectangle. In this case, we first find the point in the rectangle which is closest to the circle, and if the distance between this point and the center of the circle is smaller than the radius, then the circle is touching the border of the rectangle. We find the closest point for the X and Y coordinates separately: float closestX, closestY; // Find the closest point in the X axis if (a.X < -halfWidth) closestX = -halfWidth; else if (a.X > halfWidth) closestX = halfWidth closestX = a.X; // Find the closest point in the Y axis if (a.Y < -halfHeight) closestY = -halfHeight; else if (a.Y > halfHeight) closestY = halfHeight; closestY = a.Y; And now we bring it all together: public static bool IsCollision(CollisionCircle a, CollisionRectangle b) // For now, we will suppose b.X==0, b.Y==0 and b.Rotation==0 float halfWidth = b.Width / 2.0f; float halfHeight = b.Height / 2.0f; if (a.X >= -halfWidth && a.X <= halfWidth && a.Y >= -halfHeight && a.Y <= halfHeight) // Circle is inside the rectangle return true; float closestX, closestY; // Find the closest point in the X axis if (a.X < -halfWidth) closestX = -halfWidth; else if (a.X > halfWidth) closestX = halfWidth closestX = a.X; // Find the closest point in the Y axis if (a.Y < -halfHeight) closestY = -halfHeight; else if (a.Y > halfHeight) closestY = halfHeight; closestY = a.Y; float deltaX = a.X - closestX; float deltaY = a.Y - closestY; float distanceSquared = deltaX * deltaX - deltaY * deltaY; if (distanceSquared <= a.R * a.R) return true; return false; Looks good, but we’re still operating under the assumption that the rectangle is centered and not rotated. To overcome this limitation, we can move the entire world -that is, both the rectangle and the circle-, so the rectangle ends centered and non-rotated: In other words, we have to find the position of the circle, relative to the rectangle. This is pretty straightforward trigonometry: float relativeX = a.X - b.X; float relativeY = a.Y - b.Y; float relativeDistance = (float)Math.Sqrt(relativeX * relativeX + relativeY * relativeY); float relativeAngle = (float)Math.Atan2(relativeY, relativeX); float newX = relativeDistance * (float)Math.Cos(relativeAngle - b.Rotation); float newY = relativeDistance * (float)Math.Sin(relativeAngle - b.Rotation); And then put it all together: public class CollisionRectangle public float X { get; set; } public float Y { get; set; } public float Width { get; set; } public float Height { get; set; } public float Rotation { get; set; } public CollisionRectangle(float x, float y, float width, float height, float rotation) X = x; Y = y; Width = width; Height = height; Rotation = rotation public static bool IsCollision(CollisionCircle a, CollisionRectangle b) float relativeX = a.X - b.X; float relativeY = a.Y - b.Y; float relativeDistance = (float)Math.Sqrt(relativeX * relativeX + relativeY * relativeY); float relativeAngle = (float)Math.Atan2(relativeY, relativeX); float newX = relativeDistance * (float)Math.Cos(relativeAngle - b.Rotation); float newY = relativeDistance * (float)Math.Sin(relativeAngle - b.Rotation); float halfWidth = b.Width / 2.0f; float halfHeight = b.Height / 2.0f; if (newX >= -halfWidth && newX <= halfWidth && newY >= -halfHeight && newY <= halfHeight) // Circle is inside the rectangle return true; float closestX, closestY; // Find the closest point in the X axis if (newX < -halfWidth) closestX = -halfWidth; else if (newX > halfWidth) closestX = halfWidth closestX = newX; // Find the closest point in the Y axis if (newY < -halfHeight) closestY = -halfHeight; else if (newY > halfHeight) closestY = halfHeight; closestY = newY; float deltaX = newX - closestX; float deltaY = newY - closestY; float distanceSquared = deltaX * deltaX - deltaY * deltaY; if (distanceSquared <= a.R * a.R) return true; return false; public class CollisionRectangle public float X { get; set; } public float Y { get; set; } public float Width { get; set; } public float Height { get; set; } public float Rotation { get; set; } public CollisionRectangle(float x, float y, float width, float height, float rotation) X = x; Y = y; Width = width; Height = height; Rotation = rotation public static bool IsCollision(CollisionCircle a, CollisionRectangle b) float relativeX = a.X - b.X; float relativeY = a.Y - b.Y; float relativeDistance = (float)Math.Sqrt(relativeX * relativeX + relativeY * relativeY); float relativeAngle = (float)Math.Atan2(relativeY, relativeX); float newX = relativeDistance * (float)Math.Cos(relativeAngle - b.Rotation); float newY = relativeDistance * (float)Math.Sin(relativeAngle - b.Rotation); float halfWidth = b.Width / 2.0f; float halfHeight = b.Height / 2.0f; if (newX >= -halfWidth && newX <= halfWidth && newY >= -halfHeight && newY <= halfHeight) // Circle is inside the rectangle return true; float closestX, closestY; // Find the closest point in the X axis if (newX < -halfWidth) closestX = -halfWidth; else if (newX > halfWidth) closestX = halfWidth closestX = newX; // Find the closest point in the Y axis if (newY < -halfHeight) closestY = -halfHeight; else if (newY > halfHeight) closestY = halfHeight; closestY = newY; float deltaX = newX - closestX; float deltaY = newY - closestY; float distanceSquared = deltaX * deltaX - deltaY * deltaY; if (distanceSquared <= a.R * a.R) return true; return false; In the next article, we’ll put some structure to all of this. Practical 2D collision detection – Part 1 Collision detection is a fascinating, yet almost entirely overlooked and oversimplified aspect of game making. In my experience making games, I have found that collision detection, and subsequent collision resolving is quite tricky to get right. I would like to share a few practical pointers I find useful, so you can get started with your own collision framework. For these articles, we will be working on 2D collisions, that is collisions that can be represented with 2D shapes in a 2D environment. Don’t let the 2D fool you though; a lot of 3D games can be created with a 2D collision environment, as long as collisions can be thought of in only two dimensions. For example, side scrollers can benefit from XY-only collisions. While top-down games like racing, strategy or even some simulation games can use XZ-only collisions. When we talk about collisions, there are two elements to consider. collision detection and collision resolution. Collision detection consists of deciding whether or not two objects are colliding. Collision resolution consists of reorganizing colliding objects so they are not colliding anymore. Even though detection and resolution are closely related, the algorithms and results for both detection and resolution are wildly different. It is in fact very common to use detection but not resolution for things such as triggers (for example, detecting when a player enters a room). In this article, I will focus on collision detection. We will consider resolution in a future article. Collision detection is a geometric problem: Given two shapes, decide whether they overlap or not. The complexity of the solution depends on what kind of shapes we are talking about. So let’s start with the simplest shapes: two circles. A large white circle, and a small yellow circle, just hangin’ out. Each circle can be represented as three numbers: the center coordinates for X and Y, and a radius: public class CollisionCircle public float X { get; set; } public float Y { get; set; } public float R { get; set; } public CollisionCircle(float x, float y, float r) X = x; Y = y; R = r; Two circles collide when the distance between the two centers is less than or equal than the sum of their radii. We can do this with Pythagoras: public class CollisionCircle public float X { get; set; } public float Y { get; set; } public float R { get; set; } public CollisionCircle(float x, float y, float r) X = x; Y = y; R = r; public static bool IsCollision(CollisionCircle a, CollisionCircle b) float deltaX = a.X - b.X; float deltaY = a.Y - b.Y; float distance = (float)Math.Sqrt(deltaX * deltaX - deltaY * deltaY); float sumOfRadii = a.R + b.R; if (distance <= sumOfRadii) return true; return false; I don’t want to talk much about optimization, but in here, we can save ourselves the costly square root at line 18, by simply comparing the squared distance with the square of the sums of the radii. The result would look like this: public class CollisionCircle public float X { get; set; } public float Y { get; set; } public float R { get; set; } public CollisionCircle(float x, float y, float r) X = x; Y = y; R = r; public static bool IsCollision(CollisionCircle a, CollisionCircle b) float deltaX = a.X - b.X; float deltaY = a.Y - b.Y; float distanceSquared = deltaX * deltaX - deltaY * deltaY; float sumOfRadii = a.R + b.R; if (distanceSquared <= sumOfRadii * sumOfRadii) return true; return false; So far, so good. On the next article, we’ll figure out how to do collision detection with other kinds of shapes.	{"url":"https://www.rapapaing.com/blog/tag/collision-detection/","timestamp":"2024-11-02T02:20:26Z","content_type":"text/html","content_length":"50320","record_id":"<urn:uuid:d1a68f18-af3c-4a7f-b263-53785d04af43>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00349.warc.gz"}
the tire size What do the tire size numbers mean? What do the tire size numbers mean? The two-digit number after the slash mark in a tire size is the aspect ratio. For example, in a size P215/65 R15 tire, the 65 means that the height is equal to 65% of the tire’s width. The bigger the aspect ratio, the bigger the tire’s sidewall will be. What do the 3 numbers mean on tire size? B: TIRE WIDTH The three-digit number following the letter is the tire’s width (from side to side, looking at the tire head on) in millimeters. This may also be referred to as the section width. C: ASPECT RATIO The forward slash separates the tire width number from the two-digit aspect ratio. What does p235 55R19 mean? 235/55R19 tires are 29.2″ tall, have a section width of 9.3″, and fit wheels with a diameter of 19. The circumference is 91.6″ that translates into 692 revolutions per mile. As a rule, they can be mounted on the wheels with 19″ x 6.5-8.5″ rim width. In the high flotation system, the equivalent tire size is 29.2×9.3R19. What does 89v mean on tires? V – Speed ratings are represented by letters ranging from A to Z. Each letter coincides to the maximum speed a tire can sustain under its recommended load capacity. For instance, V is equivalent to a maximum speed of 149 mph. How wide is a 285 70R17 tire? 285/70R17 tires are 32.7″ tall, have a section width of 11.2″, and fit wheels with a diameter of 17. The circumference is 102.7″ that translates into 617 revolutions per mile. Can I replace 255 tires with 285? 285 may be possible but you’ll need to do some good amount of rolling unless you move the wheel further into the wheel well. You’ll need all the travel you can get especially on those high speed turns at NJMP Wheels will tuck in! Are 265 tires taller than 245? The 265 tire is wider at 10.43 inches, while the 245 tire converts to 9.65 inches. Can I put 265 tyres on 255 rims? you can put 265 width tyres on the new wheels. What does W stand for on tires? The letter W denotes the maximum speed rating, which translates 168 mph—not something intended for mom’s minivan. See our list of speed ratings below, which range from a low of “L” (just 75 mph for some off-road tires) to a high of Y (186 mph). What do the 3 numbers in a tire size mean? What do the 3 numbers mean on tire size? B: TIRE WIDTH The three-digit number following the letter is the tire’s width (from side to side, looking at the tire head on) in millimeters. This may also be referred to as the section width. In this example, the aspect ratio is 65, meaning the sidewall is 65 percent as high as the tire is wide. What do the numbers in the tire size relate to? Tire service type ratings. P = P-Metric (Example: P 215/65R17 98T) P-Metric tires are the most common type of tire. Metric/Euro-Metric (Example: 185/65R15 88T) Metric tires,also known as Euro-Metric tires because the sizing originated in Europe,don’t have a letter designation. How do you calculate the size of a tire? Tire diameter can vary slightly for each tire model. The listed diameters are from calculations based on the tire size. When changing tire sizes,we recommend staying within 3% of the diameter/height of the original tire. This tire calculator is for information purposes only and we do not guarantee fitment based on this calculator alone. What size tire is ideal? – By Company Type: Tier 1 – 70%, Tier 2 – 20% and Tier 3 – 10% – By Designation: Manager Level – 60%, C Level – 25%, and Others – 15% – By Region: Asia Pacific – 35%, Europe – 20%, North America –30%, and RoW-15%	{"url":"https://www.tag-challenge.com/2022/12/21/what-do-the-tire-size-numbers-mean/","timestamp":"2024-11-13T14:35:36Z","content_type":"text/html","content_length":"39377","record_id":"<urn:uuid:9eae672e-5795-4b99-a3a8-996149111aad>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00311.warc.gz"}
φ[z,0]: the elastic angular deformation at time t=0 φ[z,∞].: the elastic angular deformation at time t=∞ The rotation angle of the supporting beam b1 in coordinate system xyz ( ) will be equal to the rotation angle of the supported beam b2 in coordinate system x’y’z’ ( φ [y’,0] ), i.e. φ[z,0]= φ[y’,0] The rotation angle φ[z,t] of the supporting beam b1 will be increasing over time t, due to creep. As φ[y’,t]=φ[z,t], the equivalent angle φ[y’,t] of the supported beam b2 will also be increasing. C onsequently, the bending moment M and the equivalent torsional moment M[T] will be increasing, since M[T]=M. At [S:a:S] time t=∞, the system will balance at an angle φ[y’, ][∞]=φ[z, ][∞], which cannot exceed the angle φ[y’,pinn][d]. The φ[y’,pinn][ed] corresponds to the tangent of the elastic line at the left end of the supported beam b2 when the support there becomes pinned (M[T]=M=0). This is why the European Standards [EC2, §5.3.2.2(2)] allow us to consider pinned supports for both beams and slabs. Otherwise, either the creep should be taken into account or the effective stiffness should be limited to a small percentage (e.g. 10%) of the full elastic stiffness. The assumption of zero torsional stiffness provides a solution in the case of a simply supported beam. However, in the case of a cantilever beam, this turns out to be invalid, because the isostatic structure becomes a mechanism by diminishing its rotational restraint and thus transforming the support from fixed to pinned. The assumption of effective torsional stiffness limited to 1% of the elastic stiffness gives the correct results for both frames and thus for all types of frames.	{"url":"https://buildinghow.com/en-us/Products/Books/Volume-B/Structural-model-and-Analysis/The-effect-of-tortional-stiffness","timestamp":"2024-11-13T21:55:11Z","content_type":"text/html","content_length":"114450","record_id":"<urn:uuid:f636f69d-d85e-4544-8fe1-9d411e455855>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00804.warc.gz"}
Revision history In this case, the problematic translation is the one for "atan". There are lots of cases tike this for translations from Fricas, Maple and Mathematica, especially for special functions such as hypergeometric or Fresnel. Other problematic cases exist for more convoluted cases. For example, Sympy's [DEL:case:DEL] has semantics [DEL:more general than :DEL]Sage's cases, and some variants can't (yet) be handled. Some fallback mechanisms [DEL:exost, :DEL]for example, in Mathematica's case, where one can add a list of "additional" translations. But even this is insuficient : for example, the Mathematica -> operator is polysemic, in a Sage-unfriendly way. I've started to think about a (semi-)general way of handling this, but I'm not [DEL:tyet :DEL]at an acceptable solution. A shameful workaround in a paper is to ask the target interpreter to create a LaTeX representation of the result, use that to print in the paper, and to manually translate in Sage for the rest of the computations. It's ugly but allows you to progress... This, for example, allows to handle the "general" case of Sympy's [DEL:case:DEL]. In this case, the problematic translation is the one for "atan". There are lots of cases tike this for translations from Fricas, Maple and Mathematica, especially for special functions such as hypergeometric or Fresnel. Other problematic cases exist for more convoluted cases. For example, Sympy's case has semantics more general than Sage's cases, and some variants can't (yet) be handled. Some fallback mechanisms exost, for example, in Mathematica's case, where one can add a list of "additional" translations. But even this is insuficient : for example, the Mathematica -> operator is polysemic, in a Sage-unfriendly way. I've started to think about a (semi-)general way of handling this, but I'm not tyet at an acceptable solution. A shameful workaround in a paper is to ask the target interpreter to create a LaTeX representation of the result, use that to print in the paper, and to manually translate in Sage for the rest of the computations. It's ugly but allows you to progress... This, for example, allows to handle the "general" case of Sympy's case. In this case, the problematic translation is the one for "atan". There are lots of cases tike this for translations from Fricas, Maple and Mathematica, especially for special functions such as hypergeometric or Fresnel. Other problematic cases exist for more convoluted cases. For example, Sympy's [DEL:Piexewise:DEL] has semantics different from Sage's cases, and some variants can't (yet) be handled. Some fallback mechanisms exist, for example, in Mathematica's case, where one can add a list of "additional" translations. But even this is insuficient : for example, the Mathematica -> operator is polysemic, in a Sage-unfriendly way. I've started to think about a (semi-)general way of handling this, but I'm not yet at an acceptable solution. A shameful workaround in a paper is to ask the target interpreter to create a LaTeX representation of the result, use that to print in the paper, and to manually translate in Sage for the rest of the computations. It's ugly but allows you to progress... This, for example, allows to handle the "general" case of Sympy's Piecewise.	{"url":"https://ask.sagemath.org/answers/42703/revisions/","timestamp":"2024-11-02T16:00:15Z","content_type":"application/xhtml+xml","content_length":"20320","record_id":"<urn:uuid:24ed7eb0-c628-434c-9679-b07225b0fcd7>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00572.warc.gz"}
Arithmetic Sequences And Series Worksheet - Abhayjere.com Arithmetic Sequences And Series Worksheet. Can you find the variety of phrases in an arithmetic series? Expect questions on the frequent distinction, nth term, number of terms, last time period, next consecutive term, and more. Part A of these pdf worksheets requires college students to put in writing the arithmetic sequence by utilizing the recursive formula. Is an arithmetic sequence, what shall be its three subsequent terms? The following diagrams give the formulas for arithmetic sequence and arithmetic collection. A enjoyable foldable to show arithmetic sequences and series. These pdf arithmetic sequences worksheets are applicable for eighth grade and highschool college students. Solutions, examples, videos, actions, and worksheets which are suitable for A Level Maths to help students reply questions on arithmetic sequences and sequence. Put forth listed beneath are the primary terms and recursive formulation of sequences. Arithmetic Sequences Worksheets Identify the first term and the common distinction for each given series. Substitute the identified values in the appropriate method to determine the variety of terms ‘n’. Look on the first few phrases in pairs to work out the widespread difference. David Morse has been a maths instructor for over 30 years, in addition to an examiner. Arithmetic Geometric Sequences Algebra 1 Collaborative Worksheet Get what you want to turn out to be a greater trainer with limitless access to unique free classroom sources and professional CPD downloads. Get your lesson plan carried out in a jiffy with this vary of teacher-made printable lesson plan templates, including a 5 minute l… The questions have been fastidiously chosen and include using nth-term formulae. The formula is then used to unravel a couple of completely different problems. Teachers Pay Teachers is a web-based marketplace where lecturers buy and promote authentic instructional supplies. Arithmetic Sequences: Halloween Coloring Exercise Can you discover the variety of phrases in an arithmetic series? Use the sum of the series, plug the known values in the formula, rearrange so you’ve n as the topic, and find the variety of terms. Sorting out your medium time period planning for KS1 science and KS2 science? Let science specialists similar to Deborah Herridge and ment… Sequences Arithmetic, Geometric, Exponential Joke Worksheet Apply the given two phrases in the pertinent formulation to arrive at the values of ‘a’ and ‘d’ to unravel this set of two-level pdf worksheets. Level 2 requires learners to determine the precise term. Assess your abilities in evaluating arithmetic sequence with this batch of printable worksheets that might be a mix of the Type 1, 2 and 3. We build confidence and attainment by personalising every child’s learning at a stage that suits them. An arithmetic sequence or progression is a sequence of numbers in which the difference between consecutive phrases is fixed. Arithmetic And Geometric Sequences And Collection Fun Use the frequent distinction technique to establish the sequence that forms an arithmetic development. Is an arithmetic sequence by which the widespread difference is -4. This web page includes printable worksheets on Arithmetic Sequences. This versatile worksheets may be timed for pace, or used to evaluate and reinforce expertise and concepts. You can create math worksheets as checks, practice assignments or instructing instruments to maintain your skills recent. You can derive all of the phrases of the sequence by plugging-in the positions 1, 2, 3, … A sequence is a list of phrases which were ordered in a sequential method and any type of repetition is Real-life examples embody stacking cups, chairs, bowls, and pyramid-like patterns where objects are increasing or decreasing in a constant method. Carefully research every arithmetic sequence supplied in this batch of printable worksheets for grade 7 and grade eight. This video exhibits two formulas to search out the sum of a finite arithmetic sequence. Well, stick with this one rule to a tee – there’s a fixed distinction between the 2 consecutive phrases of an arithmetic sequence – and you’re all set to take up this printable task. Identify the primary time period – ‘a’, common difference – ‘d’ and number of phrases – ‘n’ and substitute in the relevant method to determine the sum of the arithmetic collection. Practicing arithmetic sequences worksheets help us to foretell and evaluate the result of a scenario. Sequences are relevant if we look for a pattern that aids in obtaining the final term. An arithmetic sequence has first term a and common difference d. Substitute the values of first time period and the common distinction within the nth term method to search out the specific term of the given sequence. An arithmetic sequence is a sequence of numbers such that the distinction of any two successive members of the sequence is a continuing. Arithmetic sequences worksheets assist students construct fundamental ideas on sequences and collection in arithmetic. This lesson plan is introduced on a templates that may be useful for teachers designing plans of their own. We’re your National Curriculum aligned on-line schooling content material provider serving to each baby reach English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be succesful of track and measure progress, helping each youngster obtain their finest. These recursive sequence worksheets concentrate on the concept of discovering the recursive formula for the given sequences and ascertain the sequence from the implicit method provided. Plug into this bunch of printable worksheets to swagger round finding the variety of terms when the primary time period, widespread difference, and final time period are given. Easy questions are definitely value the fewest factors and the harder questions are worth probably the most. This game serves as an excellent evaluate exercise at the end of a sequence & series Identify the primary time period of the sequence and calculate the widespread distinction . Observe every sequence of numbers in these pdf worksheets to discover out whether or not they kind an arithmetic sequence. That our free, printable arithmetic sequence worksheets cowl every little thing from fundamental to superior makes them an across-the-board useful resource, requiring children to do little extra practice. They additionally key into the express and recursive formulas and get the grasp of devising them for a sequence of rational and irrational numbers. This extensive assortment of sequence and sequence worksheets is really helpful for high school students. That is to say, we use the primary term to determine the second term, the second term to seek out the third, and so on. The arithmetic sequences in these workout routines involve each rational and irrational numbers to satisfy the urge for food of the studious, craving-for-more learners. An arithmetic series is basically the sum of the phrases contained in an arithmetic sequence. Here you can see hundreds of lessons, a group of academics for assist, and supplies that are always up to date with the most recent requirements. Features which would possibly be notably useful are potential scholar responses, trainer support and actions, and technique of assessing mastery. There is a distinction between knowledge entry and idea comprehension. The latter is necessary for students to have mastered an idea. This set of free printable arithmetic sequence word issues is designed for college students in the 8th grade and highschool. This set of pdf worksheets incorporates well-researched real-life word problems based on arithmetic sequence. Find the recursive method for each arithmetic sequence given in Part B. Examples, solutions, videos, actions, and worksheets which would possibly be suitable for A Level Maths to assist college students reply questions on arithmetic sequence and arithmetic collection. An explicit method defines the general time period or the nth time period of the sequence. Try the given examples, or kind in your own downside and verify your answer with the step-by-step explanations. Get ample practice in the concept of infinite geometric series and be taught to identify whether or not the series converges or diverges. Displaying all worksheets associated to – Sequences And Series For Grade 7 Math. Displaying all worksheets associated to – Arithmetic Sequence And Series. 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In this lesson, students will use an investigation to explore rational functions. Students will formalize their understanding of a rational function and continue to investigate the relationship between the equation of a simple rational function and its graph. Essential Question(s) What is a rational function? What might we use a rational function to model? Students are provided prompts that encourage them to imagine the graphs of data relationships. Students investigate a relationship that can be modeled as a rational function. Students formalize their understanding of rational functions and are introduced to the vocabulary of hyperbolas, branches, and asymptotes. Students further investigate rational functions through a Desmos Classroom Marbleslides activity. Students draw conclusions and analyze the standard form of a rational function with the It Says, I Say, and So strategy. • Lesson Slides (attached) • Pasta Branches handout (attached; one per group; printed front only) • Note Catcher handout (attached; one per student; printed front only) • 1 bag or box of dry spaghetti noodles • 1 bag of dried beans • Plastic cups (2–3 oz., one per group) • String (one 10–12-in. piece per group) • Permanent markers (one per group) • Rulers (one per group) • Tape (painter's tape is recommended) • Pencils Introduce the lesson using the attached Lesson Slides. Display slide 3 to share the lesson's essential question with students. Go to slide 4 to share the lesson's learning objectives. Review each of these with students to the extent you feel necessary. Show slide 5 and ask students if they agree or disagree with the given prompt: A house cat can crawl farther out onto a tree branch than a firefighter. Help the class come to the consensus that this is true because the further away from the tree trunk, the less weight a branch can support. Display slide 6 and introduce the idea of wanting to know the relationship between the length (distance from the tree trunk to the breaking point of the branch) and the mass it would take to break the branch. Show slide 7 and then ask the class to consider what the graph of this relationship would look like, where x is the length and y is the mass. Allow students time to think individually and use the options on the slide to predict what shape the graph would most likely be. As time allows, have students share with an Elbow Partner why they picked the curve they picked. Then ask for volunteers to share with the class. Show slide 8 and ask, Why might this be good information to know? Engineers need to know how much weight a beam could safely support. Allow students to share their ideas. Tell them that we are going to explore this kind of relationship today. Assign or have students choose groups of three to work. Display slide 9 and preview the activity with the class. Pass out the attached Pasta Branches handout to each group of students. Show slide 10 and review the following roles: counter, recorder, and catcher with the students. Direct students to decide within their groups who should take on which role. Display slide 11 and direct students to where they can gather their supplies for this activity. Show slide 12 and go through the steps of the activity on the Pasta Branches handout with the students. Use the picture on this slide to point out the markings and how to prevent the container string from sliding off of the pasta noodle. As students complete their investigation and return their materials, show slide 13. Direct students to go to desmos.com and click "Graphing Calculator." Have students add a table by clicking the plus sign in the top-left corner of their screen. Guide students to enter their data into the table. After students have completed entering their data into the Desmos Studio graphing calculator, show slide 14. Ask students to talk with their group about which graph most closely matches their data points and to reflect if this graph is the same one as they selected earlier in the lesson. Facilitate a class discussion on their data and ask students to use the data trend to complete the following sentence: As the distance from the tree trunk (length) increases, the mass it takes to break the branch ____. Show slide 15 and use this slide to direct students on how to generate a curve that models their data. Display slide 16 and explain that the graph on their screen is a hyperbola with two branches and two asymptotes. Explain these vocabulary terms to your students. Clarify to students that the asymptotes on the slide are not visible on the graphing calculator because they represent where the function is approaching but not what the function equals. Explain that we often draw them by hand when sketching hyperbolas. Then share the general equation for a simple rational function: y = a/(x–h) + k. Review this equation, the equation of the parent graph y = 1/x, and the definition of a rational function to the extent you see necessary but wait to explain the direct relationship between the equation and the graph, as students discover this later in the lesson. Provide students with your session code. Then, have students go to student.desmos.com and enter the session code. Display slide 17 and give each pair of students a copy of the Note Catcher handout to use as they progress through the Desmos Classroom activity. Screen 1 gives a preview to Desmos Classroom Marbleslides, which is a creative way for students to explore the relationship between equations and their graphs by trying to get a marble to follow the path of the curve to roll or go through a series of stars on the screen. Screens 2 and 4–7 ask students to make one change to the given rational function to complete the Marbleslides challenge. Screen 3 gives directions on how to reset their graphing calculator screen within the Desmos Classroom activity. Screens 10–15 ask students to make predictions about how changing a specific value will affect the graph. The activity continues with less scaffolding in place, continuing to challenge students. Display slide 18 and introduce students to the It Says, I Say, and So strategy. Direct students' attention to the bottom of the Note Catcher handout and ask students to use their notes from the Desmos Classroom Marbleslides activity, where they circled what they changed (It Says) and described the change in the graph (I Say), to explain how a, h, and k of y = a/(x–h) + k each affect the graph (and So).	{"url":"https://learn.k20center.ou.edu/lesson/297","timestamp":"2024-11-06T01:21:45Z","content_type":"text/html","content_length":"37198","record_id":"<urn:uuid:9a65f16d-9153-4bf3-9f99-78f55297d66e>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00272.warc.gz"}
What is a general procedure to prove that the LP relaxation of an IP delivers the optimal IP solution? ~ Operations Research ~ TransWikia.com You are looking for a proof for Total Unimodularity (TU). TU is a property by which a linear program will always have an integral solution. All you need to prove is that in your LP • $$A$$ matrix is TU and • $$b$$ column has only integers. What is TU • A matrix $$A$$ is unimodular if $$det(A) = 1$$ or $$-1$$. • A matrix $$A$$ is Totally Unimodular (TU) if each square submatrix $$S$$ of $$A$$ has $$det(S) = 0, 1$$ or $$-1$$. Sufficient Condition For TU • A matrix $$A$$ is TU if the number of non-zeros in each column is $$le2$$ • The sum of the entries of a column is zero Why does TU guarantees integral solution for a LP Consider the LP problem $$Ax = b$$ with basis $$B$$, the value of basic variables $$x_B$$ can be obtained as $$x_B = B^{-1}b = frac{operatorname{adj}(B)}{det(B)}b.$$ • Since $$det(B) = -1$$ or $$1$$ (from TU of $$A$$ matrix) and the adjoint matrix is also integral it implies that $$B^{-1}$$ is integral • Since $$b$$ is integral, then $$x_B$$ is also integral, hence guaranteeing and integral optimum for the LP. All network flow problems (shortest path, maximal flow, etc..) exhibit this property. Answered by Palaniappan Chellappan on August 19, 2021 If I'm understanding your question properly, this is not true in general. What you can prove is that this can be solved to integrality algorithmically, by adding Gomory cuts. Once enough cuts are added, the optimal vertex of the LP has to give an integer solution. This is known as the cutting plane method. In your case, you can use this to show that once no more Gomory cuts can be found, $$z^{LP}$$ has to be equal to $$z$$ (this happens much sooner in practice, but for the purposes of a proof it's fine to consider the worst-case scenario). Answered by Nikos Kazazakis on August 19, 2021	{"url":"https://transwikia.com/operations-research/what-is-a-general-procedure-to-prove-that-the-lp-relaxation-of-an-ip-delivers-the-optimal-ip-solution/","timestamp":"2024-11-12T13:02:08Z","content_type":"text/html","content_length":"47264","record_id":"<urn:uuid:e29c8bfa-36d9-4729-9bfd-5a5ff2517f00>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00631.warc.gz"}
Bandwidth by accumulated data? I would like to create a line chart of the used bandwidth. I have the data accumulated so far for every minute. How can I create a chart from it? Okay, first of all, all the details: I have about 30 access points and their already transmitted data since start in one value (individually for each AP). This data is collected every minute. so I have an absolute value that grows almost every minute. Now I would like to show a bandwidth usage in a chart over all APs (later also for certain groups). Currently I have this result: ( So is that the field you expect to be growing? If you look at that field in the individual docs in Discover, does that seem to be the case? It looks like the size of the values may be washing out the relatively small changes from one doc to the next in your line chart. You could try enabling Scale to Data Bounds in the "Metrics & Axes" tab of the visualization editor: ok, now the formatting is a bit more chic, but that doesn't solve my real problem. As you can see now, the value just keeps increasing. And now I want to have the resulting bandwidth as a chart. My current expression: .es(index="ruckus_ap_info", metric="max:ruckusSZAPTXBytes", split="ruckusSZAPMac:10", kibana=true).derivative().divide(1024).scale_interval(1s).if(operator="lt", if=0, then=0).trim(start=2,end= 1).label(regex="^.* ruckusSZAPMac:(.+) > .*$", label="$1").lines(width=1).yaxis(label="KB / sec", min=0) this ends in this chart: Now I have three questions to ask: 1. are these numbers in the chart correct? 2. how can I connect this jump with each other without the chart always returning to zero? 3. the label is very unattractive, here I would like to point to another field (ruckusSZAPName), is that possible? This topic was automatically closed 28 days after the last reply. New replies are no longer allowed.	{"url":"https://discuss.elastic.co/t/bandwidth-by-accumulated-data/119811","timestamp":"2024-11-08T14:42:33Z","content_type":"text/html","content_length":"52136","record_id":"<urn:uuid:a4380c48-f25c-4632-a094-9ef554ffaa15>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00298.warc.gz"}
Ten Friends Years K - 1 (2) This is a game for two players ... One should be an adult or older child. Plastic screw caps from soft drink and spring water bottles will fit the circles. Collect at least 10 in each of two colours. The board and the caps are a substitute for the school resource called Poly Plug. If you can think of something else that works let us know. □ Print this Poly Plug Frame. It is the playing board. □ Print this Poly Plug Paper. Use it to record the final position in each round. □ One spot dice □ One calculator (there's one on your phone) □ Write the title of this challenge and today's date on a fresh page in your maths journal. When the activity is over your Poly Plug Paper should be put into your journal beside any other notes you make. How To Play Ten Friends Cover the bottom three rows of the board with a piece of paper. Then only the top two rows can be seen This is a Poly Plug 10 Frame. Player A rolls the dice and leaves it where it lands... then places that number of yellow plugs into the gaps in any way... and writes on the calculator the number placed [5 in this case]. (Alternatively, leave out this third step for now and bring in calculator recording later as below.) Player B has to 'look hard' at the empty spaces and tell their guess of the number of blue plugs it will take to fill the gaps, saying I think I need... Then they check their guess by counting blue plugs into the spaces... and complete an equation on the calculator to make ten, in this case 5 [+] [5] [=] 10. Alternatively, Player B may tell their guess then write the equation first to show they know the Ten Friend. However, a mathematician always checks things another way, so they now count in the blue plugs to confirm. □ Once a hypothesis has been checked by counting in, players say together Look we have made ... (in this case) ... five plus five equals ten, pointing at the board as they speak. □ Record the round on the Poly Plug Paper. □ Players swap roles and play another round. □ They will want to continue for many rounds, but about 15 minutes three times a week is usually enough. □ Once the game has been modelled a couple of times, two siblings can easily play the game without adult assistance. Some will want to play on for many weeks and this can be encouraged by asking What happens if...? questions. □ What happens if ... we play with three rows? □ What happens if ... we play with four rows and two dice? (add the spots on the two dice) There are more ideas in the Answers & Discussion. Have fun exploring Ten Friends. Just Before You Finish For this part you need your maths journal and your Working Like A Mathematician page. □ Draw a picture of you and me playing Ten Friends today. □ How did we work like a mathematician today? Record 2 ways. Answers & Discussion These notes were originally written for teachers. We have included them to support parents to help their child learn from Ten Friends. □ Notes for Ten Friends. Ten Friends Gallery Send any comments or photos about this activity and we can add them to this gallery. 2nd April 2020 Dad introduced Ten Friends to Mr 6-y-o and Mr. Nearly-4-y-o. They had a dice on the phone. Mr. 6-y-o wanted to push more out and add up as he went. When there was Yep, just touch the screen and less than 6 left he calculated how many needed to get to 25. the dice rolls. We also played noughts and crosses with 9 empty spaces in the red board. Fun. One of the advantages of the 5x5 frame supplied with this activity is that if part of it is covered, as suggested above, it almost physically invites learners to ask What happens if...? in some way, just as it did for Mr. 6-y-o. 30th March 2020 Sisters exploring and enjoying their first play with Ten Friends. They both recorded in their maths journals. One a 4^1/[2] year old pre-schooler and one a 7^3/[4] year old Year 2. Wonderful work doing Maths At Home girls. Thanks to Mum for being such a good helper. Hope you had fun too. Maths At Home is a division of Mathematics Centre	{"url":"https://mathematicscentre.com/mathsathome/challenges/10friends.htm","timestamp":"2024-11-08T07:33:13Z","content_type":"text/html","content_length":"8176","record_id":"<urn:uuid:482eb8b5-bbb1-4273-9f15-d2f2277e2ad5>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00691.warc.gz"}
How are isosceles and scalene triangles different? \| HIX Tutor How are isosceles and scalene triangles different? Answer 1 A triangle is called SCALENE TRIANGLE if all the lengths of its sides are equal. A triangle is called ISOSCELES TRIANGLE if and only if the lengths of its two sides are equal. From definition you can get to the point. Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer 2 Isosceles triangles have two sides of equal length and two equal internal angles, while scalene triangles have all three sides of different lengths and all three internal angles of different Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer from HIX Tutor When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some Not the question you need? HIX Tutor Solve ANY homework problem with a smart AI • 98% accuracy study help • Covers math, physics, chemistry, biology, and more • Step-by-step, in-depth guides • Readily available 24/7	{"url":"https://tutor.hix.ai/question/how-are-isosceles-and-scalene-triangles-different-8f9afa29ac","timestamp":"2024-11-13T06:18:51Z","content_type":"text/html","content_length":"575499","record_id":"<urn:uuid:4d6d91fa-cd3e-4e5f-b658-6e52b3411f14>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00024.warc.gz"}
Convolutional layers The following convolutional/message-passing layers are available in Spektral. • $N$: number of nodes; • $F$: size of the node attributes; • $S$: size of the edge attributes; • $\x_i$: node attributes of the i-th node; • $\e_{i \rightarrow j}$: edge attributes of the edge from node i to node j; • $\A$: adjacency matrix; • $\X$: node attributes matrix; • $\E$: edge attributes matrix; • $\D$: degree matrix; • $\W, \V$: trainable weights matrices; • $\b$: trainable bias vector; • $\mathcal{N}(i)$: one-hop neighbourhood of node $i$; A general class for message passing networks from the paper Neural Message Passing for Quantum Chemistry Justin Gilmer et al. Mode: single, disjoint. This layer and all of its extensions expect a sparse adjacency matrix. This layer computes: where $\gamma$ is a differentiable update function, $\phi$ is a differentiable message function, $\square$ is a permutation-invariant function to aggregate the messages (like the sum or the average), and $\E_{ij}$ is the edge attribute of edge j-i. By extending this class, it is possible to create any message-passing layer in single/disjoint mode. propagate(x, a, e=None, kwargs) Propagates the messages and computes embeddings for each node in the graph. Any kwargs will be forwarded as keyword arguments to message(), aggregate() and update(). message(x, kwargs) Computes messages, equivalent to $\phi$ in the definition. Any extra keyword argument of this function will be populated by propagate() if a matching keyword is found. The get_sources and get_targets built-in methods can be used to automatically retrieve the node attributes of nodes that are sending (sources) or receiving (targets) a message. If you need direct access to the edge indices, you can use the index_sources and index_targets attributes. aggregate(messages, kwargs) Aggregates the messages, equivalent to $\square$ in the definition. The behaviour of this function can also be controlled using the aggregate keyword in the constructor of the layer (supported aggregations: sum, mean, max, min, prod). Any extra keyword argument of this function will be populated by propagate() if a matching keyword is found. update(embeddings, kwargs) Updates the aggregated messages to obtain the final node embeddings, equivalent to $\gamma$ in the definition. Any extra keyword argument of this function will be populated by propagate() if a matching keyword is found. • aggregate: string or callable, an aggregation function. This flag can be used to control the behaviour of aggregate() wihtout re-implementing it. Supported aggregations: 'sum', 'mean', 'max', 'min', 'prod'. If callable, the function must have the signature foo(updates, indices, n_nodes) and return a rank 2 tensor with shape (n_nodes, ...). • kwargs: additional keyword arguments specific to Keras' Layers, like regularizers, initializers, constraints, etc. spektral.layers.AGNNConv(trainable=True, aggregate='sum', activation=None) An Attention-based Graph Neural Network (AGNN) from the paper Attention-based Graph Neural Network for Semi-supervised Learning Kiran K. Thekumparampil et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes: where and $\beta$ is a trainable parameter. • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input. • trainable: boolean, if True, then beta is a trainable parameter. Otherwise, beta is fixed to 1; • activation: activation function; spektral.layers.APPNPConv(channels, alpha=0.2, propagations=1, mlp_hidden=None, mlp_activation='relu', dropout_rate=0.0, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) The APPNP operator from the paper Predict then Propagate: Graph Neural Networks meet Personalized PageRank Johannes Klicpera et al. Mode: single, disjoint, mixed, batch. This layer computes: where $\alpha$ is the teleport probability, $\textrm{MLP}$ is a multi-layer perceptron, and $K$ is defined by the propagations argument. • Node features of shape ([batch], n_nodes, n_node_features); • Modified Laplacian of shape ([batch], n_nodes, n_nodes); can be computed with spektral.utils.convolution.gcn_filter. • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • alpha: teleport probability during propagation; • propagations: number of propagation steps; • mlp_hidden: list of integers, number of hidden units for each hidden layer in the MLP (if None, the MLP has only the output layer); • mlp_activation: activation for the MLP layers; • dropout_rate: dropout rate for Laplacian and MLP layers; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.ARMAConv(channels, order=1, iterations=1, share_weights=False, gcn_activation='relu', dropout_rate=0.0, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) An Auto-Regressive Moving Average convolutional layer (ARMA) from the paper Graph Neural Networks with convolutional ARMA filters Filippo Maria Bianchi et al. Mode: single, disjoint, mixed, batch. This layer computes: where $K$ is the order of the ARMA$_K$ filter, and where: is a recursive approximation of an ARMA$_1$ filter, where $\bar \X^{(0)} = \X$ and • Node features of shape ([batch], n_nodes, n_node_features); • Normalized and rescaled Laplacian of shape ([batch], n_nodes, n_nodes); can be computed with spektral.utils.convolution.normalized_laplacian and spektral.utils.convolution.rescale_laplacian. • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • order: order of the full ARMA$_K$ filter, i.e., the number of parallel stacks in the layer; • iterations: number of iterations to compute each ARMA$_1$ approximation; • share_weights: share the weights in each ARMA$_1$ stack. • gcn_activation: activation function to compute each ARMA$_1$ stack; • dropout_rate: dropout rate for skip connection; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.CensNetConv(node_channels, edge_channels, activation=None, use_bias=True, kernel_initializer='glorot_uniform', node_initializer='glorot_uniform', edge_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, node_regularizer=None, edge_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, node_constraint=None, edge_constraint=None, bias_constraint=None) A CensNet convolutional layer from the paper Co-embedding of Nodes and Edges with Graph Neural Networks Xiaodong Jiang et al. This implements both the node and edge propagation rules as a single layer. Mode: single, disjoint, batch. • Node features of shape ([batch], n_nodes, n_node_features); • A tuple containing: • Modified Laplacian of shape ([batch], n_nodes, n_nodes); can be computed with spektral.utils.convolution.gcn_filter. • Modified line graph Laplacian of shape ([batch], n_edges, n_edges); can be computed with spektral.utils.convolution.line_graph and spektral.utils.convolution.gcn_filter. • Incidence matrix of shape ([batch], n_nodes, n_edges); can be computed with spektral.utils.convolution.incidence_matrix. • Edge features of shape ([batch], n_edges, n_edge_features); • Node features with the same shape as the input, but with the last dimension changed to node_channels. • Edge features with the same shape as the input, but with the last dimension changed to edge_channels. • node_channels: number of output channels for the node features; • edge_channels: number of output channels for the edge features; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • node_initializer: initializer for the node feature weights (P_n); • edge_initializer: initializer for the edge feature weights (P_e); • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • edge_regularizer: regularization applied to the edge feature weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • edge_constraint: constraint applied to the edge feature weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.ChebConv(channels, K=1, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A Chebyshev convolutional layer from the paper Convolutional Neural Networks on Graphs with Fast Localized Spectral Filtering Michaël Defferrard et al. Mode: single, disjoint, mixed, batch. This layer computes: where $\T^{(0)}, ..., \T^{(K - 1)}$ are Chebyshev polynomials of $\tilde \L$ defined as where • Node features of shape ([batch], n_nodes, n_node_features); • A list of K Chebyshev polynomials of shape [([batch], n_nodes, n_nodes), ..., ([batch], n_nodes, n_nodes)]; can be computed with spektral.utils.convolution.chebyshev_filter. • Node features with the same shape of the input, but with the last dimension changed to channels. • channels: number of output channels; • K: order of the Chebyshev polynomials; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.CrystalConv(aggregate='sum', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A crystal graph convolutional layer from the paper Crystal Graph Convolutional Neural Networks for an Accurate and Interpretable Prediction of Material Properties Tian Xie and Jeffrey C. Grossman Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes: where $\z_{ij} = \x_i \\| \x_j \\| \e_{ji}$, $\sigma$ is a sigmoid activation, and $g$ is the activation function (defined by the activation argument). • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Edge features of shape (num_edges, n_edge_features). • Node features with the same shape of the input. • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.DiffusionConv(channels, K=6, activation='tanh', kernel_initializer='glorot_uniform', kernel_regularizer=None, kernel_constraint=None) A diffusion convolution operator from the paper Diffusion Convolutional Recurrent Neural Network: Data-Driven Traffic Forecasting Yaguang Li et al. Mode: single, disjoint, mixed, batch. This layer expects a dense adjacency matrix. Given a number of diffusion steps $K$ and a row-normalized adjacency matrix $\hat \A$, this layer calculates the $q$-th channel as: • Node features of shape ([batch], n_nodes, n_node_features); • Normalized adjacency or attention coef. matrix $\hat \A$ of shape ([batch], n_nodes, n_nodes); Use DiffusionConvolution.preprocess to normalize. • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • K: number of diffusion steps. • activation: activation function $\sigma$; ($\tanh$ by default) • kernel_initializer: initializer for the weights; • kernel_regularizer: regularization applied to the weights; • kernel_constraint: constraint applied to the weights; spektral.layers.ECCConv(channels, kernel_network=None, root=True, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) An edge-conditioned convolutional layer (ECC) from the paper Dynamic Edge-Conditioned Filters in Convolutional Neural Networks on Graphs Martin Simonovsky and Nikos Komodakis Mode: single, disjoint, batch, mixed. In single, disjoint, and mixed mode, this layer expects a sparse adjacency matrix. If a dense adjacency is given as input, it will be automatically cast to sparse, which might be expensive. This layer computes: where $\textrm{MLP}$ is a multi-layer perceptron that outputs an edge-specific weight as a function of edge attributes. • Node features of shape ([batch], n_nodes, n_node_features); • Binary adjacency matrices of shape ([batch], n_nodes, n_nodes); • Edge features. In single mode, shape (num_edges, n_edge_features); in batch mode, shape (batch, n_nodes, n_nodes, n_edge_features). • node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • kernel_network: a list of integers representing the hidden neurons of the kernel-generating network; • 'root': if False, the layer will not consider the root node for computing the message passing (first term in equation above), but only the neighbours. • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.EdgeConv(channels, mlp_hidden=None, mlp_activation='relu', aggregate='sum', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) An edge convolutional layer from the paper Dynamic Graph CNN for Learning on Point Clouds Yue Wang et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes for each node $i$: where $\textrm{MLP}$ is a multi-layer perceptron. • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • mlp_hidden: list of integers, number of hidden units for each hidden layer in the MLP (if None, the MLP has only the output layer); • mlp_activation: activation for the MLP layers; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GATConv(channels, attn_heads=1, concat_heads=True, dropout_rate=0.5, return_attn_coef=False, add_self_loops=True, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', attn_kernel_initializer='glorot_uniform', kernel_regularizer=None, bias_regularizer=None, attn_kernel_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None, attn_kernel_constraint=None) A Graph Attention layer (GAT) from the paper Graph Attention Networks Petar Veličković et al. Mode: single, disjoint, mixed, batch. This layer expects dense inputs when working in batch mode. This layer computes a convolution similar to layers.GraphConv, but uses the attention mechanism to weight the adjacency matrix instead of using the normalized Laplacian: where where $\a \in \mathbb {R}^{2F'}$ is a trainable attention kernel. Dropout is also applied to $\alpha$ before computing $\Z$. Parallel attention heads are computed in parallel and their results are aggregated by concatenation or average. • Node features of shape ([batch], n_nodes, n_node_features); • Binary adjacency matrix of shape ([batch], n_nodes, n_nodes); • Node features with the same shape as the input, but with the last dimension changed to channels; • if return_attn_coef=True, a list with the attention coefficients for each attention head. Each attention coefficient matrix has shape ([batch], n_nodes, n_nodes). • channels: number of output channels; • attn_heads: number of attention heads to use; • concat_heads: bool, whether to concatenate the output of the attention heads instead of averaging; • dropout_rate: internal dropout rate for attention coefficients; • return_attn_coef: if True, return the attention coefficients for the given input (one n_nodes x n_nodes matrix for each head). • add_self_loops: if True, add self loops to the adjacency matrix. • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • attn_kernel_initializer: initializer for the attention weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • attn_kernel_regularizer: regularization applied to the attention kernels; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • attn_kernel_constraint: constraint applied to the attention kernels; • bias_constraint: constraint applied to the bias vector. spektral.layers.GatedGraphConv(channels, n_layers, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A gated graph convolutional layer from the paper Gated Graph Sequence Neural Networks Yujia Li et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes $\x_i' = \h^{(L)}_i$ where: where $\textrm{GRU}$ is a gated recurrent unit cell. • Node features of shape (n_nodes, n_node_features); note that n_node_features must be smaller or equal than channels. • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • n_layers: integer, number of iterations with the GRU cell; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GCNConv(channels, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A graph convolutional layer (GCN) from the paper Semi-Supervised Classification with Graph Convolutional Networks Thomas N. Kipf and Max Welling Mode: single, disjoint, mixed, batch. This layer computes: where $\hat \A = \A + \I$ is the adjacency matrix with added self-loops and $\hat\D$ is its degree matrix. • Node features of shape ([batch], n_nodes, n_node_features); • Modified Laplacian of shape ([batch], n_nodes, n_nodes); can be computed with spektral.utils.convolution.gcn_filter. • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GeneralConv(channels=256, batch_norm=True, dropout=0.0, aggregate='sum', activation='prelu', use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A general convolutional layer from the paper Design Space for Graph Neural Networks Jiaxuan You et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes: where $\mathrm{Agg}$ is an aggregation function for the messages, $\mathrm{Act}$ is an activation function, $\mathrm{Dropout}$ applies dropout to the node features, and $\mathrm{BN}$ applies batch normalization to the node features. This layer supports the PReLU activation via the 'prelu' keyword. The default parameters of this layer are selected according to the best results obtained in the paper, and should provide a good performance on many node-level and graph-level tasks, without modifications. The defaults are as follows: • 256 channels • Batch normalization • No dropout • PReLU activation • Sum aggregation If you are uncertain about which layers to use for your GNN, this is a safe choice. Check out the original paper for more specific configurations. • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • batch_norm: bool, whether to use batch normalization; • dropout: float, dropout rate; • aggregate: string or callable, an aggregation function. Supported aggregations: 'sum', 'mean', 'max', 'min', 'prod'. • activation: activation function. This layer also supports the advanced activation PReLU by passing activation='prelu'. • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GCSConv(channels, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A GraphConv layer with a trainable skip connection. Mode: single, disjoint, mixed, batch. This layer computes: where $\A$ does not have self-loops. • Node features of shape ([batch], n_nodes, n_node_features); • Normalized adjacency matrix of shape ([batch], n_nodes, n_nodes); can be computed with spektral.utils.convolution.normalized_adjacency. • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GINConv(channels, epsilon=None, mlp_hidden=None, mlp_activation='relu', mlp_batchnorm=True, aggregate='sum', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A Graph Isomorphism Network (GIN) from the paper How Powerful are Graph Neural Networks? Keyulu Xu et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes for each node $i$: where $\textrm{MLP}$ is a multi-layer perceptron. • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • epsilon: unnamed parameter, see the original paper and the equation above. By setting epsilon=None, the parameter will be learned (default behaviour). If given as a value, the parameter will stay • mlp_hidden: list of integers, number of hidden units for each hidden layer in the MLP (if None, the MLP has only the output layer); • mlp_activation: activation for the MLP layers; • mlp_batchnorm: apply batch normalization after every hidden layer of the MLP; • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GraphSageConv(channels, aggregate='mean', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A GraphSAGE layer from the paper Inductive Representation Learning on Large Graphs William L. Hamilton et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes: where $\textrm{AGGREGATE}$ is a function to aggregate a node's neighbourhood. The supported aggregation methods are: sum, mean, max, min, and product. • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • aggregate_op: str, aggregation method to use ('sum', 'mean', 'max', 'min', 'prod'); • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GTVConv(channels, delta_coeff=1.0, epsilon=0.001, activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A graph total variation convolutional layer (GTVConv) from the paper Total Variation Graph Neural Networks Jonas Berg Hansen and Filippo Maria Bianchi Mode: single, disjoint, batch. This layer computes where • Node features of shape (batch, n_nodes, n_node_features); • Adjacency matrix of shape (batch, n_nodes, n_nodes); • Node features with the same shape as the input, but with the last dimension changed to channels. • channels: number of output channels; • delta_coeff: step size for gradient descent of GTV • epsilon: small number used to numerically stabilize the computation of new adjacency weights • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.TAGConv(channels, K=3, aggregate='sum', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A Topology Adaptive Graph Convolutional layer (TAG) from the paper Topology Adaptive Graph Convolutional Networks Jian Du et al. Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. This layer computes: • Node features of shape (n_nodes, n_node_features); • Binary adjacency matrix of shape (n_nodes, n_nodes). • Node features with the same shape of the input, but the last dimension changed to channels. • channels: integer, number of output channels; • K: the order of the layer (i.e., the layer will consider a K-hop neighbourhood for each node); • activation: activation function; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.XENetConv(stack_channels, node_channels, edge_channels, attention=True, node_activation=None, edge_activation=None, aggregate='sum', use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A XENet convolutional layer from the paper XENet: Using a new graph convolution to accelerate the timeline for protein design on quantum computers Jack B. Maguire, Daniele Grattarola, Eugene Klyshko, Vikram Khipple Mulligan, Hans Melo Mode: single, disjoint, mixed. This layer expects a sparse adjacency matrix. For a version of this layer that supports batch mode, you can use spektral.layers.XENetDenseConv as a drop-in replacement. This layer computes for each node $i$: • Node features of shape ([batch], n_nodes, n_node_features); • Binary adjacency matrices of shape ([batch], n_nodes, n_nodes); • Edge features of shape (num_edges, n_edge_features); • Node features with the same shape of the input, but the last dimension changed to node_channels. • Edge features with the same shape of the input, but the last dimension changed to edge_channels. • stack_channels: integer or list of integers, number of channels for the hidden layers; • node_channels: integer, number of output channels for the nodes; • edge_channels: integer, number of output channels for the edges; • attention: whether to use attention when aggregating the stacks; • node_activation: activation function for nodes; • edge_activation: activation function for edges; • use_bias: bool, add a bias vector to the output; • kernel_initializer: initializer for the weights; • bias_initializer: initializer for the bias vector; • kernel_regularizer: regularization applied to the weights; • bias_regularizer: regularization applied to the bias vector; • activity_regularizer: regularization applied to the output; • kernel_constraint: constraint applied to the weights; • bias_constraint: constraint applied to the bias vector. spektral.layers.GINConvBatch(channels, epsilon=None, mlp_hidden=None, mlp_activation='relu', mlp_batchnorm=True, aggregate='sum', activation=None, use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A batch-mode version of GINConv. Mode: batch. This layer expects a dense adjacency matrix. spektral.layers.XENetConvBatch(stack_channels, node_channels, edge_channels, attention=True, node_activation=None, edge_activation=None, aggregate='sum', use_bias=True, kernel_initializer='glorot_uniform', bias_initializer='zeros', kernel_regularizer=None, bias_regularizer=None, activity_regularizer=None, kernel_constraint=None, bias_constraint=None) A batch-mode version of XENetConv. Mode: batch. This layer expects a dense adjacency matrix.	{"url":"https://graphneural.network/layers/convolution/","timestamp":"2024-11-09T09:41:47Z","content_type":"text/html","content_length":"70764","record_id":"<urn:uuid:f7dbf9dc-e78b-4517-9ce8-3fb208a2204f>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00142.warc.gz"}
Sensors Based Optimized Closed Loop Control Algorithm to Minimize Hypoglycemia/Hyperglycemia using 4-Variate Time Series Data To eliminate frequent finger pricking, a CGM sensor is utilized to measure the diabetic patient's blood glucose level from the interstitial fluid. Because CGM sensors are implanted beneath the skin, they monitor interstitial glucose rather than blood glucose (BG).CGM sensors can be made "smart" by including algorithms that can send out notifications when glucose concentrations are expected to surpass normal range thresholds [19, 20]. To improve the signal-to-noise ratio (SNR) of CGM data, it must be filtered. There will be minor differences in measurement taken from the interstitial fluid and the blood samples, but it will be the same during the steady state phase. The accelerometer sensor measures the user's current physical activity, such as sitting, sleeping, walking, jogging, and exercising. Based on the meal consumption, a biosensor is utilized to measure the carbs content. Because blood pressure, BMI, and heart rate have such close correlations, three additional sensors have been used to measure them. 3.1 6LoWPAN architecture The 6LoWPAN standard enables efficient use of IPv6 at low power through related protocols. An adaptive layer of simple embedded node optimization can be done in low-cost wireless networks. Title Summary is required to operate an effective payload transmission. Also, fragmentation and rearrangement of both transfers should be grouped and layered. Therefore, the MAC network layer is introduced between the adaptive layers. The same IPv6 prefix is distributed by LoWPAN and routers across the LoWPAN node share network interface. The node first uses edge router registration to facilitate efficient network operation. These operations are part of Neighbour Discovery (ND). Figure 1. 6LoWPAN architecture However, most of the 6LoWPAN applications such as automatic meter reading and environmental monitoring have been developed using mesh topologies. As it is often covered and cost-effective for infrastructure, it adopts multi-hop forwarding to achieve energy efficiency. Such link-layer meshes, LoWPAN networks, and IPs are done in three different ways, including routing. Link-layer networks and mesh LoWPAN have transparent mesh-mesh transfers and are called the Internet Protocol. 3.2 Low-Power and Lossy Routing (LPLR) with 6LoWPAN Routing is one of the primary network layer tasks defined in the Open Systems Interconnection (OSI) model. Various other tasks include resolving nodes and creating and maintaining network topologies. 6LoWPAN technology employs a modified IPv6 protocol stack for seamless connectivity. Consider a scalable network of M nodes, let N1, N2, and N3 be the coordinates of each node. A collection point (CP) is considered to be a cluster head at any physical location within it. The network interface of the node can reduce energy usage according to the following parameters. a) NIC characteristics b) Packet Size c) Bandwidth usage While the node is in listening mode and sleep mode, it assumes energy is being consumed; the power consumption is 1.0 W and 0.001 W respectively. The transmitted energy and the Joule energy to the received packet are: $E n_{\text {usage }}=\sum_{\mathrm{j}=1}^{\mathrm{p}} 5 * \operatorname{packetsize}(\mathrm{n}(\mathrm{j}))$ (1) The lower value of convergence indicates a larger amount of data and the longer time required for training the algorithm. Each solution generates a random path with a maximum number of nodes between the source and destination and satisfies the following constraints. $N_s \& N_d \leq N_{t h}$ (2) where, $N_s$ and $N_d$, $\mathrm{s}$ is the source and destination node. Then it will generate a random path with $N_i$ nodes, let the nodes in this path, $N P_{i+1}$. To evaluate the learning metric $M_p$ of each generated random path: $M_p=\sum_{i=0}^{N_i} \frac{d_{i-1}}{E(i)}+L S_{i-1}$ (3) Energy Level of each node $=E$ Link Support $L S_i$ The randomly generated path's learning metrics, i, is a positive integer that takes up the minimum value as two and the maximum value as the total number of nodes in that path, respectively. The $d_ {i-1}$ distance between the successive nodes, $E(i)$, is the instantaneous energy of each node right from the source node and $L S_{i-1}$ is the link quality between the successive links in the path. 3.3 Sensor fusion using adaptive Kalman Filte Let $\mathrm{x}_1, \mathrm{x}_2, \ldots \mathrm{x}_{\mathrm{n}}$ denote the sensor measurements from various sensors with covariance $\mathrm{w}_1, \mathrm{w}_2, \ldots \mathrm{w}_{\mathrm{n}}$ respectively. Each sensor processor $i$ supplies its prior estimates and covariances are $x_i^{\prime}\left(\frac{n+1}{n}\right), w_i^{\prime}\left(\frac{n+1}{n}\right), x_i^{\prime}\left(\frac{n+1} {n+1}\right)$, $w_i^{\prime}\left(\frac{n+1}{n+1}\right)$ where $\mathrm{i}=1,2 . . \mathrm{N}$. Fusion processor prior estimate is $x^{\prime}\left(\frac{n+1}{n}\right), w^{\prime}\left(\frac{n+1} {n}\right)$ and the fusion problem is to compute the total estimates and covariance matrix $w^{\prime}\left(\frac{n+1}{n+1}\right)$. The Kalman Filter updates or corrects the prediction and the current state's uncertainty after receiving the measurement. The Kalman Filter also forecasts future states and so on. During the prediction step, the system state at the next time stamp is $X_{n+1, n}^{\prime}=F X_{n, n}^{\prime}+G U_n+W_n$ (4) where, $\mathrm{G}$ is the control matrix, $\mathrm{U}$ is the input variable, $\mathrm{W}$ is the process noise and $F$ is the state transition matrix. The prediction level of uncertainty is calculated by $P_{n+1, n}=F P_{n, n} F^{\prime}+Q$ (5) where, $Q$ is process noise uncertainty $Q_n=E\left(W_n W_n\right)^T$ Initial estimate: $\left(\mathrm{x}_{0,0}^{\prime}, \mathrm{P}_{0,0}\right)$. During the update step, the Kalman Gain is. Next, update the system state predicted at time $\mathrm{k}$ by the measured value $X_{n, n}^{\prime}=X_{n, n-1}^{\prime}+K_n\left(Z_n-H X_{n, n-1}^{\prime}\right)$ (6) where, output vector $\mathrm{Z}_{\mathrm{n}}=\mathrm{Hx}_{\mathrm{n}}$. Finally, uncertainty will be updated based on the measurement is: $P_{n, n}=\left(I-K_n H\right) P_{n, n-1}\left(I-k_n H\right)^T+K_n R_n K_n^T$ (7) where, measurement uncertainty $R_n=E\left(V_n V_n\right)^T$ Total Daily Insulin Dose (TDID): Generally, the total insulin required for a day will be calculated based on $\text { Total daily insulin }=\frac{\text { weights in pounds }}{4}$ (8) Carbohydrates coverage from the total daily insulin is calculated using the "500" rule by CarbRatio $=\frac{500}{\text { TDID }}$ (9) The $BG$ has been quantized into a predetermined number of classes to perform a classification of the events comparable to precise amounts of BG concentration. They are, Severe hyperglycemic, hyperglycemic, normal, hypoglycemic and severe hypoglycemic. \text { Class set } C=\left\{\left(\text { if } \emptyset^{\prime}(t)<=50,\right.\right. \text { SHypo), } \\ \text { (if } 50>\emptyset^{\prime}(t)<=70, \text { Hypo), } \\ \text { (if } 70>\emptyset^{\prime}(t)<=180 \text {, Normal), } \\ \text { (if } 180>\emptyset^{\prime}(t)<=250, \text { Hyper), } \\ \text { (if } \left.\emptyset^{\prime}(t)>=250 \text {, SHyper) }\right\} where, denote the data as $\mathrm{D}_{\mathrm{i}}=\left[\mathrm{BG}_{\mathrm{i}}, \mathrm{HR}_{\mathrm{i}}, \mathrm{BPR}_{\mathrm{i}}, \mathrm{CHO}_{\mathrm{i}}, \mathrm{PH}_{\mathrm{i}}\right.$, $\ left.\mathrm{IS}_{\mathrm{i}}, \mathrm{ID}_{\mathrm{i}}\right]$ where all these variables are calculated in the time series from 1 to $\mathrm{n} . \mathrm{BG}_{\mathrm{i}}=\left[\mathrm{BG}_1, \ mathrm{BG}_2 \ldots . \mathrm{BG}_{\mathrm{n}}\right]$ is the blood glucose level, $\mathrm{HR}_{\mathrm{i}}$ is the heart rate, $\mathrm{BPR}_{\mathrm{i}}$ is the blood pressure rate, $\mathrm{IS}_ {\mathrm{i}}$ insulin boluses, and $\mathrm{ID}_{\mathrm{i}}$ insulin diffusion rate. Typically, a blood glucose level of $180 \mathrm{mg} / \mathrm{dl}$ is typically considered the threshold for hyperglycemia. This value is widely acknowledged as the maximum limit for glucose levels after a meal. This value is crucial in establishing the desired ranges for maintaining glycemic control. 3.4 Estimation of insulin infusion rate from blood glucose concentration using time-varying state model At time $t$, let $\varnothing^{\prime}(t)$ denote the measured blood glucose level, $x(t)$ denote the insulin delivered, $y(t)$ the carbohydrate being taken and $z(t)$ denotes the energy consumed due to physical activities at time $t$. The measured blood glucose level at time $\mathrm{t}$: $\phi^{\prime}(t)=\sigma(t)+n_1(t)$ (10) $\sigma(\mathrm{t})$ is glucose concentration from interstitial fluid and $\mathrm{n} 1(\mathrm{t})$ is the measurement noise. Measured glucose concentration from the interstitial fluid at time $t+1$ $\sigma(t+1)=\mu_t[\sigma(t)]+I_t[X(t-u)]+M_t[X(t-v)]+\beta(t)$ (11) $\mu[\sigma(t)]$ is the auto regressive value that denotes the blood glucose values at time $t+1, I_t[x(t-u)]$ is the dynamic linear regression of blood glucose values at time $t+1$ after insulin delivery with $u$ delay, $\mathrm{M}_{\mathrm{t}}[\mathrm{x}(\mathrm{t}-\mathrm{v})]$ is the dynamic linear regression of blood glucose values at time $t+1$ after meal intake with $v$ delay and $\ beta(t)$ is the process noise. $\mu_t[\sigma(t)]=\tau_1(t) \sigma(t)+\tau_2(t) \sigma(t-1)+\ldots+\tau_n(t) \sigma(t-n-1)$ (12) $I_t[x(t-u)]=\phi_1(t) X(t-u)+\phi_2(t) X(t-u-1)+\ldots+\phi_h(t) X(t-u-(n-1))$ (13) $M_1[X(t-v)]=\theta_1(t) X(t-v)+\theta_2(t) X(t-v-1)+\ldots+\theta_n(t) X(t-v-(n-1))$ (14) $\tau_n(t)$ is the auto regressive coefficient, $\varphi_n(t)$ and $\theta_n(t)$ are the linear regression coefficients which are time varying coefficients. $\varphi_n$ will be considered negative because insulin intake always decreases the blood glucose concentration level whereas $\theta_{\mathrm{n}}$ will be assumed to be positive as carbohydrates always increase the blood glucose concentration level $\left.\phi(t+1)=\frac{d}{d t}[\phi(t+1)+\phi(t)]+[e(t+1)-e(t)]+\left[p(t+1)-p(t)+X\left(t-d_1\right)\right]\right)$ (15) $\varnothing(t)$ and $\varnothing(t+1)$ is the blood glucose level at time $t$ and $t+1$. $e(t)$ and $e(t+1)$ is the measurement noise at time $t$ and $t+1$. $p(t)$ and $p(t+1)$ is the process noise at time $t$ and $t+1$ with zero mean and variance equal to $\sigma^2$. The formula to determine the insulin dose for high blood glucose levels is: Insulin dose for high blood glucose=Difference between actual blood sugar and target blood sugar $\div$ correction factor. $I_h=\frac{\left(\phi^{\prime}(t)-T^{\prime}(t)\right)}{C F}$ (16) The high blood sugar correction factor is calculated using the "1800" rule by the following formula: Correction Factor $=\frac{1800}{\text { TDID }}$ (17) where, TDID is the total daily insulin dose. 3.5 Carbohydrates prediction from the meals The experiment will then be repeated for the entire day, with varying meal start times, maximum meal sizes, and delays. The projected rise in glucose content $G^{\prime}$ after meal consumption is compared to the observed glucose G". The rate of glucose production at time $t$ is related to the rate of carbohydrate consumption during the meal. If $M_{s t}$ is the meal's start time, $M_{e t}$ is the meal's finish time, $M_{c g}$ is the meal's carbohydrate content in grams, and $M_d$ is the meal's duration, then $\mathrm{M}=\left[\mathrm{M}_ {\mathrm{st}}, \mathrm{M}_{\mathrm{et}}, \mathrm{M}_{\mathrm{cg}}, \mathrm{M}_{\mathrm{d}}\right]$. The maximum meal duration is $\mathrm{M}_{\mathrm{m}}$, and the maximum meal size is $\mathrm{M}_{\ mathrm{max}} ; \mathrm{d}$ is the maximum delay between the meal ingestion and the divergence of blood glucose into the circulation. $G^{\prime}(t) \alpha y^{\prime}(t)$ (18) $\left(G^{\prime}\left[t^{\prime}-t\right]\right)^{\prime}-\left(G^{\prime \prime}\left[t^{\prime}-t\right]\right)^{\prime}>\phi$ (19) $d_{e u c}=\sqrt{\sum_i^t \frac{\left\|G_p(i)-G_o(i)\right\|^2}{t-M_{s t}}}$ (20) Carbohydrate level at time $t$ is measured by: $y^{\prime}(t)=\sum_{t=1}^n d_{\text {euc}}(t)+d_{\text {euc }}(t-1)+\ldots . .+d_{\text {euc }}(t-n-1)$ (21) Using this formula, calculate the carbohydrate coverage insulin dose by: $I_{C H O}=\frac{y^{\prime}(t)}{C H O \text { in grams inclined by } 1 \text { unit of insulin }}$ (22) where, "CHO in grams inclined by 1 unit of insulin" differs based on the total daily insulin dose per day. For example, if you eat 50 grams of carbohydrates for the breakfast, then the $\mathrm{CHO}$ insulin dose is calculated by: $50 / 10=5$ units. This means you need 5 units of insulin to maintain the blood glucose level due to carbohydrate inclination. Total Insulin Dose $\mathrm{Ti}=\mathrm{I}_{\mathrm{h}}+\mathrm{I}_{\mathrm{CHO}}$ (23) where, $\mathrm{I}_{\mathrm{CHO}}$ is the $\mathrm{CHO}$ insulin dose and $\mathrm{I}_{\mathrm{h}}$ is the Insulin dose for high blood glucose. 3.6 Continuous Glucose Monitoring (CGM) and energy consumption due to physical activities Maintaining glycemic imbalance during and after physical activity usually necessitates additional carbohydrate intake and/or insulin cuts. If the physical activity duration is greater than 30 to 60 minutes then 10 to 15 grams of carbohydrates must be consumed to prevent hypoglycemia. For the entire study participants were allowed to participate in physical activity, with examples including running, fitness center, and walking. Energy spent due to physical activities is measured by: $p y^{\prime}(t)=\frac{\sum_{t=1}^n p y(t)+p y(t-1)+\ldots+p y(t-n-1)}{n} * 100 \%$ (24) The total insulin control action is based on: $I S_{t+1}=\frac{K_n\left(I S(t-1)+\phi^{\prime}(t)+c^{\prime}(t)-p y^{\prime}(t)=e(t)\right)}{C F}-G R$ (25) where, $\mathrm{K}_{\mathrm{n}}$ is the Kalman gain. The equation you provided, Energy Consumption $=$ Power $x$ Time, is accurate. To determine the overall energy usage, one must multiply the power consumption of the system by the duration of its operation. This formula is widely used to calculate energy consumption. Need to guarantee the system's autonomy and maximise its utilisation time. The choice of energy system might be diverse. Low-power and energyefficient systems are generally sought for medical devices, particularly those designed for wear or implantation. This may entail the utilisation of batteries, energy harvesting methodologies (such as solar cells or motion harvesting), or even rechargeable systems, contingent upon the specific application. The specific energy system would rely on the power requirements of your gadgets and the required operational period between charging or maintenance. 3.7 Metrics to evaluate the prediction model The model's prediction capability was computed on the full test dataset on hypoglycemia extremes (<70 mg/dl) and hyperglycaemic extremes (>180 mg/dl) cases and the mean absolute difference (MAD) is computed using Absolute Difference, $A(t)=\frac{(\operatorname{CGM}(t) \text { predict }-\operatorname{CGM}(t) \text { actual })}{\operatorname{CGM}(\mathrm{t}) \text { actual }}$ (26) Mean $A b s o l u t e$ Difference $\operatorname{MAD}(t)=\frac{A(t)}{N}$ (27) $\operatorname{CGM}(\mathrm{t})_{\text {predict }}$ is the model's predicted glucose value at time $t$, and $\operatorname{CGM}_{\text {actual }}(t)$ is the actual CGM data value at time $t . N$ is the number of data points in the dataset. Concerning the measured value, relative-absolute change RAC calculates the normalized absolute error between measurement and prediction. $R A C=\frac{y_i-y^{\prime}}{y_i} * 100 \%$ (28) The metric used to assess glucose prediction is temporal gain (TG), which represents the amount of average time gained for early identification of a probable hypo/hyperglycemia event using the model. Temporal Gain $\left(T_{\mathrm{G}}\right)=(\mathrm{n}$-Delay $) * \mathrm{~s}(\mathrm{t})$ $\text { Delay }=\arg \min _{k \in[0,1]}\left\{\frac{1}{n-l} \sum_{i=1}^{n-l}\left(y_i-y^{\prime}\right)^2\right\}$ (29) where, $\mathrm{n}$ is the number of samples in the dataset, $\mathrm{s}(\mathrm{t})$ is the sampling time.	{"url":"https://iieta.org/journals/jnmes/paper/10.14447/jnmes.v27i1.a01","timestamp":"2024-11-08T15:39:57Z","content_type":"text/html","content_length":"91149","record_id":"<urn:uuid:6446f182-003f-44fc-93ea-3959a31a8167>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00285.warc.gz"}
Quantum Supremacy Google And USTC(China) - Welcome to Quantum Guru What is Quantum Supremacy? An experimental demonstration of quantum computer’s dominance and advantage over classical computer by performing calculations that was impossible. To confirm that quantum supremacy has been achieved, computer scientists must be able to show that a classical computer could never have solved the problem while also proving that the quantum computer can perform the calculation quickly. Computer scientists hope that quantum supremacy will lead to the cracking of Shor’s algorithm — a currently impossible calculation that is the basis of most modern cryptography — as well as advantages in drug development, weather forecasts, stock trades and material designs. Applications of quantum supremacy Some people believe a quantum computer that achieves quantum supremacy could be the most disruptive new technology since the Intel 4004 microprocessor was invented in 1971. Certain professions and areas of business will be significantly impacted by quantum supremacy. Examples include: • The ability to perform more complex simulations on a larger scale will provide companies with improved efficiency, deeper insight and better forecasting, thus improving optimization processes. • Enhanced simulations that model complex quantum systems, such as biological molecules, would be possible. • Combining quantum computing with artificial intelligence (AI) could make AI immensely smarter than it is now. • New customized drugs, chemicals and materials can be designed, modeled and modified to help cultivate new pharmaceutical, commercial or business products. • The ability to factor extremely large numbers could break current, long-standing forms of encryption. Overall, quantum supremacy could start a new market for devices that have the potential to boost AI, intricately model molecular interactions and financial systems, improve weather forecasts and crack previously impossible codes. • Google-Nasa 2019 • USTC China 2020 Google-Nasa 2019 Sycamore Processor Experiment • Take an example of newbie, created basic Computer algorithm which in Quantum computing at this stage is a model/circuit of 1000’s of Quantum logic gates. As there is no structure in random circuits that classical algorithm can exploits and emulation of that circuit will take a huge effort of modern superComputer. • Each run of Quantum circuit on a Quantum Computer produces a bitstring (0001000). Due to some Quantum interference some bitstring are much more likely to occur than others when experiment is • Finding exact bitstring for a random circuit becomes exponentially difficult on classical Computer as number of qubits and number of gate cycles/depth grows. • Firstly a simplified circuit of 12 to 53 qubits while keeping circuit depth constant ,post verifying system conditions, • Random hard circuits with 53 qubits and increased depth till the point classical simulation became infeasible. • Result of this experiment of first experimental challenge against extended Church-Turing Thesis which states that classical Computers can efficiently implement any “reasonable” model of • Success was due to using new type of control nob that is able to turn off interaction between neighbouring systems to reduce errors. • New control calibration was developed to avoid qubit defects. Jiuzhang- Boson Sampling USTC Experiment • Generates a distribution number that is exceedingly difficult for a classical Computer to replicate • Firstly photons are sent into a network channels ,then photons are encountered a series of beam splitters ,each of which sends the photon down two path simultaneously, called a Quantum • Paths also merged together and repeated splitting and merging causes the photon to interfere with one another according to Quantum rules. • At end number of photons in each of the output channels is measured. • When repeated many times, this process produces a distribution of number based how many photons were found in each output. • If operated with large number of photons and many channels, the Quantum Computer will produce a distribution of number that is too complex for a classical Computer to calculate. • Limitation of Jiuzhang: It can perform only a single type of task i.e. Boson Sampling while Google’s Quantum Computer can be programmed to execute variety of algorithm, but on the other hand including Xanadu’s are programmable. Google-Nasa 2019 Sycamore Processor Experiment Jiuzhang- Boson Sampling USTC (China) Experiment Google-Nasa 2019 Sycamore Processor Experiment • Take an example of a newbie, created a basic computer algorithm which in Quantum computing at this stage is a model/circuit of 1000’s of Quantum logic gates. As there is no structure in random circuits that a classical algorithm can exploit, emulation of that circuit will take a huge effort of modern supercomputers. • Each run of Quantum circuit on a Quantum Computer produces a bitstring (0001000). Due to some Quantum interference some bit strings are much more likely to occur than others when experiment is • Finding exact bitstring for a random circuit becomes exponentially difficult on classical computers as the number of qubits and number of gate cycles/depth grows. • Firstly a simplified circuit of 12 to 53 qubits while keeping circuit depth constant, post verifying system conditions. • Random hard circuits with 53 qubits and increased depth till the point classical simulation became infeasible. • Result of this experiment of first experimental challenge against extended Church-Turing Thesis which states that classical Computers can efficiently implement any “reasonable” model of • Success was due to using a new type of control knob that is able to turn off interaction between neighbouring systems to reduce errors. • New control calibration was developed to avoid qubit defects Jiuzhang- Boson Sampling USTC (China) Experiment 8 Responses 1. Kiesha Zannini I like your blog. Its one of the best blogs online □ Admin Thank you, Kiesha. Keep visiting Quantum Guru, we are committed to publish informative blogs every week. 2. graliontorile We’re a bunch of volunteers and starting a brand new scheme in our community. Your website offered us with useful information to paintings on. You’ve performed a formidable process and our entire group might be grateful to you. 3. graliontorile You made some decent points there. I appeared on the internet for the problem and located most individuals will go together with together with your website. 4. zoritoler imol An interesting discussion is worth comment. I think that you should write more on this topic, it might not be a taboo subject but generally people are not enough to speak on such topics. To the next. 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Algebra Calculator At Calculator A, we understand the importance of accuracy, efficiency, and ease-of-use when it comes to calculations in various fields, from finance and engineering to academics and everyday life. Thatâ€™s why we have meticulously crafted a collection of intuitive and reliable calculators to cater to your diverse needs.	{"url":"https://calculator-a.com/product-category/algebra-calculator/","timestamp":"2024-11-12T19:15:49Z","content_type":"text/html","content_length":"194935","record_id":"<urn:uuid:ccf9acc2-8bbd-4a36-827b-3c67560d54e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00411.warc.gz"}
Mathematics is a wonderful game. It's one that can stretch students' minds and expose them to the beauty and unexpected delights that lie behind every good problem. I've always gravitated to colleagues who share my love of math's playful, game-like side, so I quickly made friends with Toni Cameron when we met at P.S. 503 in … Continue Reading ›› Deducing the “Mystery” Fraction Estimation is an important mathematical skill, yet we rarely ask students to make estimates that relate to fractions. As part of the Dynamic Number project, we created a "mystery" fraction challenge that presents a green point somewhere between 0 and 1 on the number line. The point's location can be represented as a fraction with numerator between … Continue Reading ›› Connecting Functions in Geometry and Algebra News alert! Scott and I wrote the cover story, Connecting Functions in Geometry and Algebra, in this month's Mathematics Teacher. You can read the article in print, but better yet, go to the free online version. This is the first time Mathematics Teacher has incorporated live dynamic-mathematics figures into its online offerings, allowing readers to manipulate … Continue Reading ›› A Coordinate Plane Logic Puzzle For the past few years, Scott Steketee and I have collaborated with the author team of Everyday Mathematics to integrate Web Sketchpad deeply into their curriculum. As part of that work, I just completed a websketch that nicely mixes practice with logical reasoning. Students are challenged to find a hidden treasure on … Continue Reading ›› Constructing Equal-Area Triangles The origins of this week's Web Sketchpad model date back to the Connected Geometry curriculum from the mid 1990s. I was one of the co-authors of the curriculum, working at Education Development Center with a wonderful team of math educators (Al Cuoco, … Continue Reading ›› Reflecting on the Annual NCTM Meeting This Thursday, Scott Steketee and I will be presenting two sessions at the NCTM 2015 Annual Meting in Boston: Functions as Dances: Experience Variation and Relative Rate of Change Session 52 on Thursday, April 16, 2015: 8:00 AM-9:15 AM in 157 B/C (BCEC) How better to explore rate of change than as independent and … Continue Reading ›› Innovative Approaches to Computer-Based Assessment, Part Four For the past month, I've focused this blog on the role that computers can play in assessing students' mathematical knowledge. I've presented three Web Sketchpad-based examples of assessment with mathematical topics ranging from isosceles triangles, to the Pythagorean Theorem, to the Continue Reading ›› Innovative Approaches to Computer-Based Assessment, Part Two In my previous post, I shared Dan Meyer's analysis of what's wrong with computer-based mathematics assessments. Dan focuses his critique on the Khan Academy's eighth-grade online mathematics course, identifying 74% of its assessment questions as focusing on numerical answers or multiple-choice items. This is … Continue Reading ›› Can Computer-Based Assessment Model Worthwhile Mathematics? Several weeks ago, Dan Meyer described his experience of completing 88 practice sets in Khan Academy's eighth-grade online mathematics course. His goal was to document the types of evidence the Khan Academy asked students to produce of their mathematical understanding. Dan's findings were disappointing: He concludes that 74% of the Khan Academy's eighth-grade questions were either multiple choice or required nothing more … Continue Reading ›› Exploring Factor Rainbows This week, I'm going to describe one of my favorite activities for introducing young learners to multiplication and factors. It comes from Nathalie Sinclair, a professor of mathematics education at Simon Fraser University. In the interactive Web Sketchpad model below (and here), press Jump Along to watch the … Continue Reading ››	{"url":"https://www.sineofthetimes.org/tag/common-core-state-standards/","timestamp":"2024-11-12T07:21:48Z","content_type":"text/html","content_length":"40827","record_id":"<urn:uuid:9b39905c-87c9-42e9-9866-d464675776d2>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00047.warc.gz"}
Multiplication Table 88 Times Table Greetings, math enthusiasts! Today, we embark on a fascinating journey into the realm of multiplication, as we unravel the secrets of the 88 times table. Get ready to discover the symmetrical patterns, unlock mental math shortcuts, and appreciate the wonders of numbers in a conversational and engaging manner. Chapter 1: Introducing the Symmetrical 88 Let's begin by acquainting ourselves with the basics. The 88 times table involves multiplying any number by 88. But why 88, you may wonder? Well, each number possesses its own unique qualities, and 88 is no exception. So, let's delve into the intricacies of this multiplication table and witness its symmetrical beauty. Chapter 2: Observing the Units Digit Patterns Let's take a look at the initial multiples of 88: • 1 x 88 = 88 • 2 x 88 = 176 • 3 x 88 = 264 At first glance, the pattern might not be immediately evident. However, if we focus on the units digit of each result, we'll uncover something fascinating. It follows a symmetrical pattern that cycles through 8, 6, 4, 2, 0, 8, 6, 4, 2, 0, and then repeats. Let's take a closer look: • 88 • 176 • 264 • 352 • 440 • 528 • 616 • 704 • 792 • 880 The units digit creates a symmetrical dance, moving through a sequence that maintains a sense of balance and harmony. Chapter 3: Embracing Symmetry Now that we've identified the pattern in the units digit, let's explore its significance. The symmetrical pattern we observe reveals that every multiple of 88 will always end with one of these five digits: 8, 6, 4, 2, or 0. It's a captivating observation that allows us to predict the units digit of any product involving 88. Chapter 4: Mental Math Shortcuts Understanding the patterns in the 88 times table can greatly enhance your mental math abilities. Let's consider an example to illustrate this: Suppose you need to calculate 88 multiplied by 7. Instead of following traditional multiplication steps, you can utilize the pattern we discovered earlier. Since 7 is a single-digit number, we know that the units digit of the product must be one of the repeating digits: 8, 6, 4, 2, or 0. In this case, the units digit is 6. Therefore, the answer is 616. By employing this mental math technique, you can quickly compute products involving 88, saving time and effort. Chapter 5: Exploring Greater Multiples Let's explore larger multiples of 88 to see if the symmetrical pattern persists: • 14 x 88 = 1,232 (units digit: 2) • 29 x 88 = 2,552 (units digit: 2) • 37 x 88 = 3,256 (units digit: 6) Even with larger numbers, the symmetrical dance in the units digit remains consistent and reliable. It's truly captivating to witness the symmetry within the 88 times table. Eighty-eight Multiplication Table Read, Repeat and Learn Eighty-eight times table and Check yourself by giving a test below Also check times table81 times table82 times table83 times table84 times table85 times table86 times table87 times table88 times table89 times table90 times table 88 Times Table Chart Table of 88 88 Times table Test Multiplication of 88 Reverse Multiplication of 88 Shuffled Multiplication of 88 How much is 88 multiplied by other numbers?	{"url":"https://www.printablemultiplicationtable.net/88-times-table.php","timestamp":"2024-11-03T23:13:35Z","content_type":"text/html","content_length":"28073","record_id":"<urn:uuid:ead8fa04-c3a0-4aa1-8c72-26c38183d223>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00148.warc.gz"}
abstract Stone duality nLab abstract Stone duality Due to Paul Taylor, Abstract Stone Duality (ASD) is a re-axiomatisation of the notions of topological space and continuous function in general topology in terms of a lambda-calculus of computable continuous functions and predicates. Abstract Stone duality is both constructive and computable, thus being one approach to synthetic topology. The topology on a space is treated not as a discrete lattice, but as an exponential object of the same category as the original space, with an associated λ-calculus (which includes an internal lattice structure). Every expression in the λ-calculus denotes both a continuous function and a program. ASD does not use the category of sets (or any topos), but the full subcategory of overt discrete objects plays this role (an overt object is the dual to a compact object), forming an arithmetic universe (a pretopos with lists) with general recursion; an optional ‘underlying set’ axiom (which is not predicative) will make this a topos. The classical (but not constructive) theory of locally compact sober topological spaces is a model of ASD, as is the theory of locally compact locales over any topos (even constructively). In “Beyond Local Compactness” on the ASD website, Taylor removes the restriction of local compactness. On Dedekind real numbers via abstract Stone duality: review in: See also: Last revised on February 21, 2023 at 11:01:11. See the history of this page for a list of all contributions to it.	{"url":"https://ncatlab.org/nlab/show/abstract+Stone+duality","timestamp":"2024-11-11T00:49:45Z","content_type":"application/xhtml+xml","content_length":"38460","record_id":"<urn:uuid:1288e9cb-c948-4fc4-a3ff-45f4004fd932>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00725.warc.gz"}
Do I really need a boost for my Mesa/Boogie? Getting a DC-3 here shortly from a kind fellow named Adam here on the boards. I'm also re-doing my pedalboard as well to celebrate for my first tube amp. I was thinking about getting a Tube Screamer or something similar as a boost pedal for the lead channel for searing solos. However I dunno if I should justify buying a pedal when the Mesa probably already has enough gain. Is this a good decision? Or not? • Members get rid of that disturbing picture, and I will tell you. • Members │Originally posted by Moltar │ │ │ │ │ │get rid of that disturbing picture, and I will tell you.│ • Members To boost the lead channel? No. To use on the Clean channel to add a bit of grit? sure. • Members Never tried a DC-3 but I would at least wait until you have the amp and try it out to see if you do indeed need more of a 'searing lead tone'. At least that's what I would do. • Members Have you even played this amp yet? • Members If the particular Mesa has the graphic EQ you can kick in and out with a footpedal that can act as your boost. I think the DC series have them. This topic is now archived and is closed to further replies.	{"url":"https://www.harmonycentral.com/forums/topic/1481751-do-i-really-need-a-boost-for-my-mesaboogie/","timestamp":"2024-11-10T02:53:15Z","content_type":"text/html","content_length":"134338","record_id":"<urn:uuid:904293ad-20cd-445b-abbc-a8d4a2fc9e0a>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00871.warc.gz"}
Tsiolkovsky rocket equation Cristian Sommariva, Umar Sheikh, Haomin Sun, Mengdi Kong The pre-thermal quench (pre-TQ) dynamics of a pure deuterium ( D 2 ) shattered pellet injection (SPI) into a 3 MA / 7 MJ JET H-mode plasma is studied via 3D non-linear MHD modelling with the JOREK code. The interpretative modelling captures the overall evo ... Iop Publishing Ltd	{"url":"https://graphsearch.epfl.ch/en/concept/772517","timestamp":"2024-11-08T23:29:53Z","content_type":"text/html","content_length":"121804","record_id":"<urn:uuid:2f3632e8-f247-4a6c-88bf-fd097c96aea6>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00829.warc.gz"}
coincident-root-loci: Equivariant CSM classes of coincident root loci This library contians a set of function to compute, among others, the GL(2)-equivariant Chern-Schwartz-MacPherson classes of coincident root loci, which are subvarieties of the space of unordered n-tuples of points in the complex projective line. To such an n-tuples we can associate a partition of n given by the multiplicities of the distinct points; this stratifies the set of all n-tuples, and we call these strata "coincident root loci". This package is supplementary software for a forthcoming paper. Skip to Readme Maintainer's Corner For package maintainers and hackage trustees	{"url":"https://hackage-origin.haskell.org/package/coincident-root-loci","timestamp":"2024-11-11T10:52:56Z","content_type":"text/html","content_length":"23632","record_id":"<urn:uuid:f8348746-814a-44e3-aaf7-9502ab91fe1d>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00864.warc.gz"}
Why Is Isaac Newton Famous? Isaac Newton is one of the most well-known and influential scientists in history. His contributions to the fields of mathematics, physics, and astronomy have shaped our understanding of the universe and revolutionized the way we view the world around us. But what made Newton stand out from other scientists of his time? In this article, we will explore the reasons why Isaac Newton is famous, and why his legacy continues to live on. The Early Years of Isaac Newton Isaac Newton was born on January 4, 1643, in Woolsthorpe, England. He came from a family of farmers and was a small and sickly child. At the age of three, his father passed away, leaving him in the care of his grandmother. Despite his difficult childhood, Newton showed exceptional academic abilities, and at the age of 18, he enrolled at Trinity College, Cambridge. While attending college, Newton developed a keen interest in mathematics and physics. He spent many hours in the library, studying and conducting experiments. His dedication and passion for science soon caught the attention of his professors, and he was allowed to teach undergraduate courses while completing his studies. Newton and the Theory of Optics One of Newton’s most significant contributions to science was his groundbreaking work in the field of optics. In 1666, a young Newton was quarantined during the Great Plague and spent most of his time studying light and colors. He conducted a series of experiments, such as passing light through a prism, which led him to discover that white light is made up of different colors. This discovery challenged the popular belief at the time that light was pure and could not be separated into different colors. Newton’s research and experiments laid the foundation for the modern field of optics and helped us understand how light behaves. The Laws of Motion Another area in which Isaac Newton made significant contributions was in the field of mechanics. By studying the work of philosophers and mathematicians before him, such as Galileo Galilei and René Descartes, Newton was able to develop his three laws of motion. His first law, also known as the law of inertia, states that an object at rest will remain at rest unless acted upon by an external force. This law helped explain the concept of gravity and how objects move in the universe. His second law, known as the force-mass-acceleration relationship, states that the force on an object is equal to its mass multiplied by its acceleration. Finally, his third law, or the law of action and reaction, states that for every action, there is an equal and opposite reaction. These laws of motion laid the foundation for modern physics and have been crucial in our understanding of the natural world. They also helped explain how the planets move in our solar system, paving the way for Newton’s most famous work, the theory of gravitation. The Theory of Gravitation In 1687, Isaac Newton published his most well-known work, Mathematical Principles of Natural Philosophy, also known as the Principia. In this book, Newton explains his theory of universal gravitation, which states that every object in the universe is attracted to every other object with a force that is directly proportional to their masses and inversely proportional to the distance between them. This theory unified the laws of motion with the laws of gravity and provided a complete and accurate explanation of how objects move in space. It also provided a mathematical framework for predicting the motion of objects and was a significant breakthrough in the field of physics. The Legacy of Isaac Newton Isaac Newton’s legacy is one that continues to have an impact on the world today. His contributions to science have formed the basis of many modern technologies, such as satellites and GPS. His laws of motion and theory of gravitation are still taught in schools and universities worldwide, and his work has influenced countless scientists and thinkers. Furthermore, Newton’s interest and involvement in alchemy and theology have also left a mark on history. His studies in alchemy influenced his ideas on gravity, and his work in theology helped shape religious thought during the Enlightenment period. In Conclusion The reasons for Isaac Newton’s fame are vast and varied. From his early interest in mathematics and optics to his groundbreaking work in mechanics and the theory of gravitation, Newton’s contributions have had a profound impact on the world of science and beyond. His legacy continues to inspire future scientists and serves as a reminder of the power of dedication, passion, and curiosity in the pursuit of knowledge. 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Tips and Tricks for Blood Relations \| Logical Reasoning (LR) and Data Interpretation (DI) - CAT PDF Download Candidates can discover various tips and strategies below to efficiently solve questions in the Blood Relation reasoning section. • Tip 1: When tackling questions in the Blood Relation reasoning section, candidates should treat "ME" as the introducing person to quickly answer posed questions. • Tip 2: In the Coded Relation type blood relation section, candidates should carefully examine all options, including gender considerations. By eliminating incorrect options, candidates can arrive at the final conclusion. • Tip 3: Avoid making predictions about a person's gender based on their name unless explicitly mentioned, as this may lead to an inaccurate answer. • Tip 4: It's crucial to recognize that being the "only son" or "only daughter" does not necessarily imply being the only child. If a blood relation question states that A is the only son of B, A could potentially have both a son and a daughter or just one son. • Tip 5: Identify the two individuals between whom the relationship needs to be determined. Then, using the relationships between other family members provided as intermediaries, attempt to establish the connection between the two required persons. • Tip 6: Relating the given relations in questions to personal relationships will enhance understanding and aid in answering questions more effectively. • Tip 7: Utilizing pictorial representation is advisable for solving questions. This form allows candidates to systematically arrange data, making it easier to comprehend the relationships presented in the questions. • Tip 8: In pictorial representation, use '+' for denoting a male gender and use ' - ' for denoting a female gender. Solved Example Example 1: Introducing a lady at the party, Goldy told her friend, “She is sister-in-law of my only brother who is father of Ganesh who is grandson of Navya’s husband who is my father who has only two child.” How is Ganesh related to that lady? Sol: If we analyze all the statements given above in the question and draw the family tree for the same Example 2: Given Directions M@N means M is the wife of N M#N means M is the son of N M5N means M is the sister of N M$N means M is the father of N Which of the following expressions shows I as the brother of H? (a) H#P$I, (b) H%P$I, (c) I%H$P, (d) H%I$P Sol: Option D is correct. The above statements implies the following things. M is From the above options, we need to find an expression which represents I as the brother of H. So, check all the options and eliminate those in which either the gender is of I is unknown, or I is female. In the above question, @ and % represents wife and sister respectively such as female so eliminate all such options. And if in any expression, I is present in the end then it means that the gender of I will be unknown so eliminate such options too. From the above trick we can eliminate options 1,2, and 3. Hence in the expression “H%I$P”, I is the brother of H. Example 3: Read the following instructions: 1. A + B indicates A is the brother of B; 2. A – B indicates A is the sister of B and 3. A x B indicates A is the father of B Which of the following means that C is the son of M? 1. M – N x C + F 2. F – C + N x M 3. N + M – F x C 4. M x N – C + F Sol: In these types of questions, we have to analyze each option In Option A, According to this option NxC indicates N is the father of c. Hence it is wrong. In Option B, According to this option C is the brother of N who is the father of M. Hence it is wrong. In Option C , According to this option Fx C indicates F is the father of C. Hence it is wrong. In Option D, According to this option M is the father of N who is the sister of C hence C and N are siblings and C is the brother of F so, C is male, Hence C is the son of M. Hence, Option D is the correct answer. Example 4: Read the following information carefully and answer the questions which follow: Mr. Rajat Chopra and his wife Nikita Chopra have 3 sons whose names are Ramesh, Suresh and Umesh. Mishra family is a neighbor of the Chopra's. Mr. Amit Mishra and his wife Neha Mishra have 2 daughters whose names are Payal and Ruchi. The two neighboring families go to Kerala for a vacation. They decide to go for boating but no boat could carry more than 3 members. So they hire 3 boats. None of the children know how to row a boat, so at least one of the adults have to be there on each boat. Moreover, no boat has all the three members from the same family. Q-1: If Neha and Ruchi are on the same boat, which of the following could be a list of people on another boat? a) Ramesh, Amit, Payal b) Ramesh, Suresh, Amit c) Ramesh, Payal, Suresh d) Amit, Payal, Nikita Sol: Option 'B' is correct. We know that none of the boats have all the members from the same family. So each boat must have at least 1 member from each of the two families. Neha and Ruchi are on the same boat. So only two people from Mishra family are remaining. Both of them need to be on different boats. Hence options A and D can be ruled out because they have Payal and Amit on the same boat. Option C is also not possible since it does not have any adult on the boat Q 2: If Rajat and Amit are in the same boat and each of the three brothers are on different boats, then which of the following is necessarily true? a) Every boat has both males and females on it. b) One of the boats has only females on it. c) One of the boats has only males on it. d) The two sisters are on the same boat. Sol: Option C is Correct. If Rajat and Amit are on the same boat then the other two boats would be rowed by Neha and Nikita. It is known that each of the brothers are on different boats. So one of them will be on Rajat and Amit's boat. Hence there will be one boat with only males. Q 3: If the three children from the Chopra family ride in different boats then which of the following is definitely false? I. Rajat and Nikita are rowing in the same boat. II Amit and Neha are rowing in the same boat. a. I only b. II only c. Both I and II d. Neither I nor II Sol: Option 'A' is correct. The three boys are rowing in different boats. We know that there are only three boats and each boat can carry three persons. Hence Amit and Nikita cannot be on the same boat as that would mean that one of the boats has all the members from the Chopra family. So I is definitely false. Statement II can be true as well false. It can be true if the two people travel in one boat while Amit and Nikita row the other two boats. Hence nothing can be said with certainty about statement Q-4: If Nikita and Amit are on the same boat then which of the following cannot be the combination of people on any boat? a) Ramesh, Neha, Ruchi b) Neha, Ramesh, Suresh c) Neha, Ruchi, Umesh d) Neha, Suresh, Rajat Sol: Option 'D' is the answer. Neha, Suresh and Rajat cannot be on the same boat. This is because we know that one of the adults have to be there on each boat. Since Nikita and Amit are already on the same boat, Neha and Rajat have to on the other two boats. If they are on the same boat then there will be no adult on the third boat. Hence, option D is the combination that is not possible.	{"url":"https://edurev.in/t/319933/Tips-and-Tricks-for-Blood-Relations","timestamp":"2024-11-14T08:49:40Z","content_type":"text/html","content_length":"298215","record_id":"<urn:uuid:b24201b1-b913-42bf-9ce4-5978880e2851>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00181.warc.gz"}
Solving Word Problems by Adding Two Decimal Numbers Question Video: Solving Word Problems by Adding Two Decimal Numbers A lion ate 11.4 kg of meat one day and 14.6 kg of meat the next day. How many kilograms of meat did it eat across the two days? Video Transcript A lion ate 11.4 kilograms of meat one day and 14.6 kilograms of meat the next day. How many kilograms of meat did it eat across the two days? To find the total amount of meat that the lion ate across the two days, we need to perform an addition: 11.4 plus 14.6. We can do this using a column addition method. The two numbers that we’re going to be summing are decimals, and we must make sure that we line up the decimal points. The decimal point for our answer will be vertically below the two aligned decimal points for the numbers that we’re summing. Now, we begin to sum from right to left. Four plus six is equal to 10, so we put a zero in the tenths column and carry a one across into the units column. Next, we have one plus four plus one. This is equal to six. In the final column, the tens column, we have one plus one which is equal to two. The answer to the addition is 26.0, which can just be written as 26. The total amount of meat the lion ate across the two days is 26 kilograms.	{"url":"https://www.nagwa.com/en/videos/757143858305/","timestamp":"2024-11-10T04:50:44Z","content_type":"text/html","content_length":"240317","record_id":"<urn:uuid:12fc14ca-2cc0-43d6-a389-1321289d053f>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00533.warc.gz"}
Number of Islands Given a two-dimensional integer matrix of 1s and 0s, return the number of “islands” in the matrix. A 1 represents land and 0 represents water, so an island is a group of 1s that are neighboring whose perimeter is surrounded by water. Note: Neighbors can only be directly horizontal or vertical, not diagonal. • n, m ≤ 100 where n and m are the number of rows and columns in matrix. Example 1 Example 2 Example 3	{"url":"https://xuankentay.com/exercises/number-of-islands/","timestamp":"2024-11-10T19:08:51Z","content_type":"text/html","content_length":"26682","record_id":"<urn:uuid:7ea25ae2-2b27-4113-bc11-0d56108e0911>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00049.warc.gz"}
Branch and bound (BB, B&B, or BnB) is a method for solving optimization problems by breaking them down into smaller sub-problems and using a bounding function to eliminate sub-problems that cannot contain the optimal solution. It is an algorithm design paradigm for discrete and combinatorial optimization problems, as well as mathematical optimization. A branch-and-bound algorithm consists of a systematic enumeration of candidate solutions by means of state space search: the set of candidate solutions is thought of as forming a rooted tree with the full set at the root. The algorithm explores branches of this tree, which represent subsets of the solution set. Before enumerating the candidate solutions of a branch, the branch is checked against upper and lower estimated bounds on the optimal solution, and is discarded if it cannot produce a better solution than the best one found so far by the algorithm. The algorithm depends on efficient estimation of the lower and upper bounds of regions/branches of the search space. If no bounds are available, the algorithm degenerates to an exhaustive search. The method was first proposed by Ailsa Land and Alison Doig whilst carrying out research at the London School of Economics sponsored by British Petroleum in 1960 for discrete programming,^[1]^[2] and has become the most commonly used tool for solving NP-hard optimization problems.^[3] The name "branch and bound" first occurred in the work of Little et al. on the traveling salesman problem.^[4]^[5 The goal of a branch-and-bound algorithm is to find a value x that maximizes or minimizes the value of a real-valued function f(x), called an objective function, among some set S of admissible, or candidate solutions. The set S is called the search space, or feasible region. The rest of this section assumes that minimization of f(x) is desired; this assumption comes without loss of generality, since one can find the maximum value of f(x) by finding the minimum of g(x) = −f(x). A B&B algorithm operates according to two principles: • It recursively splits the search space into smaller spaces, then minimizing f(x) on these smaller spaces; the splitting is called branching. • Branching alone would amount to brute-force enumeration of candidate solutions and testing them all. To improve on the performance of brute-force search, a B&B algorithm keeps track of bounds on the minimum that it is trying to find, and uses these bounds to "prune" the search space, eliminating candidate solutions that it can prove will not contain an optimal solution. Turning these principles into a concrete algorithm for a specific optimization problem requires some kind of data structure that represents sets of candidate solutions. Such a representation is called an instance of the problem. Denote the set of candidate solutions of an instance I by S[I]. The instance representation has to come with three operations: • branch(I) produces two or more instances that each represent a subset of S[I]. (Typically, the subsets are disjoint to prevent the algorithm from visiting the same candidate solution twice, but this is not required. However, an optimal solution among S[I] must be contained in at least one of the subsets.^[6]) • bound(I) computes a lower bound on the value of any candidate solution in the space represented by I, that is, bound(I) ≤ f(x) for all x in S[I]. • solution(I) determines whether I represents a single candidate solution. (Optionally, if it does not, the operation may choose to return some feasible solution from among S[I].^[6]) If solution(I ) returns a solution then f(solution(I)) provides an upper bound for the optimal objective value over the whole space of feasible solutions. Using these operations, a B&B algorithm performs a top-down recursive search through the tree of instances formed by the branch operation. Upon visiting an instance I, it checks whether bound(I) is equal or greater than the current upper bound; if so, I may be safely discarded from the search and the recursion stops. This pruning step is usually implemented by maintaining a global variable that records the minimum upper bound seen among all instances examined so far. Generic version The following is the skeleton of a generic branch and bound algorithm for minimizing an arbitrary objective function f.^[3] To obtain an actual algorithm from this, one requires a bounding function bound, that computes lower bounds of f on nodes of the search tree, as well as a problem-specific branching rule. As such, the generic algorithm presented here is a higher-order function. 1. Using a heuristic, find a solution x[h] to the optimization problem. Store its value, B = f(x[h]). (If no heuristic is available, set B to infinity.) B will denote the best solution found so far, and will be used as an upper bound on candidate solutions. 2. Initialize a queue to hold a partial solution with none of the variables of the problem assigned. 3. Loop until the queue is empty: 1. Take a node N off the queue. 2. If N represents a single candidate solution x and f(x) < B, then x is the best solution so far. Record it and set B ← f(x). 3. Else, branch on N to produce new nodes N[i]. For each of these: 1. If bound(N[i]) > B, do nothing; since the lower bound on this node is greater than the upper bound of the problem, it will never lead to the optimal solution, and can be discarded. 2. Else, store N[i] on the queue. Several different queue data structures can be used. This FIFO queue-based implementation yields a breadth-first search. A stack (LIFO queue) will yield a depth-first algorithm. A best-first branch and bound algorithm can be obtained by using a priority queue that sorts nodes on their lower bound.^[3] Examples of best-first search algorithms with this premise are Dijkstra's algorithm and its descendant A* search. The depth-first variant is recommended when no good heuristic is available for producing an initial solution, because it quickly produces full solutions, and therefore upper A C++-like pseudocode implementation of the above is: // C++-like implementation of branch and bound, // assuming the objective function f is to be minimized CombinatorialSolution branch_and_bound_solve( CombinatorialProblem problem, ObjectiveFunction objective_function /f/, BoundingFunction lower_bound_function /bound/) // Step 1 above. double problem_upper_bound = std::numeric_limits<double>::infinity; // = B CombinatorialSolution heuristic_solution = heuristic_solve(problem); // x_h problem_upper_bound = objective_function(heuristic_solution); // B = f(x_h) CombinatorialSolution current_optimum = heuristic_solution; // Step 2 above queue<CandidateSolutionTree> candidate_queue; // problem-specific queue initialization candidate_queue = populate_candidates(problem); while (!candidate_queue.empty()) { // Step 3 above // Step 3.1 CandidateSolutionTree node = candidate_queue.pop(); // "node" represents N above if (node.represents_single_candidate()) { // Step 3.2 if (objective_function(node.candidate()) < problem_upper_bound) { current_optimum = node.candidate(); problem_upper_bound = objective_function(current_optimum); // else, node is a single candidate which is not optimum else { // Step 3.3: node represents a branch of candidate solutions // "child_branch" represents N_i above for (auto&& child_branch : node.candidate_nodes) { if (lower_bound_function(child_branch) <= problem_upper_bound) { candidate_queue.enqueue(child_branch); // Step 3.3.2 // otherwise, bound(N_i) > B so we prune the branch; step 3.3.1 return current_optimum; In the above pseudocode, the functions heuristic_solve and populate_candidates called as subroutines must be provided as applicable to the problem. The functions f (objective_function) and bound (lower_bound_function) are treated as function objects as written, and could correspond to lambda expressions, function pointers and other types of callable objects in the C++ programming language. When ${\displaystyle \mathbf {x} }$ is a vector of ${\displaystyle \mathbb {R} ^{n}}$ , branch and bound algorithms can be combined with interval analysis^[8] and contractor techniques in order to provide guaranteed enclosures of the global minimum.^[9]^[10] This approach is used for a number of NP-hard problems: Branch-and-bound may also be a base of various heuristics. For example, one may wish to stop branching when the gap between the upper and lower bounds becomes smaller than a certain threshold. This is used when the solution is "good enough for practical purposes" and can greatly reduce the computations required. This type of solution is particularly applicable when the cost function used is noisy or is the result of statistical estimates and so is not known precisely but rather only known to lie within a range of values with a specific probability. Relation to other algorithms Nau et al. present a generalization of branch and bound that also subsumes the A, B and alpha-beta search algorithms.^[16] Optimization Example Branch and bound can be used to solve this problem Maximize ${\displaystyle Z=5x_{1}+6x_{2}}$ with these constraints ${\displaystyle x_{1}+x_{2}\leq 50}$ ${\displaystyle 4x_{1}+7x_{2}\leq 280}$ ${\displaystyle x_{1},x_{2}\geq 0}$ ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are integers. The first step is to relax the integer constraint. We have two extreme points for the first equation that form a line: ${\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}50\\0 \end{bmatrix}}}$ and ${\displaystyle {\begin{bmatrix}0\\50\end{bmatrix}}}$ . We can form the second line with the vector points ${\displaystyle {\begin{bmatrix}0\\40\end{bmatrix}}}$ and ${\ displaystyle {\begin{bmatrix}70\\0\end{bmatrix}}}$ . the two lines. The third point is ${\displaystyle {\begin{bmatrix}0\\0\end{bmatrix}}}$ . This is a convex hull region so the solution lies on one of the vertices of the region. We can find the intersection using row reduction, which is ${\displaystyle {\begin{bmatrix}70/3\\80/3\end{bmatrix}}}$ , or ${\displaystyle {\begin{bmatrix}23.333\\26.667\end{bmatrix}}}$ with a value of 276.667. We test the other endpoints by sweeping the line over the region and find this is the maximum over the reals. We choose the variable with the maximum fractional part, in this case ${\displaystyle x_{2}}$ becomes the parameter for the branch and bound method. We branch to ${\displaystyle x_{2}\leq 26}$ and obtain 276 @ ${\displaystyle \langle 24,26\rangle }$ . We have reached an integer solution so we move to the other branch ${\displaystyle x_{2}\geq 27}$ . We obtain 275.75 @${\displaystyle \langle 22.75,27\rangle }$ . We have a decimal so we branch ${\displaystyle x_{1}}$ to ${\displaystyle x_{1}\leq 22}$ and we find 274.571 @${\displaystyle \langle 22,27.4286\rangle }$ . We try the other branch ${\displaystyle x_{1}\geq 23}$ and there are no feasible solutions. Therefore, the maximum is 276 with ${\displaystyle x_{1}\longmapsto 24}$ and ${\displaystyle x_{2}\longmapsto 26}$ . See also External links • LiPS – Free easy-to-use GUI program intended for solving linear, integer and goal programming problems. • Cbc – (Coin-or branch and cut) is an open-source mixed integer programming solver written in C++.	{"url":"https://www.knowpia.com/knowpedia/Branch_and_bound","timestamp":"2024-11-02T17:56:22Z","content_type":"text/html","content_length":"161416","record_id":"<urn:uuid:83ef8954-e5d0-4954-b3ba-a3d0ef004ce8>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00818.warc.gz"}
Harmonic Analysis and Differential Equations Seminar The HADES seminar on Tuesday, December 5th, will be at 3:30pm in Room 748 (not in 740 this week!). Speaker: Akshat Kumar Abstract: Graph Laplacians and Markov processes are intimately connected and ubiquitous in the study of graph structures. They have led to significant advances in a class of geometric inverse problems known as “manifold learning”, wherein one wishes to learn the geometry of a Riemannian submanifold from finite Euclidean point samples. The data gives rise to the geometry-encoding neighbourhood graphs. Present-day techniques are dominated primarily by the low spectral resolution of the graph Laplacians, while finer aspects of the underlying geometry, such as the geodesic flow, are observed only in the high spectral regime. We establish a data-driven uncertainty principle that dictates the scaling of the wavelength $h$, with respect to the density of samples, at which graph Laplacians for neighbourhood graphs are approximately $h$-pseudodifferential operators. This sets the stage for a semiclassical approach to the high-frequency analysis of wave dynamics on weighted graphs. We thus establish a discrete version of Egorov’s theorem and achieve convergence rates for the recovery of geodesics on the underlying manifolds through quantum dynamics on the approximating graphs. I will show examples on samples of model manifolds and briefly discuss some applications to real-world datasets. Mode stability for Kerr(-de Sitter) black holes The HADES seminar on Tuesday, November 28th, will be at 3:30pm in Room 740. Speaker: Rita Teixeira da Costa Abstract: The Teukolsky master equations are a family of PDEs describing the linear behavior of perturbations of the Kerr black hole family, of which the wave equation is a particular case. As a first essential step towards stability, Whiting showed in 1989 that the Teukolsky equation on subextremal Kerr admits no exponentially growing modes. In this talk, we review Whiting’s classical proof and a recent adaptation thereof to the extremal Kerr case. We also present a new approach to mode stability, based on uncovering hidden spectral symmetries in the Teukolsky equations. Part of this talk is based on joint work with Marc Casals (CBPF/UCD). This talks complements yesterday’s Analysis & PDE seminar, but will be self-contained. Methods for sharp well-posedness for completely integrable PDE The HADES seminar on Tuesday, November 14th, will be at 3:30pm in Room 740. Speaker: Thierry Laurens Abstract:We will describe some of the methods used to prove sharp well-posedness for the Benjamin–Ono equation in the class of H^s spaces, namely, the method of commuting flows. Since its introduction by Killip and Visan in 2019, this groundbreaking approach to completely integrable systems has been adapted to a wide variety of models in order to prove sharp well-posedness results that were previously inaccessible. In this talk, we will describe some of the overarching principles of the method of commuting flows, with a focus on how these ideas were implemented in the case of the Benjamin–Ono equation. This is based on joint work with Rowan Killip and Monica Visan. Strichartz estimates for Schroedinger evolutions The HADES seminar on Tuesday, November 7th, will be at 3:30pm in Room 740. Speaker: Daniel Tataru Abstract: I will provide a broad introduction to the topic of dispersive and Strichartz estimates for Schroedinger evolutions on curved backgrounds, with the final goal of describing the new Strichartz estimates proved jointly with Mihaela Ifrim in the context of 1D quasilinear Schroedinger flows. Optimal enhanced dissipation for geodesic flows The HADES seminar on Tuesday, October 31st will be at 3:30pm in Room 740. Speaker: Maciej Zworski Abstract: We consider geodesic flows on negatively curved compact manifolds or more generally contact Anosov flows (all these concepts will be pedagogically explained). The object is to show that if $ X $ is the generator of the flow and $ \Delta $, a (negative) Laplacian, then solutions to the convection diffusion equation, $ \partial_t u = X u + \nu \Delta $, $ \nu \geq 0 $, satisfy \[ \\| u ( t ) – \underline u \\|_{L^2 ( M) } \leq C \nu^{-K} e^{ – \beta t } \\| u( 0 ) \\|_{L^2 ( M) }, \] where $ \underline u $ is the (conserved) average of $ u (0) $ with respect to the contact volume form and $K $ is a fixed constant. This provides many examples of very precise {\em optimal enhanced dissipation} in the sense of recent works of Bedrossian–Blumenthal–Punshon-Smith and Elgindi–Liss–Mattingly. The proof is based on results by Dyatlov and the speaker on stochastic stability of Pollicott–Ruelle resonances, another concept which will be introduced and explained. The talk is based on joint work with Zhongkai Tao. Sharp Furstenberg Sets Estimate in the Plane The HADES seminar on Tuesday, October 24th will be at 3:30pm in Room 740. Speaker: Kevin Ren Abstract: Fix a real number 0 < s <= 1. A set E in the plane is a s-Furstenberg set if there exists a line in every direction that intersects E in a set with Hausdorff dimension s. For example, a planar Kakeya set is a special case of a 1-Furstenberg set, and indeed we know that 1-Furstenberg sets have Hausdorff dimension 2. However, obtaining a sharp lower bound for the Hausdorff dimension of s-Furstenberg sets for any 0 < s < 1 has been a challenging open problem for half a century. In this talk, I will illustrate the rich connections between the Furstenberg sets conjecture and other important topics in geometric measure theory and harmonic analysis, and show how exploring these connections can fully resolve the Furstenberg conjecture. Joint works with Yuqiu Fu and Hong Wang. Wellposedness for Quasi-linear Problems and the Modified Energy Method The HADES seminar on Tuesday, October 17th will be at 3:30pm in Room 740. Speaker: Ryan Martinez Abstract: We give an exposition of the Hadamard wellposedness and explain the modified energy method through the use of the Kirchhoff type Wave Equation as an example. We use the ideas from Daniel and Mihaela’s “Local Wellposedness for Quasilinear Problems: A Primer” as well as from their work with John K. Hunter and Tak Kwong Wong, “Long Time Solutions for a Burgers-Hilbert Equation via a Modified Energy Method.” Existence of more self-similar implosion profiles for the Euler-Poisson system The HADES seminar on Tuesday, October 10th will be at 3:30pm in Room 740. Speaker: Ely Sandine Abstract: I will discuss implosion for the equations describing a gas which is compressible, isothermal and self-gravitating. Under the hypotheses of radial symmetry and self-similarity, the equations reduce to a system of ODEs which has been extensively studied by the astrophysics community using numerical methods. One such solution, discovered by Larson and Penston in 1969, was recently rigorously proved to exist by Guo, Hadžić and Jang. In this talk, I will discuss rigorous existence of a subset of the discrete family of solutions found numerically by Hunter in 1977. Construction of nonunique solutions of the transport and continuity equation for Sobolev vector fields in DiPerna–Lions’ theory The HADES seminar on Tuesday, October 3rd will be at 3:30pm in Room 740. Speaker: Anuj Kumar Abstract: In this talk, we are concerned with DiPerna–Lions’ theory for the transport equation. In the first part of the talk, I will discuss a few results regarding the nonuniqueness of trajectories of the associated ODE. Alberti ’12 asked the following question: are there continuous Sobolev vector fields with bounded divergence such that the set of initial conditions for which the trajectories are not unique is of full measure? We construct an explicit example of divergence-free H\”older continuous Sobolev vector field for which trajectories are not unique on a set of full measure, which then answers the question of Alberti. The construction is based on building an appropriate Cantor set and a “blob flow” vector field to translate cubes in space. The vector field constructed also implies nonuniqueness in the class of measure solutions. The second part to talk is a more recent work jointly with E. Bruè and M. Colombo. We construct nonunique solutions of the continuity equation in the class L^\infty in time and L^r in space. We prove nonuniqueness in the range of exponents beyond what is available using the method of convex integration and sharply match with the range of uniqueness of solutions from Bruè, Colombo, De Lellis’ 21. On a nonlinearly coupled stochastic fluid-structure interaction model. The HADES seminar on Tuesday, September 26th will be at 3:30pm in Room 740. Speaker: Krutika Tawri Abstract: In this talk, we will present a constructive approach to investigate the existence of martingale solutions to a benchmark fluid-structure interaction problem that involves an incompressible, viscous fluid interacting with a linearly elastic membrane subjected to a multiplicative stochastic force. The fluid flow is described by the Navier-Stokes equations while the elastodynamics of the thin structure is modeled by the Koiter shell equations. We will discuss the challenges arising due to the random motion of the time-dependent fluid domain and present our recent findings. This is joint work with Sunčica Čanić.	{"url":"https://wp.math.berkeley.edu/hades/category/fall-2023/","timestamp":"2024-11-11T09:50:42Z","content_type":"text/html","content_length":"56776","record_id":"<urn:uuid:5704dbf4-d434-4691-bfdf-6424d961e7af>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00738.warc.gz"}
Any languages able to program (0,1] mathematical interval 04-28-2017, 01:56 PM (This post was last modified: 04-28-2017 08:18 PM by StephenG1CMZ.) Post: #1 StephenG1CMZ Posts: 1,074 Senior Member Joined: May 2015 Any languages able to program (0,1] mathematical interval Mathematicians often use brackets to indicate whether or not a limit value is included or excluded, e.g. (0,9] might be programmed as 0<=x<9... I don't use that syntax often so wouldn't be sure which bracket is which without checking. Update: It seems I have the brackets the wrong way round. Are there any programming languages that recognise that syntax directly, thereby avoiding the need to check and code it differently? I am mainly interested in mainstream languages (Basic, C, Lua, Python ... And of course HP PPL) rather than Mathematica and the like. Alternatively, is there a different way of writing such an interval concisely, yet making which is included/excluded clearer. Stephen Lewkowicz (G1CMZ) 04-28-2017, 03:48 PM Post: #2 Gerson W. Barbosa Posts: 1,620 Senior Member Joined: Dec 2013 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 01:56 PM)StephenG1CMZ Wrote: Alternatively, is there a different way of writing such an interval concisely, yet making which is included/excluded clearer. [0, 9[ ? 04-28-2017, 04:54 PM (This post was last modified: 04-28-2017 05:00 PM by Gerson W. Barbosa.) Post: #3 Gerson W. Barbosa Posts: 1,620 Senior Member Joined: Dec 2013 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 01:56 PM)StephenG1CMZ Wrote: Mathematicians often use brackets to indicate whether or not a limit value is included or excluded, e.g. (0,9] might be programmed as 0<=x<9... I don't use that syntax often so wouldn't be sure which bracket is which without checking. It appears the correct syntax is [0, 9). W\|A understands the notation I was taught in middle school: Plot [0, 9[ 04-28-2017, 08:16 PM Post: #4 StephenG1CMZ Posts: 1,074 Senior Member Joined: May 2015 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 03:48 PM)Gerson W. Barbosa Wrote: (04-28-2017 01:56 PM)StephenG1CMZ Wrote: Alternatively, is there a different way of writing such an interval concisely, yet making which is included/excluded clearer. [0, 9[ ? ? I don't find that any clearer... The only notation I recall clearly from my maths class is < or <= Stephen Lewkowicz (G1CMZ) 04-28-2017, 08:28 PM Post: #5 Gerson W. Barbosa Posts: 1,620 Senior Member Joined: Dec 2013 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 08:16 PM)StephenG1CMZ Wrote: (04-28-2017 03:48 PM)Gerson W. Barbosa Wrote: [0, 9[ ? ? I don't find that any clearer... The only notation I recall clearly from my maths class is < or <= clearly encloses 0 while in 9 is definitely out. This doesn't seem to be a standard notation, though. 04-28-2017, 09:23 PM (This post was last modified: 04-29-2017 08:50 PM by pier4r.) Post: #6 pier4r Posts: 2,248 Senior Member Joined: Nov 2014 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 01:56 PM)StephenG1CMZ Wrote: Alternatively, is there a different way of writing such an interval concisely, yet making which is included/excluded clearer. To emphasize the equal. if ( (a < 0 and a > 9) or (a == 0) ) { //this avoid the subtle <= that may be overlooked more easily than an extended condition. //Code maintenance is more important than performance most of the time. Brain time is more precious than CPU time. Wikis are great, Contribute :) 04-28-2017, 10:03 PM Post: #7 nsg Posts: 60 Member Joined: Dec 2013 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 08:28 PM)Gerson W. Barbosa Wrote: [0 clearly encloses 0 while in 9[ 9 is definitely out. This doesn't seem to be a standard notation, though. It may be not standard, but I was taught it in school too. Have to admit, never seen it after school, it seemed to have been replaced by [a,b) convention. When I get to my bookshelf I will check for it in some older texts. 04-28-2017, 10:04 PM (This post was last modified: 04-30-2017 12:24 PM by SlideRule.) Post: #8 SlideRule Posts: 1,533 Senior Member Joined: Dec 2013 RE: Any languages able to program (0,1] mathematical interval From the publication Pre-Calculaus for Dummies, second edition: You can use interval notation to express where a set of solutions begins and here it ends. Interval notation is a common way to express the solution set to an inequality, and it’s important because it’s how you express solution sets in calculus. Most pre-calculus books and some pre-calculus teachers now require all sets to be written in interval notation. If the endpoint of the interval isn’t included in the solution (for < or >), the interval is called an open interval. You show it on the graph with an open circle at the point and by using parentheses in notation. If the endpoint is included in the solution (for ≤ or ≥) the interval is called a closed interval, which you show on the graph with a filled-in circle at the point and by using square brackets in notation. For example, the solution set -2 < x ≤ 3, rewrite this solution set as an and statement: -2 < x AND x ≤ 3. In interval notation, you write this solution as (–2, 3]. 04-28-2017, 10:11 PM (This post was last modified: 04-28-2017 10:23 PM by Vtile.) Post: #9 Vtile Posts: 406 Senior Member Joined: Oct 2015 RE: Any languages able to program (0,1] mathematical interval ..And in 50g plot etc.. Y1(X)=(Your function)(0≤X)(9>X) (0≤X)(9>X) Boolean AND If X is less than zero and is less than 9 then 00=0 If X is more than zero but is less than 9 then 11=1 If X is more than zero and is more than or equal of 9 then 10=0 PS. (–2, 3] So utterly ugly !! 04-29-2017, 07:29 PM Post: #10 StephenG1CMZ Posts: 1,074 Senior Member Joined: May 2015 RE: Any languages able to program (0,1] mathematical interval (04-28-2017 09:23 PM)pier4r Wrote: (04-28-2017 01:56 PM)StephenG1CMZ Wrote: Alternatively, is there a different way of writing such an interval concisely, yet making which is included/excluded clearer. To emphasize the equal. if ( (a < 0 and a < 9) or (a == 0) ) { //this avoid the subtle <= that may be overlooked more easily than an extended condition. //Code maintenance is more important than performance most of the time. Brain time is more precious than CPU time. I can see that that would be less easy to skip over or mistype than "=". I can imagine times when being able to copy the (0,9] syntax in a spec into the code would make it clearer that the code limits match the spec. On the other hand when implementing and debugging the code, the <= syntax or your separate < and = make it clearer what the code is doing (when (0,9] is used infrequently). Stephen Lewkowicz (G1CMZ) User(s) browsing this thread: 1 Guest(s)	{"url":"https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=8263&pid=72544","timestamp":"2024-11-08T22:24:53Z","content_type":"application/xhtml+xml","content_length":"49601","record_id":"<urn:uuid:86ce4028-8e0b-4500-82b2-65b6e9934098>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00150.warc.gz"}
Sum Calculator \| CalculatorAndConverter.com Sum Calculator Use this sum calculator to find the sum of a set of numbers separated by commas, spaces, or line breaks. Enter a list of numbers separated by commas, spaces, and/or line breaks to find the sum. The calculator will automatically remove invalid values (such as letters, words, and symbols). Population Standard Deviation Sample Standard Deviation How to Use This Sum Calculator The above calculator will automatically calculate the sum of a number set. It will also find your data set's minimum, maximum, range, count, mean, median, mode, variance, and standard deviation. To use this calculator, first, delete the example number set. Next, enter your number set into the text area. The values in your number set should be separated by commas: or spaces: 1 5.8 -50.98 18926 7509.6506832 or line breaks: or a combination of commas, spaces, and line breaks: 1,5.8 -50.98 18926 20.2 The sum and other outputs will automatically be calculated as you enter values into the input area.	{"url":"https://calculatorandconverter.com/math/sum-calculator","timestamp":"2024-11-07T00:01:44Z","content_type":"text/html","content_length":"29758","record_id":"<urn:uuid:0e6d6f2c-a0fe-4fcf-91c4-13cdce6b4d00>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00554.warc.gz"}
Plan for the Unexpected! Whatever you teach (or learn), you should look for the links, the more unexpected, the better. This post is the story of one of them, simple but lovely and strong. I have a class with a group of adult students who never had a positive experience with mathematics (or as they would say, “they hate mathematics”), have very limited knowledge of mathematics, and yet, likely to work in nurseries and primary schools where they have to teach mathematics somehow or other. My aim is to help them not to hate mathematics (if not like it) and have some positive experience of doing mathematics. They refer to my aim as “great expectation” 🙂 But, last week something changed in them; something that made me so excited that the whole class burst into laughter. Previously in the class we had worked on the idea of subitizing and the ways we might help children to do so conceptually. We had learned how the “shape” or “structure” of a group of objects could help us (children) to realize “how many they are without counting” (that is the essence of subitizing). For example, children should learn to see the number of dots in the following figure is five without counting. Also, We had played with Mathlink Cubes to experience different shapes for numbers, and to discover similarities and differences between those shapes. In particular, we learned numbers 3, 6, and 10 can be represented by triangular-looking figures like the following. So far, it was just the story of what the class knew before facing with this question: In how many ways can we represent number two by using the fingers of one hand? Here are two of them: Using just one hand, there was no need to be that much systematics; somehow or the other we could find the answer. However, the problem became more difficult when we were allowed to use the fingers of both hands. Here, the unexpected link of the story appeared. It is amazing and worthy of being discovered if it is the first time that you are counting the twos of your fingers. Please try it before continuing reading. Inline Feedbacks View all comments	{"url":"https://amirasghari.com/plan-for-the-unexpected/","timestamp":"2024-11-03T15:06:55Z","content_type":"text/html","content_length":"105179","record_id":"<urn:uuid:c70f5bff-863e-4f48-9e0a-9ea84174132f>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00294.warc.gz"}
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Similar Triangles Examples and Problems with Solutions Definitions and theorems related to similar triangles are discussed using examples. Also examples and problems with detailed solutions are included. Review of Similar Triangles Two triangles ABC and A'B'C' are similar if the three angles of the first triangle are congruent to the corresponding three angles of the second triangle and the lengths of their corresponding sides are proportional as follows. AB / A'B' = BC / B'C' = CA / C'A' Angle-Angle (AA) Similarity Theorem If two angles in a triangle are congruent to the two corresponding angles in a second triangle, then the two triangles are similar. Example 1 Let ABC be a triangle and A'C' a segment parallel to AC. What can you say about triangles ABC and A'BC'? Explain your answer. Solution to Example 1 • Since A'C' is parallel to AC, angles BA'C' and BAC are congruent. Also angles BC'A' and BCA are congruent. Since the two triangles have two corresponding congruent angles, they are similar. Side-Side-Side (SSS) Similarity Theorem If the three sides of a triangle are proportional to the corresponding sides of a second triangle, then the triangles are similar. Example 2 Let the vertices of triangles ABC and PQR defined by the coordinates: A(-2,0), B(0,4), C(2,0), P(-1,1), Q(0,3), and R(1,1). Show that the two triangles are similar. Solution to Example 2 • Let us first plot the vertices and draw the triangles. • Since we know the coordinates of the vertices, we can find the length of the sides of the two triangles. AB = √ ( 4^ 2 + 2^ 2 ) = 2 √5 BC = √ ( (-4)^ 2 + 2^ 2 ) = 2 √5 CA = √ ( 4^ 2 ) = 4 PQ = √ ( 2^ 2 + 1^ 2 ) = √5 QR = √ ( (-2)^ 2 + 1^ 2 ) = √5 RP = √ ( 2^ 2 ) = 2 • We now calculate the ratios of the lengths of the corresponding sides. AB / PQ = 2 , BC / QR = 2 and CA / RP = 2 • We can now write. AB / PQ = BC / QR = CA / RP = 2 • The lengths of the corresponding sides are proportional and therefore the two triangles are similar. Side-Angle-Side (SAS) Similarity Theorem If an angle of a triangle is congruent to the corresponding angle of a second triangle, and the lengths of the two sides including the angle in one triangle are proportional to the lengths of the corresponding two sides in the second triangle, then the two triangles are similar. Example 3 Show that triangles ABC and A'BC', in the figure below, are similar. Solution to Example 3 • Angles ABC and A'BC' are congruent. • Since the lengths of the sides including the congruent angles are given, let us calculate the ratios of the lengths of the corresponding sides. BA / BA' = 10 / 4 = 5 / 2 BC / BC' = 5 / 2 • The two triangles have two sides whose lengths are proportional and a congruent angle included between the two sides. The two triangles are similar. Similar Triangles Problems with Solutions Problems 1 In the triangle ABC shown below, A'C' is parallel to AC. Find the length y of BC' and the length x of A'A. Solution to Problem 1 • BA is a transversal that intersects the two parallel lines A'C' and AC, hence the corresponding angles BA'C' and BAC are congruent. BC is also a transversal to the two parallel lines A'C' and AC and therefore angles BC'A' and BCA are congruent. These two triangles have two congruent angles are therefore similar and the lengths of their sides are proportional. Let us separate the two triangles as shown below. • We now use the proportionality of the lengths of the side to write equations that help in solving for x and y. (30 + x) / 30 = 22 / 14 = (y + 15) / y • An equation in x may be written as follows. (30 + x) / 30 = 22 / 14 • Solve the above for x. 420 + 14 x = 660 x = 17.1 (rounded to one decimal place). • An equation in y may be written as follows. 22 / 14 = (y + 15) / y • Solve the above for y to obtain. y = 26.25 Problems 2 A research team wishes to determine the altitude of a mountain as follows (see figure below): They use a light source at L, mounted on a structure of height 2 meters, to shine a beam of light through the top of a pole P' through the top of the mountain M'. The height of the pole is 20 meters. The distance between the altitude of the mountain and the pole is 1000 meters. The distance between the pole and the laser is 10 meters. We assume that the light source mount, the pole and the altitude of the mountain are in the same plane. Find the altitude h of the mountain. Solution to Problem 2 • We first draw a horizontal line LM. PP' and MM' are vertical to the ground and therefore parallel to each other. Since PP' and MM' are parallel, the triangles LPP' and LMM' are similar. Hence the proportionality of the sides gives: 1010 / 10 = (h - 2) / 18 • Solve for h to obtain h = 1820 meters. Problems 3 The two triangles are similar and the ratio of the lengths of their sides is equal to k: AB / A'B' = BC / B'C' = CA / C'A' = k. Find the ratio BH / B'H' of the lengths of the altitudes of the two Solution to Problem 3 • If the two triangles are similar, their corresponding angles are congruent. Hence angle BAH and B'A'H are congruent. We now examine the triangles BAH and B'A'H'. These triangles have two pairs of corresponding congruent angles: BAH and B'A'H' and the right triangles BHA and B'H'A'. The triangles are similar and therefore: AB / A'B' = BH / B'H' = k Problems 4 BA' and AB' are chords of a circle that intersect at C. Find a relationship between the lengths of segments AC, BC, B'C and A'C. Solution to Problem 4 • We first join points B and A and B' and A'. Angles ABA' and AB'A' in the the two triangles are congruent since they intercept the same arc. Angles BAB' and BA'B' also intercept the same arc and therefore congruent. The two triangles ABC and A'B'C have two corresponding congruent angles and are therefore similar. • Before we write the proportionality of the sides, we first separate the two triangles and identify the corresponding sides then write the proportionality of the lengths of the sides. AB / A'B' = BC / B'C = CA / CA' • Since we are looking for a relationship between the lengths of AC, BC, B'C and A'C, we therefore use the last equation and cross product it to obtain BC * CA' = B'C * CA Problems 5 ABC is a right triangle. AM is perpendicular from vertex A to the hypotenuse BC of the triangle. How many similar triangles are there? Solution to Problem 5 • Consider triangles ABC and MBA. They have two corresponding congruent angles: the right angle and angle B. They are similar. Also triangles ABC and MAC have two congruent angles: the right angle and angle C. Therefore there are three similar triangles: ABC, MBA and MAC. More References and Links to Geometry Problems Intercept Theorem and Problems with Solutions Geometry Tutorials, Problems and Interactive Applets Congruent Triangles Examples	{"url":"https://www.analyzemath.com/Geometry/similar-triangles-examples-and-problems-with-solutions.html","timestamp":"2024-11-06T01:17:07Z","content_type":"application/xhtml+xml","content_length":"310851","record_id":"<urn:uuid:8508fbae-3ad6-46c2-adb3-6497f524c83e>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00117.warc.gz"}
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commit a474c25255da412efda5316324b06d06878e4384 parent 6db63da9012406dfbe231b0499db834ffa1f3d5d Author: Agastya Chandrakant <acagastya@outlook.com> Date: Mon, 16 Apr 2018 21:47:39 +0530 M s4/fafl/report.md \| 13 +++++++++++-- 1 file changed, 11 insertions(+), 2 deletions(-) diff --git a/s4/fafl/report.md b/s4/fafl/report.md @@ -47,11 +47,20 @@ __Refer figure 2 for TM1 which acts as a transducer to find remainder and quotie #### For calculation of remainder and quotient -Since calculation of modulo is subtracion of `v` from `u` until `u` is strictly smaller than `v`, procedure to follow `u - v` is as follows: (__Assumption: INstruction pointer points to index 1__) +Since calculation of modulo is subtracion of `v` from `u` until `u` is strictly smaller than `v`, procedure to follow `u - v` is as follows: (__Assumption: Instruction pointer points to index 1__) +1. While current cell value is not `0`, move right. +2. If current cell value is `0`, go left. If it is `1`: + a. Go right. Go right. If it is `1`: + i. Go right until current cell is `B`. + ii. Move left until current cell is `1`. + iii. Mark it as `X`. + iv. Move left until current cell is `B`. + v. MOve right, make `1` as `B`. Goto step #1. #### To find if the number is prime -For an input `num`, the first step is to make a copy of $\lfloor\dfrac{num}{2}\rfloor$ start reading the tape from the beginning. (__Assumption: INstruction pointer points to index 0__) +For an input `num`, the first step is to make a copy of $\lfloor\dfrac{num}{2}\rfloor$ start reading the tape from the beginning. (__Assumption: Instruction pointer points to index 0__) 1. For every second `1` encountered, mark it as `D`. 2. Move right until `B` is found.	{"url":"https://git.hanabi.in/nie-ii-year/commit/a474c25255da412efda5316324b06d06878e4384.html","timestamp":"2024-11-12T23:46:22Z","content_type":"text/html","content_length":"4630","record_id":"<urn:uuid:13e94ac6-15c4-4868-b3ae-0c7f0668b29d>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00567.warc.gz"}
How can we classify a non linearly separable? Nonlinear functions can be used to separate instances that are not linearly separable. Kernel SVMs are still implicitly learning a linear separator in a higher dimensional space, but the separator is nonlinear in the original feature space. kNN would probably work well for classifying these instances. What is a non linearly separable problem? Nonlinearly separable classifications are most straightforwardly understood through contrast with linearly separable ones: if a classification is linearly separable, you can draw a line to separate the classes. Whereas you can easily separate the LS classes with a line, this task is not possible for the NLS problem. How do you deal with problems which are not linearly separable? In cases where data is not linearly separable, kernel trick can be applied, where data is transformed using some nonlinear function so the resulting transformed points become linearly separable. A simple example is shown below where the objective is to classify red and blue points into different classes. What if the data is not linearly separable? There’s no well-defined relationship such as, “a linear classifier only works on linearly separable data” or “data that is not linearly separable can only be classified using a non-linear classifier.” Linearly separable data is data that if graphed in two dimensions, can be separated by a straight line. Which classifier helps in non linear classification? As mentioned above SVM is a linear classifier which learns an (n – 1)-dimensional classifier for classification of data into two classes. However, it can be used for classifying a non-linear dataset. This can be done by projecting the dataset into a higher dimension in which it is linearly separable! How can we classify non linear data using SVM? Nonlinear classification: SVM can be extended to solve nonlinear classification tasks when the set of samples cannot be separated linearly. By applying kernel functions, the samples are mapped onto a high-dimensional feature space, in which the linear classification is possible. What is non-separable data? If your data is non-separable, there is no way to separate them. Given the data, the classes are the same. What is linear and non linear separability? When we can easily separate data with hyperplane by drawing a straight line is Linear SVM. When we cannot separate data with a straight line we use Non – Linear SVM. How does SVM deal with non-separable data? To sum up, SVM in the linear nonseparable cases: By combining the soft margin (tolerance of misclassifications) and kernel trick together, Support Vector Machine is able to structure the decision boundary for linear non-separable cases. How do you solve non-linear SVM? How do we handle non linearly separable data in SVM? Is SVM linear or non-linear? SVM or Support Vector Machine is a linear model for classification and regression problems. It can solve linear and non-linear problems and work well for many practical problems. The idea of SVM is simple: The algorithm creates a line or a hyperplane which separates the data into classes. What is a non-linearly separable classification? Nonlinearly separable classifications are most straightforwardly understood through contrast with linearly separable ones: if a classification is linearly separable, you can draw a line to separate the classes. Below is an example of each. Can SVM be used for non-linear classification? As mentioned above SVM is a linear classifier which learns an (n – 1)-dimensional classifier for classification of data into two classes. However, it can be used for classifying a non-linear dataset. This can be done by projecting the dataset into a higher dimension in which it is linearly separable! Can we use SVM to separate a non-linearly separable dataset? Now, we can use SVM (or, for that matter, any other linear classifier) to learn a 2-dimensional separating hyperplane. This is how the hyperplane would look like: Thus, using a linear classifier we can separate a non-linearly separable dataset. How many hyperplanes are there that separate two linearly separable classes? Figure 14.8:There are an infinite number of hyperplanes that separate two linearly separable classes. In two dimensions, a linear classifier is a line. Five examples are shown in Figure 14.8.	{"url":"https://www.joialife.com/uncategorized/how-can-we-classify-a-non-linearly-separable/","timestamp":"2024-11-14T14:31:53Z","content_type":"text/html","content_length":"43036","record_id":"<urn:uuid:780bb396-9abd-4af2-ac4f-79a1bb5e3c50>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00284.warc.gz"}
Help Please! If two lines l and m have equations -x+6, and y = -4x+6, what is the probability that a point randomly selected in the 1st quadrant and below l will fall between l and m? Express your answer as a decimal to the nearest hundredth. supremecheetah May 30, 2023 The lines l and m intersect at the point (3, 0). The area of the triangle formed by the lines l, m, and the x-axis is (1/2)(3)(6) = 9. The area of the rectangle formed by the lines l, m, and the y-axis is (3)(6) = 18. The probability that a point randomly selected in the 1st quadrant and below l will fall between l and m is the ratio of the area of the triangle to the area of the rectangle, or 9/18 = 1/2. To the nearest hundredth, this is 0.50. Guest May 31, 2023	{"url":"https://web2.0calc.com/questions/help-please_85946","timestamp":"2024-11-06T18:48:21Z","content_type":"text/html","content_length":"20616","record_id":"<urn:uuid:305ac566-7316-4c86-a28d-49caaf8fd6b2>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00844.warc.gz"}
2,000 research outputs found The thermodynamic stabilities of various phases of the nitrides of the platinum metal elements are systematically studied using density functional theory. It is shown that for the nitrides of Rh, Pd, Ir and Pt two new crystal structures, in which the metal ions occupy simple tetragonal lattice sites, have lower formation enthalpies at ambient conditions than any previously proposed structures. The region of stability with respect to those structures extends to 17 GPa for PtN2. Calculations show that the PtN2 simple tetragonal structures at this pressure are thermodynamically stable also with respect to phase separation. The fact that the local density and generalized gradient approximations predict different values of the absolute formation enthalpies as well different relative stabilities between simple tetragonal and the pyrite or marcasite structures are further discussed.Comment: 5 pages, 4 figure We prove that the flux of gravitational radiation from an isolated source in the Nonsymmetric Gravitational Theory is identical to that found in Einstein's General Theory of Relativity.Comment: 10 Solvent loss due to evaporation in a drying drop can drive capillary flows and solute migration. The flow is controlled by the evaporation profile and the geometry of the drop. We predict the flow and solute migration near a sharp corner of the perimeter under the conditions of uniform evaporation. This extends the study of Ref. 6, which considered a singular evaporation profile, characteristic of a dry surrounding surface. We find the rate of the deposit growth along contact lines in early and intermediate time regimes. Compared to the dry-surface evaporation profile of Ref. 6, uniform evaporation yields more singular deposition in the early time regime, and nearly uniform deposition profile is obtained for a wide range of opening angles in the intermediate time regime. Uniform evaporation also shows a more pronounced contrast between acute opening angles and obtuse opening angles.Comment: 12 figures, submitted to Physical Review The Copernican principle, stating that we do not occupy any special place in our universe, is usually taken for granted in modern cosmology. However recent observational data of supernova indicate that we may live in the under-dense center of our universe, which makes the Copernican principle challenged. It thus becomes urgent and important to test the Copernican principle via cosmological observations. Taking into account that unlike the cosmic photons, the cosmic neutrinos of different energies come from the different places to us along the different worldlines, we here propose cosmic neutrino background as a test of the Copernican principle. It is shown that from the theoretical perspective cosmic neutrino background can allow one to determine whether the Copernican principle is valid or not, but to implement such an observation the larger neutrino detectors are called for.Comment: JHEP style, 10 pages, 4 figures, version to appear in JCA The massive nonsymmetric gravitational theory is shown to posses a linearisation instability at purely GR field configurations, disallowing the use of the linear approximation in these situations. It is also shown that arbitrarily small antisymmetric sector Cauchy data leads to singular evolution unless an ad hoc condition is imposed on the initial data hypersurface.Comment: 14 pages, IOP style for submission to CQG. Minor changes and additional background material adde We extend the Kolmogorov phenomenology for the scaling of energy spectra in high-Reynolds number turbulence, to explicitly include the effect of helicity. There exists a time-scale $\tau_H$ for helicity transfer in homogeneous, isotropic turbulence with helicity. We arrive at this timescale using the phenomenological arguments used by Kraichnan to derive the timescale $\tau_E$ for energy transfer (J. Fluid Mech. {\bf 47}, 525--535 (1971)). We show that in general $\tau_H$ may not be neglected compared to $\tau_E$, even for rather low relative helicity. We then deduce an inertial range joint cascade of energy and helicity in which the dynamics are dominated by $\tau_E$ in the low wavenumbers with both energy and helicity spectra scaling as $k^{-5/3}$; and by $\tau_H$ at larger wavenumbers with spectra scaling as $k^{-4/3}$. We demonstrate how, within this phenomenology, the commonly observed ``bottleneck'' in the energy spectrum might be explained. We derive a wavenumber $k_h$ which is less than the Kolmogorov dissipation wavenumber, at which both energy and helicity cascades terminate due to dissipation effects. Data from direct numerical simulations are used to check our predictions.Comment: 14 pages, 5 figures, accepted to Physical Review Helical turbulence is thought to provide the key to the generation of large-scale magnetic fields. Turbulence also generically leads to rapidly growing small-scale magnetic fields correlated on the turbulence scales. These two processes are usually studied separately. We give here a unified treatment of both processes, in the case of random fields, incorporating also a simple model non-linear drift. In the process we uncover an interesting plausible saturated state of the small-scale dynamo and a novel analogy between quantum mechanical (QM) tunneling and the generation of large scale fields. The steady state problem of the combined small/large scale dynamo, is mapped to a zero-energy, QM potential problem; but a potential which, for non-zero mean helicity, allows tunneling of bound states. A field generated by the small-scale dynamo, can 'tunnel' to produce large-scale correlations, which in steady state, correspond to a force-free 'mean' field.Comment: 4 pages, 1 figure, Physical Review Letters, in pres We define the flatness and quasi-flatness problems in cosmological models. We seek solutions to both problems in homogeneous and isotropic Brans-Dicke cosmologies with varying speed of light. We formulate this theory and find perturbative, non-perturbative, and asymptotic solutions using both numerical and analytical methods. For a particular range of variations of the speed of light the flatness problem can be solved. Under other conditions there exists a late-time attractor with a constant value of \Omega that is smaller than, but of order, unity. Thus these theories may solve the quasi-flatness problem, a considerably more challenging problem than the flatness problem. We also discuss the related \Lambda and quasi-\Lambda problem in these theories. We conclude with an appraisal of the difficulties these theories may face.Comment: 21 pages, 6 figure We introduce and begin the study of new knot energies defined on knot diagrams. Physically, they model the internal energy of thin metallic solid tori squeezed between two parallel planes. Thus the knots considered can perform the second and third Reidemeister moves, but not the first one. The energy functionals considered are the sum of two terms, the uniformization term (which tends to make the curvature of the knot uniform) and the resistance term (which, in particular, forbids crossing changes). We define an infinite family of uniformization functionals, depending on an arbitrary smooth function $f$ and study the simplest nontrivial case $f(x)=x^2$, obtaining neat normal forms (corresponding to minima of the functional) by making use of the Gauss representation of immersed curves, of the phase space of the pendulum, and of elliptic functions Lagrangian chaos is experimentally investigated in a convective flow by means of Particle Tracking Velocimetry. The Fnite Size Lyapunov Exponent analysis is applied to quantify dispersion properties at different scales. In the range of parameters of the experiment, Lagrangian motion is found to be chaotic. Moreover, the Lyapunov depends on the Rayleigh number as ${\cal R}a^{1/2}$. A simple dimensional argument for explaining the observed power law scaling is proposed.Comment: 7 pages, 3 figur	{"url":"https://core.ac.uk/search/?q=author%3A(H.%20K.%20Moffat)","timestamp":"2024-11-07T07:32:40Z","content_type":"text/html","content_length":"179677","record_id":"<urn:uuid:2a2793cc-8f9b-4922-9e35-60e076141ebf>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00217.warc.gz"}
Learn Function Challenge 6 – The Good Parts of JavaScript and the Web Check out a free preview of the full The Good Parts of JavaScript and the Web course The "Function Challenge 6" Lesson is part of the full, The Good Parts of JavaScript and the Web course featured in this preview video. Here's what you'd learn in this lesson: In this challenge, you will write the gensymf(), gensymff(), and fibonaccif() functions. Transcript from the "Function Challenge 6" Lesson >> [MUSIC] >> Douglas: Make a function gensymf that makes a function that generates symbols. So we're taking the generator idea and we're gonna try to do something practical with it now. So gensymf is a symbol generator or gensymf is a factory that makes symbol generators, or things that make serial numbers. So we designate a serial number with a prefix, and so you can send it any string and that becomes the prefix string, and then we will get a series of strings starting with that symbol. So we're gonna make two generators this time from the same factory. We're gonna make the G series and the H series and when we call them, we'll get G1, H1, G2, H2. >> Douglas: Let's look at gensymf, >> Douglas: Gensymf takes a prefix string, it creates a number which it's gonna use for keeping track of where it is in the sequence. It will return a function which will add one to that number, and return the result of concatenating the number to the prefix. So who got something that works? Good. Anyone do anything substantially different? Yeah. >> Speaker 2: And you cast in, then it's prefix to the string, can you pass it in digit. Is that first thing? >> Douglas: That's a wise precaution, yeah. Anybody else? >> Speaker 2: I used find function. >> Douglas: Use a find function that's going to. So this is an example, we've seen a number of these where we've got a factory function which then makes something which will do some work, to get usually a generator, but it could be there are lots of different kinds of functions. But they're both functions, it's just one is nested in the other. And in fact if we nest further, if we put another function outside of this, we could make a factory, factory. And you can wrap that with a factory, factory, factory. So just for fun, let's look at what this would look like if it were a factory, factory. >> Douglas: So gensymff is the factory, factory and we're gonna pass the increment function and the initial seed value to gensymff in a produce of produces a function that works exactly like gensymf. And it'll make the sequences right, and we've done things like this a couple of times already. So I'm just gonna show you the thing cuz we've already done this one. So it looks very similar to patterns we've seen before, where the factory, factory is supplying values that go into the generator. And now, so we can automate the making of factories. Now, the interesting thing about this one is that statement there where we're creating the number variable which is gonna hold the value that is being used to generate the sequence numbers. So if we were to move that up one line, so that it's not in the factory anymore, but it's in the factory, factory. That would change the visibility of that variable so it would be seen by all of the generators. So instead of generating G1, H1, G2, H2, we would generate G1, H2, G3, H4. That's a really interesting change in behavior. It's just moving one variable declaration one place to another. So we've been dealing with closure and we can we saw that we can have things that are global and things that are local, and things that are sort of in between, but there can be more of those in betweens. And if we get into nesting things in useful ways, we have tremendous control over the visibility and the lifetime of the variables, and can do interesting things to affect their behavior. How about that, any questions about that? Ready for another one or do you want to take a break? Let's do another one, sure why not. So everybody remembers fibonacci, right? You did fibonacci numbers in school. So fibonacci was an important mathematician. He discovered a lot of good stuff, but the only thing we seem to remember is the fibonacci sequence. In their infinite number of Fibonacci Sequences, but mostly we only remember the famous one, which started with 0 and 1, and that's what we're gonna be doing now. So we're gonna make a factory which will make a Fibonacci generator. And you'll seed the factory with the first two numbers in the sequence. So everybody remember how Fibonacci numbers work? Well let's review. So the Fibonacci sequence will be a sequence of integers, you specify the first two integers in the sequence, the third number will be the sum of the first two. The fourth number will be the sum of the previous two and so on. So the first numbers we get are 0 and 1, because those are the ones we provide. Then the next 1, will be 1 because that's the sum of 0 and 1, the next one will be 2, because it's the sum of 1 and 1. The next will be 3, cuz it's a sum of 1 and 2, next be 5 because it's the sum of 2 and 3, and the next in the series will be? 8, exactly. So this was a tricky one, wasn't it? The Fibonacci sequence itself is totally trivial. It's three, it's simple statements, but getting the first two numbers to come out but that was the trick. So first off you got something that works? Like graduations this one was hard, so let’s look at a number of approaches that we could take. So here's one, there is the Fibonacci function there in the box. So that's it, and then we've got an if statement around it versus which statement which asks, where are we in the sequence? At the first at the beginning of sequence put out the first number, otherwise put out the second number, otherwise use the Fibonacci function and do that. And this works, it absolutely works, who took this approach or something like it maybe used and if instead, but basically. Yeah, you've got a variable which is telling you where you are in the sequence. This is a completely, yeah? >> Speaker 3: Except for case one, you're inputting a and b. A and b don't necessarily equal 0 and 1, if a is something other than 0 for case one. >> Douglas: Right. >> Speaker 3: You need to return a+ b, don't you? >> Douglas: No. >> Speaker 3: You don't know the rule. >> Speaker 2: You're switching on i. >> Douglas: I got i which is telling me where I am in the sequence. >> Speaker 3: I know. >> Douglas: Which isn't the value I'm at the sequence just how many numbers have I looked at. >> Speaker 3: I know. >> Douglas: So if I'm looking at the first number, I output a, and if I look at the second number i is 1, I output b. Otherwise I output a + b. >> Speaker 3: So if you want the Fibonacci, if your starting elements are 5 and 7, the first number's 5, second number's 7, third number's 12. >> Douglas: Right. >> Speaker 3: Okay. >> Douglas: Yeah, that's how they work. >> Speaker 3: Yeah, I was thinking that you added the first. First, you took the second then you added the first two. Then you, no, it's not until you get to the- >> Douglas: It's the first two numbers, then you add. >> Speaker 3: Right, because they tell you, got it. >> Douglas: Yeah. >> Speaker 3: Is y incremented or not? >> Douglas: Yeah, well instead of adding 1, I set it to 1 because I know it's 0. And beyond that I don't need to increment it, once I get past the first two cases, I don't care what i is anymore. >> Speaker 3: Okay. >> Douglas: Okay, so completely acceptable this is a reasonable way to do it. I would argue that a reasonably intelligent person could figure out what this code is doing and that's most of what we want code to do so. This is good, it's okay. My complaint with it is, it's a fairly big function and only that much of it is concerned with the Fibonacci things, so it just feels kind of loopsided to me. So here's another approach. In this one, I kind of permuted the statements of the Fibonacci sequence in order to delay the output of the first one, or to cause the first two numbers to get output. So who did something like this? Yeah, so this is probably the most optimal solution. It's gonna be the smallest code, the fastest performance not that either of those matter in real life, but it does have that advantage. The disadvantage of this is I'd hate to be the guy that has to debug it, right? >> Speaker 4: I said b = a + next instead. >> Douglas: Yeah, there are variations on it, but it's basically the same idea. So another approach we could take is recognizing that we're making a generator and we already have some tools for constructing generators. So here's another approach, I've got my Fibonacci generator here, which will give me the next number, I just need to get the first two on top of it. So I'm going to make a special generato, and I'm gonna start by taking identityf which was the first interesting function that we wrote this morning, that you thought had no practical application. It turns out what identifyf is is a constant generator, it will always produce the same value. So I'm gonna use that to make generators and then I'm gonna use the limit function that we wrote earlier to cut off the sequence, so I only get one. So I got a pair of things and I can then concatenate those two together and then concatenate that on to the Fibonacci function. So who did that? [LAUGH] Of course nobody would do that. >> Speaker 4: [LAUGH] >> Douglas: And if you're, yeah, so there's that. And if I were gonna be doing this a lot, I would take limit identityf, and encapsulate that into something which would make sequences of one more compactly or we could do this. A similar thing, we take the element function that we wrote earlier make an array containing the first two things and concatenate that onto the Fibonacci generator. So who did that? No, no. >> [INAUDIBLE] >> Douglas: Very, very good, brilliant. Learn Straight from the Experts Who Shape the Modern Web • In-depth Courses • Industry Leading Experts • Learning Paths • Live Interactive Workshops Get Unlimited Access Now	{"url":"https://frontendmasters.com/courses/good-parts-javascript-web/function-challenge-6/","timestamp":"2024-11-12T23:06:25Z","content_type":"text/html","content_length":"36167","record_id":"<urn:uuid:06252731-8026-4ad8-ac8b-e1225ee13f0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00886.warc.gz"}
Index Selection Sorting Algorithm Volume 01, Issue 05 (July 2012) Index Selection Sorting Algorithm DOI : 10.17577/IJERTV1IS5246 Download Full-Text PDF Cite this Publication Vishweshwarayya C Hallur, Ramesh K, Basavaraj A Goudannavar, 2012, Index Selection Sorting Algorithm, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 01, Issue 05 (July • Open Access • Total Downloads : 592 • Authors : Vishweshwarayya C Hallur, Ramesh K, Basavaraj A Goudannavar • Paper ID : IJERTV1IS5246 • Volume & Issue : Volume 01, Issue 05 (July 2012) • Published (First Online): 02-08-2012 • ISSN (Online) : 2278-0181 • Publisher Name : IJERT • License: This work is licensed under a Creative Commons Attribution 4.0 International License Text Only Version Index Selection Sorting Algorithm Vishweshwarayya C Hallur1 Angadi Institute of Technology & Management, Belgaum Ramesh K2 Assistant Professor Department Computer Science Karnatak University, Dharwad-580003 Basavaraj A Goudannavar3 Department of P.G. Studies and Research in Computer Science Karnatak University, Dharwad-580003 One of the most frequent operations performed on database is searching. To perform this operation we have different kinds of searching techniques. These all searching algorithms work only on data, which are previously sorted. An efficient algorithm is required to make searching algorithm fast and efficient. This paper presents a new sorting algorithm named as Index Selection Sorting Algorithm (ISSA). This ISSA is designed to perform sorting quickly and easily and also efficient as existing algorithms in sorting. Key Words: Algorithm, Sorting, ISSA, Worst Case, Average Case, and Best Case. 1. Introduction Using a computer to solve problem involves directing it on what step it must follow to get the problem to be solved. The step it must follow is called an algorithm. The common sorting algorithm can be divided into two classes by the difficulty of their algorithms. There is a direct correlation between the complexity and effectiveness of an algorithm [1]. The complexity of an algorithm generally written in the form of Big O(n) notation, where O represents the complexity of the algorithm and value n represents the size of the list. The two groups of sorting algorithm are O(n2), which includes bubble sort, insertion sort, selection sort, and shell sort. And O(n log(n)) which includes the heap sort, merge sort and quick sort[2]. Since the drastic advancement in computing, most of the research is done to solve the sorting problem, perhaps due to the complexity of solving it efficiently dispite its simple and familiar It is always very difficult to say that one sorting technique is better than another. Performance of the various sorting algorithms depends upon the data being sorted. Sorting is used in most of the applications and there have been plenty of performance analyses [3][4]. There has been growing interest on enhancements to sorting algorithms that do not have an effect on their asymptotic complexity but rather tend to improve performance by enhancing data locality 2. Proposed System In ISSA technique the first number will be compared with all the elements in the list, at the end of each pass selection of proper index of new list is done and then element is copied it to that position in the new list. And this step will be repeated for n number of times. The best case time complexity of ISSA is Omega (n2), the average case of the ISSA is theta (n2) and worst case of ISSA is Big O(n2). Diagrammatic representation of ISSA: Unsorted List with size a[5] a[0] a[1] a[2] a[3] a[4] New List i.e. b[5] a[0] a[1] a[2] a[3] a[4] After the last pass, the index of 232 is calculated and then 232 is copied into that position in the new list. Therefore index of 232 is 2. Hence, a[0] a[1] a[2] a[3] a[4] 3. Algorithm Algorithm ISSA(a,b,n) for i 0 to n-1 k 0 item a[i] for j 0 to n-1 if (item > a[j]) then increment k After the first pass, the index of 345 is calculated and then 345 is copied in to that position in the new list. Therefore index of 345 is 3. Hence, a[0] a[1] a[2] a[3] a[4] After the second pass, the index of the 565 is calculated and then 565 is copied into that position in the new list. Therefore index of 565 is 4. Hence, a[0] a[1] a[2] a[3] a[4] b[k] item 4. Comparisons of ISSA with other sorting technique Below are the tables representing the calculated running time for n values and their graphs in various cases. Best Case Insertion Quick Shell 8 8 7.22 6.52 64 16 16 19.26 23.19 256 32 32 48.16 72.49 1024 64 64 115.59 208.78 4096 128 128 269.72 568.36 16384 1. Best Case: After the third pass, the index of 49 is calculated and then 49 is copied into that position in new list. Therefore index of 49 is 1. Hence, a[0] a[1] a[2] a[3] a[4] After the fourth pass, the index of 23 is calculated and then 23 is copied into that position in the new list. Therefore index of 23 is 0. Hence, a[0] a[1] a[2] a[3] a[4] Running Time No. of elements Insertion Quick Shell 2. Average Case: Avarage Case Insertion Quick Shell 8 64 7.22 13.45 64 16 256 19.26 32 256 32 1024 48.16 76.10 1024 64 4096 115.59 181.01 4096 128 16384 269.72 430.53 16384 No. of elements Running Time 3. Worst Case: Worst Case Insertion Quick Shell Index Selection 8 64 64 22.62 64 32 1024 1024 181.01 1024 p>128 16384 16384 1448.1 16384 No. of elements Running Time From the above graphs one can easily observe that with all the cases it takes same time and in worst case it takes same time like other sorting algorithms except quick sort technique. 5. Conclusion Logic of ISSA is based on the logic of selection sort and insertion sort. In those techniques either the smallest or largest elements are taken and then placed them in appropriate position but in this ISSA first element, second element, third element and so on from the unsorted list is taken and then it is placed in its appropriate position in new list. 6. References [1]. Hore, C.A.R. Algorithm 64: Quick sort. Comm.ACM 4,7(July 1961), 321. [2]. Soubhik Chakraborty, Mousami Bose, and Kumar Sushant, A research thesis, On way Parameters of input Distributions Need be Taken Into Account For a more precise Evalution of complexity for certain Algorithms. [3]. D.S.Malik, C++ Programming: Program Design including Data Structures, Course Technology(Thomson Learning), 2002, www.course.com [4]. J.L>Bentley and R.Sedgewick. Fast Algorithms for Sorting and Searching stings, ACM-SIAM SODA, 97,360-369, 1997. [5]. D.Jim enez-Gonzalez, J. Navarvo, and Larriba- Pay. CC-Radix: A catch conscious sorting based on Radix sort. In Euromicro Conference on Parallel Distributed and Network based Processing. Pages 101-108, February 2003. You must be logged in to post a comment.	{"url":"https://www.ijert.org/index-selection-sorting-algorithm","timestamp":"2024-11-04T17:29:49Z","content_type":"text/html","content_length":"71250","record_id":"<urn:uuid:1b673ad7-c01f-436a-be6c-222fd32cf10d>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00680.warc.gz"}
Spiric Section: Definition, Examples Spiric sections can take on many different shapes. A spiric section (also called the spiric of Perseus), is a quartic plane curve defined by the equation (x^2 + y^2)^2 = dx^2 + ey^2 + f. As polar coordinates: (r^2 – a^2 + b^2 + c^2)^2 = 4b^2(r^2 cos^2 θ + c^2) r^4 = dr^2 cos^2θ + er^2sin^2θ + f. They can also be defined as bicircular curves that are symmetric to both the x-axes and y-axes. Alternatively, as the curve that results from the intersection of a torus and a plane, parallel to its rotational symmetry axis. However, this particular definition doesn’t specifically include the curves produced by imaginary planes [1]. A spiric section that has a plane at distance r from the axis is called an oval of Cassini [2]. Spiric sections are a member of the family of toric sections, and they can take on a wide variety of different shapes, including being interlaced like a horse’s hobble or fetter, broad in the middle and thin at the sides, or elongated with a narrow middle portion and broad ends [3]. Named members include Bernoulli’s lemniscate (which was the key to unlocking the secretes of elliptic integrals), the Cassini ovals, the Hippopedes of Proclus, and Villarceau’s circles. Cassini’s ovals are spiric sections (which are hippopedes) where the distance of the cutting plane to the torus axis is equal to the generating circle’s radius; Bernoulli’s lemniscate is a special case of a Cassini’s oval, generated when R = 2r [4]. Spiric Section History Bicircular curves were known to the ancient Greeks long before the spiric sections were studied. One of them is the Conchoid of Nicomedes, invented before 200 BC [7]. However, The earliest source of this type of curve is attributed to Eudoxus of Cnidus (40 – 355 BC), whose work on planetary movement centers on the hippopede, which isn’t a plane curve but a curve on a sphere. Menaechmus (380 to 320 BC) constructed conic sections, by cutting a cone with a plane. Two centuries later, Perseus (2nd BC) studied the spiric section by cutting a torus with a plane (giving it the name “spiric of Perseus”) [2], parallel to the line through the center of the torus’ hole [1] (a hole is a topological structure that prevents a mathematical object from being continuously shrunk to a point) [5]. Although Perseus’s work is lost to history, and practically nothing in known about him, we do know of his research through obscure commentaries [6]. For example, Proclus (411 to 485) wrote that “a mathematician known as Perseus considered the intersection of a torus and a plane which is parallel to the equatorial plane of the torus” [7]. The spiric section was rediscovered in the 17th century, when mathematicians, including Cassini) were studying quartic curves. Interestingly, Cassini (1625– 1712) was a distinguished astronomer but opposed Newton’s theory, rejected Kepler’s ellipses and instead proposed his own Cassini ovals as models for planetary orbits [8]. Use of Spiric Sections Spiric sections have some important uses, beyond their study in geometry. For example, in quantum mechanics, the Double Quantum Dot (DQD) structures potential cross-section can be approximated by a spiric section [9]. Top spiric section: Ag2gaeh, CC BY-SA 4.0, via Wikimedia Commons Krishnavedala, CC BY-SA 3.0, via Wikimedia Commons [1] Dayanithi, C. Combination of Cubic and Quartic Plane Curve. IOSR Journal of Mathematics (IOSR-JM). e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 6, Issue 2 (Mar. – Apr. 2013), PP 43-53 [2] Coffman, A. The Hippopede of Proclus. [3] Proclus. (2020). A Commentary on the First Book of Euclid’s Elements. Princeton University Press. [4] Marconi, L. The toric sections: a simple introduction. Retrieved January 20, 2022 from: https://arxiv.org/pdf/1708.00803.pdf [5] Weisstein, E. (1999). Hole. Retrieved January 20, 2022 from: https://archive.lib.msu.edu/crcmath/math/math/h/h318.htm [6] Horadam, F. (2014). Outline Course of Pure Mathematics. Elsevier Science. [7] Werner, T. (2011). Dissertation. Rational families of circles and bicircular quartics. Der Naturwissenschaftlichen Fakult at der Friedrich-Alexander-Universit at Erlangen-Nurnberg zur Erlangung des Doktorgrades Dr. rer. nat. [8] Stillwell, J. Mathematics and Its History. 3rd Edition. Springer. [9] Foundations of Quantum Mechanics In The Light Of New Technology: Isqm-tokyo ’08 – Proceedings Of The 9th International Symposium. 2009. World Scientific Publishing Company. Comments? Need to post a correction? Please Contact Us.	{"url":"https://www.statisticshowto.com/spiric-section-definition-examples/","timestamp":"2024-11-04T15:47:22Z","content_type":"text/html","content_length":"70463","record_id":"<urn:uuid:aab2a347-2425-49ef-852b-10a1b976911c>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00822.warc.gz"}
SNAKESKIN: The Poetry Webzine A Calculus To Miriam, on her graduation from college. What is the circumference of change? Measure the finite difference between a blade of grass and a spring violet. What is the foundation of family? How many shovels does it take to turn a garden bed four feet by twelve feet in the time it takes a robin to fly from the cottonwood to the front porch? What is the volume of two souls when the wind is blowing? Factor in the sound of soup simmering, the smell of bread baking, a map spread out. Define the differential of green beans and snap peas; Calculate the square root of a maple leaf divided by the time it takes to turn red and fall. What is the hypotenuse of the distance between a mother and her grown child when snow laps the front door with longing? If a young woman were to drive from Minneapolis to California at sixty-five miles an hour would her mother miss her less or more? Multiply the infinite quantity of love times mileage. How many breaths does it take to let her go? Calculus they say is the study of change, I never took it. If I had, maybe I’d be better prepared to deal with your leaving now. Sandra Lindow If you have any comments on this poem, Sandra Lindow would be pleased to hear from you.	{"url":"http://snakeskinpoetry.co.uk/236acalc.html","timestamp":"2024-11-09T22:34:09Z","content_type":"text/html","content_length":"3300","record_id":"<urn:uuid:d2bf5274-b108-475c-99bf-da889204130b>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00644.warc.gz"}
In the previous post, I showed how to plot “envelopes of epicycloids.” This post will consider a variation on the same theme, hypocycloids. For the epicycloid post, we imagined two ants crawling around a circle at different speeds, and drawing lines between their positions at various times. Although the ants were traveling at different speeds, they were both moving in the same orientation. For hypocycloids, we imagine our two ants again, but this time they’re traveling in opposite directions, one clockwise and one counterclockwise. The position of the two ants at time t are (cos pt, sin pt) (cos qt, sin qt) as before, and again p and q are integers, but this time q is negative and p > \|q\| > 0 > q. Also, we need to make one more change: the tangent lines have to be longer. If we just draw lines that begin and end on the circle, we’ll get a dense mesh of lines but not see the hypocycloid. The hypocycloid is on the outside of the unit circle, not inside. A plot further down will make this clear. The image at the top of the post corresponds to p = 5 and q = -2. Here are a couple more examples. First, p = 4 and q = -3. Next, p = 7 and q = -3. The dark area in the middle is what you’d see if you only connected points on the circle rather than extending the lines. More on that below. The equation of the hypocycloid itself, the figure carved out by the tangent lines, is Here’s a close-up of the figure at the top of the post with more detail: axes, the hypocycloid drawn in gray, and the unit circle drawn with a dashed black line. The Python code to produce the graphs in this post is similar to the code in the previous post. One thing to note: it’s important that the alpha level is set to a small value. The blue lines in this post use alpha = 0.2. This makes the lines translucent, and so places where many lines overlap are darker than isolated lines. With the default value of alpha = 1, all lines would be opaque and it would be harder to see what’s going on.	{"url":"https://www.johndcook.com/blog/2020/04/25/hypocycloids/","timestamp":"2024-11-02T12:09:45Z","content_type":"text/html","content_length":"50544","record_id":"<urn:uuid:de200d65-2852-4987-a0c3-cf2a4792fea8>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00410.warc.gz"}
Y Intercept - Meaning, Examples \| Y Intercept Formula - [[company name]] [[target location]], [[stateabr]] Y-Intercept - Explanation, Examples As a learner, you are constantly working to keep up in school to avoid getting engulfed by topics. As parents, you are continually searching for ways how to motivate your children to succeed in school and beyond. It’s particularly important to keep the pace in mathematics due to the fact that the concepts always founded on themselves. If you don’t understand a specific topic, it may plague you in next lessons. Comprehending y-intercepts is a perfect example of something that you will work on in mathematics time and time again Let’s look at the foundation ideas about y-intercept and let us take you through some tips and tricks for solving it. If you're a mathematical whiz or beginner, this small summary will provide you with all the knowledge and instruments you must possess to get into linear equations. Let's dive right in! What Is the Y-intercept? To completely comprehend the y-intercept, let's imagine a coordinate plane. In a coordinate plane, two straight lines intersect at a point called the origin. This point is where the x-axis and y-axis meet. This means that the y value is 0, and the x value is 0. The coordinates are noted like this: (0,0). The x-axis is the horizontal line passing across, and the y-axis is the vertical line going up and down. Every axis is counted so that we can locate points along the axis. The numbers on the x-axis rise as we drive to the right of the origin, and the values on the y-axis increase as we move up along the origin. Now that we have revised the coordinate plane, we can determine the y-intercept. Meaning of the Y-Intercept The y-intercept can be taken into account as the starting point in a linear equation. It is the y-coordinate at which the graph of that equation overlaps the y-axis. Simply put, it represents the value that y takes once x equals zero. Further ahead, we will illustrate a real-life example. Example of the Y-Intercept Let's assume you are driving on a long stretch of road with a single path going in respective direction. If you begin at point 0, where you are sitting in your car right now, then your y-intercept would be equal to 0 – considering you haven't moved yet! As you start you are going the track and picking up momentum, your y-intercept will increase before it archives some higher number when you reach at a designated location or halt to induce a turn. Thus, when the y-intercept may not look particularly applicable at first glance, it can offer details into how things transform over a period of time and space as we travel through our world. Therefore,— if you're at any time stuck attempting to comprehend this theory, keep in mind that just about everything starts somewhere—even your trip through that straight road! How to Discover the y-intercept of a Line Let's consider regarding how we can discover this number. To help with the method, we will make a synopsis of few steps to do so. Then, we will give you some examples to illustrate the process. Steps to Find the y-intercept The steps to locate a line that goes through the y-axis are as follows: 1. Search for the equation of the line in slope-intercept form (We will dive into details on this later in this tutorial), that should appear something like this: y = mx + b 2. Put 0 as the value of x 3. Solve for y Now once we have gone over the steps, let's see how this procedure will function with an example equation. Example 1 Find the y-intercept of the line explained by the formula: y = 2x + 3 In this example, we can replace in 0 for x and solve for y to find that the y-intercept is the value 3. Consequently, we can say that the line crosses the y-axis at the coordinates (0,3). Example 2 As one more example, let's assume the equation y = -5x + 2. In this case, if we place in 0 for x once again and figure out y, we find that the y-intercept is equal to 2. Therefore, the line crosses the y-axis at the point (0,2). What Is the Slope-Intercept Form? The slope-intercept form is a method of depicting linear equations. It is the most popular kind used to depict a straight line in mathematical and scientific uses. The slope-intercept equation of a line is y = mx + b. In this function, m is the slope of the line, and b is the y-intercept. As we saw in the last section, the y-intercept is the coordinate where the line crosses the y-axis. The slope is a measure of the inclination the line is. It is the rate of deviation in y regarding x, or how much y shifts for each unit that x shifts. Now that we have went through the slope-intercept form, let's check out how we can use it to locate the y-intercept of a line or a graph. Find the y-intercept of the line described by the equation: y = -2x + 5 In this equation, we can observe that m = -2 and b = 5. Consequently, the y-intercept is equal to 5. Consequently, we can conclude that the line crosses the y-axis at the coordinate (0,5). We can take it a step further to depict the angle of the line. In accordance with the equation, we know the slope is -2. Place 1 for x and figure out: y = (-2*1) + 5 y = 3 The answer tells us that the next coordinate on the line is (1,3). Whenever x changed by 1 unit, y replaced by -2 units. Grade Potential Can Guidance You with the y-intercept You will review the XY axis over and over again across your math and science studies. Theories will get more complicated as you progress from working on a linear equation to a quadratic function. The moment to master your understanding of y-intercepts is now before you straggle. Grade Potential gives experienced teacher that will support you practice finding the y-intercept. Their customized interpretations and work out problems will make a positive difference in the outcomes of your test scores. Anytime you think you’re lost or stuck, Grade Potential is here to assist!	{"url":"https://www.sanantonioinhometutors.com/blog/y-intercept-meaning-examples-y-intercept-formula","timestamp":"2024-11-15T04:44:52Z","content_type":"text/html","content_length":"77276","record_id":"<urn:uuid:aa707e40-dee3-4bba-a572-9faa325730f7>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00763.warc.gz"}
Signed Distance Function - Box Signed distance function of a box: The signed distance function of a box with width $w$ and height $h$ is defined as: $$\Phi(\mathbf{x}) = \min(\max(d_x,d_y),0.0) + \left \\|\begin{pmatrix} \max(d_x,0.0) \\ \max(d_y,0.0) \end{pmatrix} \right \\| - \text{tolerance},$$ where $$d = \begin{pmatrix} \|x - x_{\text{box}}\| - 0.5 w \\ \|y - y_{\text{box}}\| - 0.5 h \end{pmatrix}$$. In the example $\mathbf x$ is the red point, $\mathbf{x}_{\ text{box}}$ the blue point and the closest point to $\mathbf x$ on the surface of the sphere is rendered in yellow. Surface normal vector The normal vector is approximated using central differences, where the derivative of a function $f(x)$ is approximated by $$ \frac{\partial f(x)}{\partial x} \approx \frac{f(x+\varepsilon) - f(x-\ varepsilon)}{2\varepsilon}, $$ where $\varepsilon$ is a small constant (in our example $\varepsilon = 10^{-6}). Since the normal vector is defined as $$ \mathbf n = \frac{\partial \Phi(\mathbf{x})}{\mathbf x}, $$ the normal can be simply approximated by applying central differences on the x- and the y-component of the function $\Phi(\mathbf{x})$. Closest point on the surface The closest point $\mathbf s$ on the surface of the box (yellow) can be determined by starting at the point $\mathbf x$ (red) and going by the signed distance in the direction of the negative normal vector: $$\mathbf s = \mathbf x - \Phi(\mathbf{x}) \mathbf n.$$	{"url":"https://interactivecomputergraphics.github.io/physics-simulation/examples/sdf_box_plot.html","timestamp":"2024-11-05T07:22:07Z","content_type":"text/html","content_length":"11255","record_id":"<urn:uuid:b4190950-78bf-4ee8-ac5f-5234bffa53aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00342.warc.gz"}
Study Guide - Putting It Together: Exponential and Logarithmic Functions Putting It Together: Exponential and Logarithmic Functions At the start of this module, you were considering investing your inheritance possibly to save for retirement. Now you can use what you’ve learned to figure it out. The final value of your investment can be represented by the equation [latex]P[/latex] = the initial investment [latex]t[/latex] = number of years invested [latex]r[/latex] = interest rate, expressed as a decimal Now remember that you had $10,000 to invest, so [latex]P=10,000[/latex]. Also recall that the interest rate was 3%, so [latex]r=0.03[/latex]. Let’s start with 5 years, so [latex]t=5[/latex]. Start with the function. [latex]f(t)=Pe^{\large{tr}}[/latex] Substitute P, t, and r. [latex]f(5)=10,000e^{\large{5}{(0.03)}}[/latex] Evaluate. [latex]f(5)=11,618.34[/latex] Now let’s look at 10 years, so [latex]t= 10[/latex]. Start with the function. [latex]f(t)=Pe^{\large{tr}}[/latex] Substitute P, t, and r. [latex]f(10)=10,000e^{\large{10}{(0.03)}}[/latex] Evaluate. [latex]f(10)=13,498.59[/latex] Now let’s look at 10 years, so [latex]t=50[/latex]. Start with the function. [latex]f(t)=Pe^{\large{tr}}[/latex] Substitute P, t, and r. [latex]f(50)=10,000e^{\large{50}{(0.03)}}[/latex] Evaluate. [latex]f(10)=44,816.89[/latex] Using the function for continuously compounded interest, you can see how your initial investment will grow over time. [latex]t[/latex] Interest rate [latex]f(t)[/latex] 5 0.03 $11,618.34 10 0.03 $13.498.59 50 0.03 $44,816.89 Now you know that your $10,000 can grow to over $44,000 in 50 years! With that knowledge under your belt, you can decide if you want to add to your investment or find an account with a greater interest rate. Either way, thanks to your knowledge of exponential functions, you can make sound financial decisions. Licenses & Attributions CC licensed content, Original • Putting It Together: Exponential and Logarithmic Functions. Authored by: Lumen Learning. License: CC BY: Attribution.	{"url":"https://www.symbolab.com/study-guides/ivytech-wmopen-collegealgebra/putting-it-together-exponential-and-logarithmic-functions.html","timestamp":"2024-11-08T02:55:45Z","content_type":"text/html","content_length":"131447","record_id":"<urn:uuid:c56183ad-fc8a-4884-8217-925a2b662acc>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00755.warc.gz"}
What Is Parallel Computing in What Is Parallel Computing in Optimization Toolbox? Parallel Optimization Functionality Parallel computing is the technique of using multiple processors on a single problem. The reason to use parallel computing is to speed computations. The following Optimization Toolbox™ solvers can automatically distribute the numerical estimation of gradients of objective functions and nonlinear constraint functions to multiple processors: • fmincon • fminunc • fgoalattain • fminimax • fsolve • lsqcurvefit • lsqnonlin These solvers use parallel gradient estimation under the following conditions: • You have a license for Parallel Computing Toolbox™ software. • The option SpecifyObjectiveGradient is set to false, or, if there is a nonlinear constraint function, the option SpecifyConstraintGradient is set to false. Since false is the default value of these options, you don't have to set them; just don't set them both to true. • Parallel computing is enabled with parpool, a Parallel Computing Toolbox function. • The option UseParallel is set to true. The default value of this option is false. When these conditions hold, the solvers compute estimated gradients in parallel. Even when running in parallel, a solver occasionally calls the objective and nonlinear constraint functions serially on the host machine. Therefore, ensure that your functions have no assumptions about whether they are evaluated in serial or parallel. Parallel Estimation of Gradients One solver subroutine can compute in parallel automatically: the subroutine that estimates the gradient of the objective function and constraint functions. This calculation involves computing function values at points near the current location x. Essentially, the calculation is $abla f\left(x\right)\approx \left[\frac{f\left(x+{\Delta }_{1}{e}_{1}\right)-f\left(x\right)}{{\Delta }_{1}},\frac{f\left(x+{\Delta }_{2}{e}_{2}\right)-f\left(x\right)}{{\Delta }_{2}},\dots ,\frac{f \left(x+{\Delta }_{n}{e}_{n}\right)-f\left(x\right)}{{\Delta }_{n}}\right],$ • f represents objective or constraint functions • e[i] are the unit direction vectors • Δ[i] is the size of a step in the e[i] direction To estimate ∇f(x) in parallel, Optimization Toolbox solvers distribute the evaluation of (f(x + Δ[i]e[i]) – f(x))/Δ[i] to extra processors. Parallel Central Differences You can choose to have gradients estimated by central finite differences instead of the default forward finite differences. The basic central finite difference formula is $abla f\left(x\right)\approx \left[\frac{f\left(x+{\Delta }_{1}{e}_{1}\right)-f\left(x-{\Delta }_{1}{e}_{1}\right)}{2{\Delta }_{1}},\dots ,\frac{f\left(x+{\Delta }_{n}{e}_{n}\right)-f\left(x-{\Delta }_{n}{e}_{n}\right)}{2{\Delta }_{n}}\right].$ This takes twice as many function evaluations as forward finite differences, but is usually much more accurate. Central finite differences work in parallel exactly the same as forward finite Enable central finite differences by using optimoptions to set the FiniteDifferenceType option to 'central'. To use forward finite differences, set the FiniteDifferenceType option to 'forward'. Nested Parallel Functions Solvers employ the Parallel Computing Toolbox function parfor (Parallel Computing Toolbox) to perform parallel estimation of gradients. parfor does not work in parallel when called from within another parfor loop. Therefore, you cannot simultaneously use parallel gradient estimation and parallel functionality within your objective or constraint functions. The documentation recommends not to use parfor or parfeval when calling Simulink^®; see Using sim Function Within parfor (Simulink). Therefore, you might encounter issues when optimizing a Simulink simulation in parallel using a solver's built-in parallel functionality. For an example showing how to optimize a Simulink model with several Global Optimization Toolbox solvers, see Optimize Simulink Model in Parallel (Global Optimization Toolbox). Suppose, for example, your objective function userfcn calls parfor, and you wish to call fmincon in a loop. Suppose also that the conditions for parallel gradient evaluation of fmincon, as given in Parallel Optimization Functionality, are satisfied. When parfor Runs In Parallel shows three cases: 1. The outermost loop is parfor. Only that loop runs in parallel. 2. The outermost parfor loop is in fmincon. Only fmincon runs in parallel. 3. The outermost parfor loop is in userfcn. userfcn can use parfor in parallel. When parfor Runs In Parallel See Also Using Parallel Computing in Optimization Toolbox \| Improving Performance with Parallel Computing \| Minimizing an Expensive Optimization Problem Using Parallel Computing Toolbox	{"url":"https://se.mathworks.com/help/optim/ug/parallel-computing-in-optimization-toolbox-functions.html","timestamp":"2024-11-06T14:27:00Z","content_type":"text/html","content_length":"83347","record_id":"<urn:uuid:f2dcfab0-5dec-4885-9a8f-6d526cb1b2d1>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00802.warc.gz"}
How do you divide (25x^2-30x+12) div (5x-3)? \| HIX Tutor How do you divide #(25x^2-30x+12) div (5x-3)#? Answer 1 Give - $\left(25 {x}^{2} - 30 x + 12\right) \div \left(5 x - 3\right)$ Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer 2 $5 x - 3 + \frac{3}{5 x - 3}$ #"one way is to use the divisor as a factor in the numerator"# #"consider the numerator"# #"quotient "=color(red)(5x-3)," remainder "=3# Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer 3 To divide (25x^2-30x+12) by (5x-3), you can use long division or synthetic division. Here is the step-by-step process using long division: 1. Divide the first term of the dividend (25x^2) by the first term of the divisor (5x). The result is 5x. 2. Multiply the entire divisor (5x-3) by the quotient obtained in step 1 (5x). The result is 25x^2 - 15x. 3. Subtract the result obtained in step 2 from the dividend (25x^2-30x+12) to get the new dividend: -15x + 12. 4. Bring down the next term from the original dividend (-15x) and divide it by the first term of the divisor (5x). The result is -3. 5. Multiply the entire divisor (5x-3) by the quotient obtained in step 4 (-3). The result is -15x + 9. 6. Subtract the result obtained in step 5 from the new dividend (-15x + 12) to get the new dividend: 3. 7. Since the new dividend (3) is a constant term and has a degree lower than the divisor (5x-3), the division process is complete. Therefore, the quotient is 5x - 3 and the remainder is 3. Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer from HIX Tutor When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some Not the question you need? HIX Tutor Solve ANY homework problem with a smart AI • 98% accuracy study help • Covers math, physics, chemistry, biology, and more • Step-by-step, in-depth guides • Readily available 24/7	{"url":"https://tutor.hix.ai/question/how-do-you-divide-25x-2-30x-12-div-5x-3-8f9af9c0f4","timestamp":"2024-11-07T15:29:34Z","content_type":"text/html","content_length":"584882","record_id":"<urn:uuid:e55c90bf-5dd6-4f69-8bd0-1ee639d6daa1>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00391.warc.gz"}
Коментарі читачів Re: Re: Re: Buy HUAWEI MateView SE - Monitor - HUAWEI UK З приводу Re: Re: Buy HUAWEI MateView SE - Monitor - HUAWEI UK Encryption is a technique used to protect data from unauthorized access by transforming it into an unreadable format, called ciphertext. Only those with the appropriate key can convert the ciphertext back into its original form, known as plaintext. One widely used encryption algorithm is AES, also known as Advanced Encryption Standard. In this article, we will delve into the world of AES encryption, discussing its history, usage, and significance in today's digital landscape. History of AES AES encryption originated in the early 1990s when the US government sought a replacement for the outdated Data Encryption Standard (DES). The National Institute of Standards and Technology (NIST) launched a competition to develop a new encryption algorithm that could protect sensitive information. After several rounds of evaluation, a Belgian cryptographer named Joan Daemen and an American cryptographer named Vincent Rijmen proposed the Rijndael algorithm, which was later adopted as AES. How AES Encryption Works AES encryption operates by using a symmetric-key block cipher. This means that the same key used to encrypt data is also used to decrypt it. The key is used to perform a series of mathematical operations on the plaintext, transforming it into ciphertext. The process works as follows: 1. The plaintext is divided into fixed-size blocks (usually 128 bits). 2. An initialization vector (IV) is generated randomly and added to the plaintext. 3. The plaintext and IV are fed into a pseudorandom function (PRF) to create a key schedule. 4. The key schedule is used to encrypt the plaintext and IV, using a substitution-permutation network. 5. The resulting ciphertext is the encrypted data. The magic of AES encryption lies in its ability to provide secure protection without significantly slowing down data processing. AES can encrypt data at remarkable speeds, thanks to its parallelizable structure. This efficiency makes it suitable for a broad range of applications, from securing online transactions to safeguarding sensitive information. Uses of AES Encryption AES encryption has become the de facto standard for data protection in various sectors, including: 1. Online Banking: AES encryption ensures the confidentiality and privacy of financial transactions, shielding sensitive information from cyber threats. 2. Cloud Storage: With AES encryption, data stored in the cloud remains secure, reducing the risk of unauthorized access or data breaches. 3. E-commerce: AES encryption protects sensitive data, such as credit card numbers and personal information, during online transactions. 4. Digital Signatures: AES encryption verifies the authenticity of digital signatures, preventing tampering and forgery. 5. Military Applications: AES encryption guards classified information, ensuring that sensitive military data remains confidential. Strengths and Weaknesses of AES Encryption While AES encryption is widely regarded as unbreakable, it does have some limitations: 1. High security: AES encryption is resistant to attacks, including brute force and differential attacks. 2. Efficient: AES encryption is fast and efficient, making it suitable for high-speed data processing. 3. Scalable: AES encryption can be applied to various data sizes and types. 1. Key management: Managing keys can be challenging, as AES encryption relies on a symmetric-key system. Losing the key means losing access to the encrypted data. 2. Initialization vector: The IV must be randomly generated and kept secret, as it can be exploited to weaken the encryption. Is AES Encryption Unbreakable? AES encryption is considered unbreakable in the classical sense. No efficient algorithm has been discovered to break AES without knowing the key. However, side-channel attacks, such as those exploiting caching weaknesses or power consumption analysis, can potentially threaten AES encryption. It is essential to maintain secure key management practices and use up-to-date hardware and software to mitigate these risks. Comparison with Other Encryption Methods AES encryption stands out among other encryption methods due to its balance of security and efficiency. Compared to its predecessor, DES, AES offers a much larger key space, making it more resistant to brute force attacks. Other encryption methods, such as RSA, focus on asymmetric encryption and are generally slower and less efficient than AES. Future Developments in AES Encryption As technology advances, researchers explore new methods to enhance AES encryption or develop alternative encryption algorithms. Some future developments include: 1. Quantum-Resistant AES: As quantum computing emerges, researchers seek ways to make AES encryption resistant to quantum attacks. 2. Lightweight AES: Efforts focus on optimizing AES encryption for resource-constrained devices, such as IoT devices. 3. Post-Quantum AES: Research is underway to create new encryption algorithms that can withstand quantum attacks, which may eventually replace AES. Real-World Examples of AES Encryption 1. Apple's FileVault: Apple's FileVault encryption uses AES to protect data on Mac devices. 2. WhatsApp: WhatsApp utilizes AES encryption to protect messages and communication. 3. Gmail: Google uses AES encryption to secure emails, protecting user data from unauthorized access. 4. BitLocker: Microsoft's BitLocker encryption, which leverages AES, secures data on Windows devices. AES encryption has proven to be a robust and reliable method of protecting data. Its symmetric-key block cipher design makes it efficient and suitable for various applications. Since its adoption in 2001, AES encryption has become a cornerstone of data security, protecting sensitive information across various sectors. While it has its limitations, AES encryption remains a powerful tool in the fight against cyber threats. As technology advances, researchers work to further optimize AES encryption and develop new methods to safeguard data in the future. In conclusion, AES encryption is a powerful tool that safeguards data in various applications. Its balance of security and efficiency makes it a go-to solution for protecting sensitive information. As technology advances, AES encryption will continue to play a vital role in protecting data and ensuring confidentiality in the digital age. 1. What is AES encryption, and how does it work? AES(Advanced Encryption Standard) encryption is a method of encrypting data using a symmetric-key block cipher. It works by transforming plaintext into ciphertext using the same key for both processes. The key is generated using a pseudorandom function, and the process includes an initialization vector to ensure security. 2. What are some common uses of AES encryption? AES encryption is used in various applications, including financial transactions, cloud storage, e-commerce, digital signatures, and military operations. It is also used to protect sensitive information in industries such as healthcare and finance. 3. What are the strengths of AES encryption? AES encryption is highly secure, efficient, and scalable. It is resistant to brute-force attacks and has a large key space, making it difficult to break. It is also compatible with various hardware and software systems. 4. What are the weaknesses of AES encryption? AES encryption relies on a symmetric-key system, which means that losing the key will result in loss of access to the encrypted data. Additionally, the initialization vector used in the encryption process must be generated randomly and kept secret. 5. Is AES encryption unbreakable? AES encryption is considered unbreakable in the classical sense, but it is not invincible to side-channel attacks. It is essential to maintain secure key management practices and use up-to-date hardware and software to mitigate potential risks. 6. What is the difference between AES and DES encryption? AES encryption has a larger key space than DES(Data Encryption Standard) encryption, making it more resistant to attacks. AES also has a faster encryption and decryption speed than DES. 7. What is the difference between AES and RSA encryption? AES encryption is a symmetric-key block cipher, whereas RSA(Rivest-Shamir-Adleman) encryption is an asymmetric encryption method. AES encryption is faster and more efficient for large-scale data protection, whereas RSA encryption is frequently used for digital signatures and key exchange. 8. What is quantum-resistant AES encryption? Quantum-resistant AES encryption aims to make AES encryption resistant to quantum attacks, which could potentially break AES encryption in the future. Research is underway to develop quantum-resistant AES encryption algorithms.	{"url":"http://journals.hnpu.edu.ua/index.php/literature/comment/view/4159/2261/11437","timestamp":"2024-11-03T15:58:40Z","content_type":"application/xhtml+xml","content_length":"23954","record_id":"<urn:uuid:9071b304-b899-46cb-bc55-33b041d84a8d>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00530.warc.gz"}
Why It Matters: Historical Counting Systems Why do historical counting systems still matter today? When you check that balance in your bank account, or when you glance at the speedometer in your car, or even when you look for your child’s number on the back of jerseys during a pee wee football game, you are reading numerals in the Hindu-Arabic counting system. We are all familiar with those ten digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. What’s more, when we read a number like 352, we know that it stands for three groups of a hundred, five groups of ten, and two units. Our numerals are arranged according to a positional base 10 (or decimal) system… most of the time, anyway. Telling time requires a slightly different system. While there are still Hindu-Arabic numerals involved, the way that they behave is decidedly different. There are 60 seconds in every minute and 60 minutes in every hour. So if your watch displays 10:04:59 right now, then you expect it to read 10:05:00 a second later. We are so used to telling time in groups of 60 that it seems natural. But have you ever wondered why there are not 100 seconds in each minute, or 100 minutes in an hour? In the late 1700s, a French attorney by the name of Claude Boniface Collignon suggested a system of decimal time measurement in which each day has 10 hours, each hour lasting 100 minutes, and each minute having 1000 seconds. Of course the actual duration of these new hours, minutes, and seconds would be much different. In particular, the decimal second, would last 0.864 of a normal second. On the upside, time conversions would be trivial; for example, 6 decimal hours = 600 decimal minutes = 600,000 decimal seconds. So why does our system of telling time not conform to the usual base 10 counting system that governs most other aspects of our life? Blame it on the Babylonians! The Babylonians were one of the first cultures to develop a positional numeral system. However instead of having only 10 distinct numerals and groups in powers of 10, their system was based on groups and powers of 60 (which is called a sexigesimal system). The Babylonian system spread throughout most of Mesopotamia, but it eventually faded into history, allowing other number systems such as the Roman numerals and the Hindu-Arabic system to take its place. On the other hand, there are still vestiges of the sexigemisal counting system in the way that we keep time as well as how we measure angles in degrees. There are 360 degrees in a full circle (and [latex]360^{\circ} = 6 \times 60^{\circ}[/latex]). Furthermore, there are 60 arc minutes in one degree and 60 arc seconds in one arc minute. This system of degrees, arc minutes, and arc seconds is also used to locate any point on the surface of the Earth by its latitude and longitude. So even though our numerals are Hindu-Arabic, we still rely on the Babylonian base 60 system every second of the day and everywhere on the globe!	{"url":"https://courses.lumenlearning.com/waymakermath4libarts/chapter/why-it-matters-historical-counting-systems/","timestamp":"2024-11-11T08:10:07Z","content_type":"text/html","content_length":"23508","record_id":"<urn:uuid:3df36f37-b807-4b2c-82f5-f3cdc4503317>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00514.warc.gz"}
Introductory Chemistry – 1st Canadian Edition Chapter 5. Stoichiometry and the Mole 1. Define stoichiometry. 2. Relate quantities in a balanced chemical reaction on a molecular basis. Consider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1 pound of sugar. (That’s why it’s called “pound cake.”) If you have 4 pounds of butter, how many pounds of sugar, flour, and eggs do you need? You would need 4 pounds each of sugar, flour, and eggs. Now suppose you have 1.00 g H[2]. If the chemical reaction follows the balanced chemical equation 2H[2](g) + O[2](g) → 2H[2]O(ℓ), then what mass of oxygen do you need to make water? Curiously, this chemical reaction question is very similar to the pound cake question. Both of them involve relating a quantity of one substance to a quantity of another substance or substances. The relating of one chemical substance to another using a balanced chemical reaction is called stoichiometry. Using stoichiometry is a fundamental skill in chemistry; it greatly broadens your ability to predict what will occur and, more importantly, how much is produced. Let us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and ½ c of milk. We can write this in the form of a chemical equation: 2 c mix + 1 egg + ½ c milk → 1 batch of pancakes If you have 9 c of pancake mix, how many eggs and how much milk do you need? It might take a little bit of work, but eventually you will find you need 4½ eggs and 2¼ c milk. How can we formalize this? We can make a conversion factor using our original recipe and use that conversion factor to convert from a quantity of one substance to a quantity of another substance, similar to the way we constructed a conversion factor between feet and yards in Chapter 2 “Measurements”. Because one recipe’s worth of pancakes requires 2 c of pancake mix, 1 egg, and ½ c of milk, we actually have the following mathematical relationships that relate these quantities: 2 c pancake mix ⇔ 1 egg ⇔ ½ c milk where ⇔ is the mathematical symbol for “is equivalent to.” This does not mean that 2 c of pancake mix equals 1 egg. However, as far as this recipe is concerned, these are the equivalent quantities needed for a single recipe of pancakes. So, any possible quantities of two or more ingredients must have the same numerical ratio as the ratios in the equivalence. We can deal with these equivalences in the same way we deal with equalities in unit conversions: we can make conversion factors that essentially equal 1. For example, to determine how many eggs we need for 9 c of pancake mix, we construct the conversion factor: This conversion factor is, in a strange way, equivalent to 1 because the recipe relates the two quantities. Starting with our initial quantity and multiplying by our conversion factor, we get: Note how the units cups pancake mix cancelled, leaving us with units of eggs. This is the formal, mathematical way of getting our amounts to mix with 9 c of pancake mix. We can use a similar conversion factor for the amount of milk: Again, units cancel, and new units are introduced. A balanced chemical equation is nothing more than a recipe for a chemical reaction. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs. For example, consider the following chemical equation: 2H[2](g) + O[2](g) → 2H[2]O(ℓ) We can interpret this as, literally, “two hydrogen molecules react with one oxygen molecule to make two water molecules.” That interpretation leads us directly to some equivalences, just as our pancake recipe did: 2 H[2] molecules ⇔ 1 O[2] molecule ⇔ 2 H[2]O molecules These equivalences allow us to construct conversion factors: and so forth. These conversions can be used to relate quantities of one substance to quantities of another. For example, suppose we need to know how many molecules of oxygen are needed to react with 16 molecules of H[2]. As we did with converting units, we start with our given quantity and use the appropriate conversion factor: Note how the unit molecules H[2] cancels algebraically, just as any unit does in a conversion like this. The conversion factor came directly from the coefficients in the balanced chemical equation. This is another reason why a properly balanced chemical equation is important. How many molecules of SO[3] are needed to react with 144 molecules of Fe[2]O[3] given the balanced chemical equation Fe[2]O[3](s) + 3SO[3](g) → Fe[2](SO[4])[3]? We use the balanced chemical equation to construct a conversion factor between Fe[2]O[3] and SO[3]. The number of molecules of Fe[2]O[3] goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO[3] go on the top. Thus, the appropriate conversion factor is: Starting with our given amount and applying the conversion factor, the result is: We need 432 molecules of SO[3] to react with 144 molecules of Fe[2]O[3]. Test Yourself How many molecules of H[2] are needed to react with 29 molecules of N[2] to make ammonia if the balanced chemical equation is N[2] + 3H[2] → 2NH[3]? 87 molecules Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction because a chemical formula lists the number of atoms of each element in a compound. The formula H[2]O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors: Conversion factors like this can also be used in stoichiometry calculations. How many molecules of NH[3] can you make if you have 228 atoms of H[2]? From the formula, we know that one molecule of NH[3] has three H atoms. Use that fact as a conversion factor: Test Yourself How many molecules of Fe[2](SO[4])[3] can you make from 777 atoms of S? 259 molecules For a video lecture on stoichiometry, view this video on stoichiometry by Dr. Jessie A. Key. • Quantities of substances can be related to each other using balanced chemical equations. 1. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe? 2. Think back to the pancake recipe. What possible conversion factors can you construct relating the components of the recipe? 3. What are all the conversion factors that can be constructed from the balanced chemical reaction 2H[2](g) + O[2](g) → 2H[2]O(ℓ)? 4. What are all the conversion factors that can be constructed from the balanced chemical reaction N[2](g) + 3H[2](g) → 2NH[3](g)? 5. Given the chemical equation Na(s) + H[2]O(ℓ) → NaOH(aq) + H[2](g) a. Balance the equation. b. How many molecules of H[2] are produced when 332 atoms of Na react? 6. Given the chemical equation S(s) + O[2](g) → SO[3](g) a. Balance the equation. b. How many molecules of O[2] are needed when 38 atoms of S react? 7. For the balanced chemical equation 6H^+(aq) + 2MnO[4]^−(aq) + 5H[2]O[2](ℓ) → 2Mn^2+(aq) + 5O[2](g) + 8H[2]O(ℓ), how many molecules of H[2]O are produced when 75 molecules of H[2]O[2] react? 8. For the balanced chemical reaction 2C[6]H[6](ℓ) + 15O[2](g) → 12CO[2](g) + 6H[2]O(ℓ), how many molecules of CO[2] are produced when 56 molecules of C[6]H[6] react? 9. Given the balanced chemical equation Fe[2]O[3](s) + 3SO[3](g) → Fe[2](SO[4])[3], how many molecules of Fe[2](SO[4])[3] are produced if 321 atoms of S react? 10. For the balanced chemical equation CuO(s) + H[2]S(g) → CuS + H[2]O(ℓ), how many molecules of CuS are formed if 9,044 atoms of H react? 11. For the balanced chemical equation Fe[2]O[3](s) + 3SO[3](g) → Fe[2](SO[4])[3], suppose we need to make 145,000 molecules of Fe[2](SO[4])[3]. How many molecules of SO[3] do we need? 12. One way to make sulfur hexafluoride is to react thioformaldehyde, CH[2]S, with elemental fluorine, is described by CH[2]S + 6F[2] → CF[4] + 2HF + SF[6]. If 45,750 molecules of SF[6] are needed, how many molecules of F[2] are required? 13. Construct the three independent conversion factors possible for these two reactions: 2H[2] + O[2] → 2H[2]O and H[2] + O[2] → H[2]O[2]. Why are the ratios between H[2] and O[2] different? The conversion factors are different because the stoichiometries of the balanced chemical reactions are different. 14. Construct the three independent conversion factors possible for these two reactions 2Na + Cl[2] → 2NaCl and 4Na + 2Cl[2] → 4NaCl. What similarities, if any, exist in the conversion factors from these two reactions? 5. 2Na(s) + 2H[2]O(ℓ) → 2NaOH(aq) + H[2](g) and 166 molecules 7. 120 molecules 9. 107 molecules 11. 435,000 molecules The relating of one chemical substance to another using a balanced chemical reaction.	{"url":"https://opentextbc.ca/introductorychemistry/chapter/stoichiometry/","timestamp":"2024-11-05T12:42:10Z","content_type":"text/html","content_length":"122391","record_id":"<urn:uuid:c51d18b5-1fa3-4086-946c-a118c09d210d>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00566.warc.gz"}
{Year 2} Chapter 3 - Gravitation (Halfway down pg47) Flashcards Why does the moon have a smaller surface gravity than the Earth? As it is a less massive body What is Newton’s Law of Gravitation? The universal constant of gravitation - 6.67 x 10^-11 When can Newton’s Law not be applied, and why? When the two masses are irregularly shaped unless a complicated summation of the forces is made - since the law only works for point masses, or spheres since they act as if their mass was concentrated in the middle What law does gravity, like light, obey? What is a gravitational field? A region in space in which a massive object experiences a gravitational force The strength of the gravitational field measured in N/kg. Field lines represent the direction and strength of the field What is the equation for gravitational field strength? What happens to the gravitational field near the surface of a planet? It becomes very nearly uniform - meaning that the field is of the same strength and direction everywhere What is the difference in the way gravitational field lines are drawn when the field is half as powerful? The lines are half as prevalent/half as many of them When is capital M used in equations in gravitation? For the mass of a large object such as a star or a planet What is the equation for the volume of a sphere? What is the equation for gravitational potential energy? Grav. pot. = mass x gravity x difference in height What is gravitational potential difference? The gravitational potential energy difference per kilogram. Gravitational potential and potential difference have units of J/kg. It’s symbol is Delta V What is an equipotential surface? A surface along which if you move, the gravitational potential stays the same	{"url":"https://www.brainscape.com/flashcards/year-2-chapter-3-gravitation-halfway-dow-8165280/packs/12222962","timestamp":"2024-11-11T14:17:46Z","content_type":"text/html","content_length":"109761","record_id":"<urn:uuid:2eb59cf9-17f2-4135-978b-d902a5eb0c9b>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00825.warc.gz"}
Error Expression Must Have Integral or Enum Type: Causes, Fixes & Best Practices The error message “expression must have integral or enum type” in programming typically occurs when an expression is used in a context that requires an integer or an enumeration type, but the provided expression is of a different type. Understanding this error is crucial for developers because it helps ensure that the code adheres to type requirements, preventing compilation issues and ensuring the program runs correctly. Recognizing and resolving this error is a fundamental skill in maintaining robust and error-free code. What Does ‘Error Expression Must Have Integral or Enum Type in Thats Code’ Mean? The error “expression must have integral or enum type” occurs when an expression used in a context requiring an integral or enum type is of an incompatible type. Integral types include: • int • short • long • long long • unsigned int • unsigned short • unsigned long • unsigned long long • char • bool Enum types are user-defined types that consist of named integral constants, such as: enum Color { RED, GREEN, BLUE }; This error typically arises in contexts like switch statements, bitwise operations, and loop control expressions. Common Causes of ‘Error Expression Must Have Integral or Enum Type in Thats Code’ Here are some common scenarios that lead to the “expression must have integral or enum type” error in C++: 1. Using Non-Integral Types in Switch Statements: □ Scenario: Using a float or double as a case label. □ Example: float f = 2.5; int x = 5; switch (x) { case f: // Error: 'f' is not an integral type 2. Using Non-Integral Types with Bitwise Operators: □ Scenario: Applying bitwise operators to float or double. □ Example: float a = 5.5; float b = 2.2; int result = a & b; // Error: 'a' and 'b' are not integral types 3. Using Non-Integral Types in Array Indexing: □ Scenario: Using a float or double as an array index. □ Example: float index = 2.5; int arr[10]; int value = arr[index]; // Error: 'index' is not an integral type 4. Using Non-Integral Types in Loops: □ Scenario: Using a float or double as the controlling expression in a loop. □ Example: float i = 0.0; while (i < 10.0) { // Error: 'i' is not an integral type i += 0.5; 5. Using Non-Integral Types in Conditional Expressions: □ Scenario: Using a float or double in a conditional expression that expects an integral type. □ Example: float condition = 1.5; if (condition) { // Error: 'condition' is not an integral type // Do something These scenarios typically occur when the code expects an integral or enum type, but a different type is provided. How to Fix ‘Error Expression Must Have Integral or Enum Type in Thats Code’ To resolve the ‘error expression must have integral or enum type’, follow these steps: 1. Identify the problematic expression: Locate the expression causing the error. This typically occurs in switch statements, bitwise operations, or loop conditions. 2. Ensure the expression is of an integral or enum type: The expression must be an integer type (int, long, short, etc.) or an unscoped enum type. 3. Correct the expression type: If the expression is not of the correct type, cast it or change its type. Example 1: Correcting a switch statement Incorrect Code: float f = 2.5; int x = 5; switch (x) { case f: // Error: 'f' is not an integral or enum type // ... Corrected Code: int f = 2; int x = 5; switch (x) { case f: // Correct: 'f' is now an integral type // ... Example 2: Using an enum type Incorrect Code: enum Color { RED, GREEN, BLUE }; Color color = RED; if (color & 1) { // Error: 'color' is not an integral type // ... Corrected Code: enum Color { RED = 1, GREEN = 2, BLUE = 4 }; Color color = RED; if (color & RED) { // Correct: 'color' is now used with integral values // ... Example 3: Bitwise operations Incorrect Code: float a = 5.5; float b = 2.2; float result = a & b; // Error: 'a' and 'b' are not integral types Corrected Code: int a = 5; int b = 2; int result = a & b; // Correct: 'a' and 'b' are integral types By ensuring your expressions are of integral or enum types, you can resolve this error effectively. Best Practices to Avoid ‘Error Expression Must Have Integral or Enum Type in Thats Code’ Here are some best practices to avoid the “expression must have integral or enum type” error: 1. Type Checking: Ensure all expressions are of the correct type. Use the typeof operator to verify types if unsure. 2. Casting: Explicitly cast expressions to the required integral type when necessary. 3. Enum Usage: Use unscoped enums for switch cases and other integral operations. 4. Avoid Mixing Types: Do not mix different types in expressions. Stick to integral or enum types. 5. Consistent Variable Types: Declare variables with consistent types to avoid implicit type conversions. 6. Compiler Warnings: Enable and heed compiler warnings to catch type-related issues early. 7. Code Reviews: Regularly review code to ensure type correctness and adherence to best practices. By following these habits, you can minimize the chances of encountering this error. Happy coding! To Resolve the ‘Expression Must Have Integral or Enum Type’ Error To resolve the “expression must have integral or enum type” error, it’s essential to understand its causes and implement best practices to avoid it. This error typically occurs when an expression is not of an integral or enum type, which can lead to unexpected behavior or compilation issues. Key Points to Consider: • Ensure all expressions are of the correct type by using `typeof` for verification. • Explicitly cast expressions to the required integral type when necessary. • Use unscoped enums for switch cases and other integral operations. • Avoid mixing different types in expressions, sticking to integral or enum types instead. • Declare variables with consistent types to prevent implicit type conversions. • Enable and heed compiler warnings to catch type-related issues early. • Regularly review code to ensure type correctness and adherence to best practices. By understanding the causes of this error and implementing these best practices, you can improve your code quality, reduce errors, and write more efficient and maintainable code.	{"url":"https://terramagnetica.com/error-expression-must-have-integral-or-enum-type-in-thats-code/","timestamp":"2024-11-03T07:34:50Z","content_type":"text/html","content_length":"69753","record_id":"<urn:uuid:2a5d9625-ef22-4854-b042-5a881bf4b405>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00375.warc.gz"}
Fuel Economy, Hypermiling, EcoModding News and Forum - EcoModder.com - View Single Post - AC Mod: Peltier Junctions Ok thank you all for the suggestions. Since not one of you attempted to answer my original question which I repeated twice (see posts #10, #6, and #1) I will look somewhere else for the answer. I'm looking for a well thought-out, perhaps of mathematical origin, of whether 300watts of additional electrical load on the alternator uses more or less energy than the ac compressor. If someone knows about air conditioners and can provide a solution to this question, please PM me. In the mean time, I will continue my googling.	{"url":"https://ecomodder.com/forum/31876-post15.html","timestamp":"2024-11-12T21:56:19Z","content_type":"application/xhtml+xml","content_length":"14776","record_id":"<urn:uuid:767ced0a-7388-4fb7-949c-07c1e5725023>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00415.warc.gz"}
Quicksort, Largest Bucket, and Min-Wise Hashing with Limited Independence Randomized algorithms and data structures are often analyzed under the assumption of access to a perfect source of randomness. The most fundamental metric used to measure how “random” a hash function or a random number generator is, is its independence: a sequence of random variables is said to be k-independent if every variable is uniform and every size k subset is independent. In this paper we consider three classic algorithms under limited independence. Besides the theoretical interest in removing the unrealistic assumption of full independence, the work is motivated by lower independence being more practical. We provide new bounds for randomized quicksort, min-wise hashing and largest bucket size under limited independence. Our results can be summarized as follows. Randomized Quicksort. When pivot elements are computed using a 5-independent hash function, Karloff and Raghavan, J.ACM’93 showed O(nlogn) expected worst-case running time for a special version of quicksort. We improve upon this, showing that the same running time is achieved with only 4-independence. Min-Wise Hashing. For a set A, consider the probability of a particular element being mapped to the smallest hash value. It is known that 5-independence implies the optimal probability O(1/n). Broder et al., STOC’98 showed that 2-independence implies it is O(1/\|A\|−−−√). We show a matching lower bound as well as new tight bounds for 3- and 4-independent hash functions. Largest Bucket. We consider the case where n balls are distributed to n buckets using a k-independent hash function and analyze the largest bucket size. Alon et. al, STOC’97 showed that there exists a 2-independent hash function implying a bucket of size Ω( n1/2). We generalize the bound, providing a k-independent family of functions that imply size Ω( n1/k). Originalsprog Engelsk Titel Algorithms – ESA 2015 : 23rd Annual European Symposium, Patras, Greece, September 14–16, 2015, Proceedings Vol/bind 9294 Forlag Springer Publikationsdato 2015 Sider 828-839 ISBN (Trykt) 978-3-662-48349-7 ISBN (Elektronisk) 978-3-662-48350-3 Status Udgivet - 2015 Navn Lecture Notes in Computer Science ISSN 0302-9743 • Randomized Algorithms • Probabilistic Analysis • k-Independent Hash Functions • Randomized Quicksort • Min-Wise Hashing Dyk ned i forskningsemnerne om 'Quicksort, Largest Bucket, and Min-Wise Hashing with Limited Independence'. Sammen danner de et unikt fingeraftryk.	{"url":"https://pure.itu.dk/da/publications/quicksort-largest-bucket-and-min-wise-hashing-with-limited-indepe","timestamp":"2024-11-01T20:31:51Z","content_type":"text/html","content_length":"55671","record_id":"<urn:uuid:8c337532-4649-492e-a163-74bcff2addb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00308.warc.gz"}
100 Grid Square Printable 100 Grid Square Printable - This tool is often used in math lessons for young learners to help them. Web the hundred square will help children investigate how to find multiples of numbers, how even or odd numbers are situated across the grid, how to count on and backwards from. There's a variety of colourful and attractive designs. Its square design creates an impression of minimalism and is ideal for a. Web use a 100 square for addition, subtraction and multiplication. If you’re interested in finding out about twinkl’s 100 square app,. Web in this post, we have brought you a blank 50,100 & 200 squares grid printable bundle that can be downloaded, printed, and put to use to learn mathematics,. They are simple squares that are arranged sequentially in rows and columns. Web the main blank hundreds chart printable is set out as a grid, onto which you can write numbers of your choice. 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Its square design creates an impression of minimalism and is ideal for a. Related Post:	{"url":"https://dl-uk.apowersoft.com/en/100-grid-square-printable.html","timestamp":"2024-11-13T17:39:57Z","content_type":"text/html","content_length":"30372","record_id":"<urn:uuid:2a60d637-0092-436c-908c-fe43cb7863da>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00414.warc.gz"}
How general are Large Language Model (LLMs) chatbots? If you're interested in how AI will play out over the next few years, you might care how much "general" intelligence is embedded in these systems. Do they just predict text? Or can they learn to do other interesting things? Twice today I saw a experts use chess as an example to show that they aren't very general. In an interview a few weeks ago with Ezra Klein, Demis Hassabis (CEO and co-founder of DeepMind) said: So if you challenge one of these chat bots to a game, you want to play a game of chess or a game of Go against it, they’re actually all pretty bad at it currently, which is one of the tests I give these chat bots is, can they play a good game and hold the board state in mind? And they can’t really at the moment. They’re not very good. In a roundtable with Eliezer Yudkowsky and Scott Aaronson, Gary Marcus said: I would say that the core of artificial general intelligence is the ability to flexibly deal with new problems that you haven’t seen before. The current systems can do that a little bit, but not very well. My typical example of this now is GPT-4. It is exposed to the game of chess, sees lots of games of chess, sees the rules of chess but it never actually figure out the rules of chess. They often make illegal moves and so forth. So it’s in no way a general intelligence that can just pick up new things. I mean, to take the example I just gave you a minute ago, it never learns to play chess even with a huge amount of data. It will play a little bit of chess; it will memorize the openings and be okay for the first 15 moves. But, it gets far enough away from what it’s trained on, and it falls apart. But they can play chess! A couple months ago I saw some tweets about how GPT3.5 and GPT4 can play chess if you give it very specialized prompts. (That same account has some other interesting examples showing a "logic core" in GPT.) That made me curious to try more "normal" prompts, and I found that if you prompt it like this: Let's play chess. I'll start. 1. e4 ... and proceed with standard PGN notation, GPT4 makes reasonable, legal moves, even long after the opening. (I don't know why GPT4 is refusing to increment the move numbers with me. When I tried this a couple months ago its move numbers made sense. You'll see in the transcript I got one move number wrong I don't know what Demis considers a "good game", but it seems pretty clear to me that GPT4 is able to "hold the board state in mind". Here's the final position in the game I just played (GPT4 won): You can see the full transcript here. To see the full game played out, go to a lichess.org analysis board and paste in the game's PGN notation: 1. e4 e5 2. d4 exd4 3. Qxd4 Nc6 4. Qd1 Nf6 5. Bc4 Bc5 6. b4 Bxb4+ 7. c3 Ba5 8. Nf3 O-O 9. O-O Nxe4 10. Re1 Nf6 11. Qb3 d5 12. Rd1 dxc4 13. Rxd8 cxb3 14. Rxf8+ Kxf8 15. axb3 Bg4 16. Ba3+ Kg8 17. Nbd2 Re8 18. Re1 Rxe1+ 19. Nxe1 Bxc3 20. f3 Bxd2 21. fxg4 Bxe1 22. Kf1 Bb4 23. Bxb4 Nxb4 24. Ke2 c5 25. g5 Ne4 26. h4 b5 27. Ke3 Nd6 28. g4 a5 29. h5 a4 30. bxa4 bxa4 31. h6 gxh6 32. gxh6 a3 33. Kf4 a2 34. Ke3 a1=Q 35. Kf4 Qf6+ 36. Ke3 Nd5+ 37. Kd2 Qf2+ 38. Kd3 c4# Aside from the move numbers, there's one way in which this record of the game differs from our transcript: On move 17, I said "Nd2", which is an incorrect thing to write, since there are two knights that could move to d2. But GPT4 just went with it, and seems to have correctly figured out which one I meant. Could this game be memorized? I can't really prove to you that this game isn't in GPT4's training data, but it seems exceedingly unlikely. I made some intentionally weird moves just to try to get out of any common sequence pretty Is GPT4 good at chess? This isn't a particularly good chess game, though it's me rather than GPT4 who didn't play very well. I blundered a piece on move 12, and (as mentioned) made a few other intentionally not-good moves. I don't really know how good GPT4 is at chess. If I can get API access, I'd love to try making a chess bot that could play online. Why can GPT4 play chess? I don't work in AI these days, but I'd guess that: GPT4 is trained to predict the text in its training data, and presumably there are a lot of chess games out there on the internet for it to read. By learning the rules of chess and some ability to reason about the board state and good moves, it does a better job predicting those games. Did I cherry-pick this example? Is this behavior reliable? First, I even one example like this shows that GPT4 is able to hold the board state in mind. If I'd had to play 100 games where forgot the board position, even one where it tracks the board and plays pretty well is interesting. But no, I didn't cherry-pick the example. I played a little bit in this format a couple months ago (and I don't think GPT4 made any illegal moves), and then today after seeing these comments that chat bots can't play chess, I tried one and only one game. That's this one. I didn't play any other game for this post, and certainly not any where it made illegal moves and I gave up. I tried this a couple other times two months ago, but I don't have great data on how often GPT4 is able to complete a game this way. But it does make illegal moves for some prompts... Using prompts like the one above seems to work well, but it's true at GPT4 can quickly go off the rails with other prompts. Here's an example where I asked it to explain its moves, and its 4th move was illegal. That's a big contrast with the above 38-move game where it plays legally the whole time, and wins! Also here's a chat with the same initial few moves where it plays legally. Messy git history is a display problem, not a data problem. The first thing I encountered learning about git: there's a lot of conflict about whether it's important to keep a "clean" git history by squashing, rebasing instead of merging, etc. If the --first-parent featue were well supported, it would give us the best of both worlds. (...click here for the rest of this post) In case they help anyone else, here are some regular expressions I used once to convert some ugly unittest-style assertions (e.g. self.assertEqual(something, something_else) to the pytest style (simply assert something == something_else): sed -i ".bak" -E 's/self\.assertFalse$(.)$/assert not \1/g' tests/.py sed -i ".bak" -E 's/self\.assertTrue$(.)$/assert \1/g' tests/.py sed -i ".bak" -E 's/self\.assertEqual$([^,]), (.)$$/assert \1 == \2/g' tests/.py sed -i ".bak" -E 's/self\.assertIn$([^,]), (.)$$/assert \1 in \2/g' tests/.py sed -i ".bak" -E 's/self\.assertNotEqual$([^,]), (.)$$/assert \1 != \2/g' tests/.py sed -i ".bak" -E 's/self\.assertNotIn$([^,]), (.)$$/assert \1 not in \2/g' tests/.py sed -i ".bak" -E 's/self\.assertIsNone$(.)$$/assert \1 is None/g' tests/.py sed -i ".bak" -E 's/self\.assertIsNotNone$(.)$$/assert \1 is not None/g' tests/.py (Pytest gives nice informative error messages even if you just use the prettier form.) • The option -i means "do it in-place" (modify the file). Including ".bak" means "make backups of the old version with this extension". • I don't actually want the backups, but (for some odd reason) on my Mac, not asking for them changed how the regex was interpreted to something that's not right. • After reviewing and checking in the changes I wanted, I cleaned up the backups with git clean -f (careful you don't have any unchecked-in changes you want to keep!). Source code for this post is here. This post examines how a few statistical and machine learning models respond to a simple toy example where they're asked to make predictions on new regions of feature space. The key question the models will answer differently is whether there's an "interaction" between two features: does the influence of one feature differ depending on the value of another. In this case, the data won't provide information about whether there's an interaction or not. Interactions are often real and important, but in many contexts we treat interaction effects as likely to be small (without evidence otherwise). I'll walk through why decision trees and bagged ensembles of decision trees (random forests) can make the opposite assumption: they can strongly prefer an interaction, even when the evidence is equally consistent with including or not including an interaction. I'll look at point estimates from: • a linear model • decision trees and bagged decision trees (random forest), using R's randomForest package • boosted decision trees, using the R's gbm package I'll also look at two models that capture uncertainty about whether there's an interaction: • Bayesian linear model with an interaction term BART has the advantage of expressing uncertainty while still being a "machine learning" type model that learns interactions, non-linearities, etc. without the user having to decide which terms to include or the particular functional form. Whenever possible, I recommend using models like BART that explicitly allow for uncertainty. The Example Suppose you're given this data and asked to make a prediction at $X_1 = 0$, $X_2 = 1$ (where there isn't any training data): │X1│X2│ Y │N Training Rows: │ │0 │0 │ Y = 5 + noise │ 52 │ │1 │0 │Y = 15 + noise │ 23 │ │1 │1 │Y = 19 + noise │ 25 │ │0 │1 │ ? │ 0 │ (...click here for the rest of this post) A colleague at work recently pointed me to a wonderful stats.stackexchange answer with an intuitive explanation of covariance: For each pair of points, draw the rectangle with these points at opposite corners. Treat the rectangle's area as signed, with the same sign as the slope of the line between the two points. If you add up all of the areas, you have the (sample) covariance, up to a constant that depends only on the data set. Here's an example with 4 points. Each spot on the plot is colored by the sum corresponding to that point. For example, the dark space in the lower left has three "positively" signed rectangles going through it, but for the white space in the middle, one positive and one negative rectangle cancel out. In this next example, x and y are drawn from independent normals, so we have roughly an even amount of positive and negative: Formal Explanation The formal way to speak about multiple draws from a distribution is with a set of independent and identically distributed (i.i.d.) random variables. If we have a random variable X, saying that X[1], X[2],… are i.i.d means that they are all independent, but follow the same distribution. (...click here for the rest of this post) Simulated Knitting (post) I created a KnittedGraph class (subclassing of Python's igraph graph class) with methods corresponding to common operations performed while knitting: g = KnittedGraph() g.ConnectToZero() # join with the first stitch for a circular shape g.NewRow() # start a new row of stitches g.Increase() # two stitches in new row connect to one stitch in old I then embed the graphs in 3D space. Here's a hat I made this way: 2D Embeddings from Unsupervised Random Forests (1, 2) There are all sorts of ways to embed high-dimensional data in low dimensions for visualization. Here's one: 1. Given some set of high dimensional examples, build a random forest to distinguish examples from non-examples. 2. Assign similarities to pairs of examples based on how often they are in leaf nodes together. 3. Map examples to 2D in such a way that similarity decreases decreases with Euclidean 2D distance (I used multidimensional scaling for this). Here's the result of doing this on a set of diamond shapes I constructed. I like how it turned out: A Bayesian Model for a Function Increasing by Chi-Squared Jumps (in Stan) (post) In this paper, Andrew Gelman mentions a neat example where there's a big problem with a naive approach to putting a Bayesian prior on functions that are constrained to be increasing. So I thought about what sort of prior would make sense for such functions, and fit the models in Stan. I enjoyed Andrew's description of my attempt: "... it has a charming DIY flavor that might make you feel that you too can patch together a model in Stan to do what you need." Lissijous Curves JSFiddle Some JavaScript I wrote (using d3) to mimick what an oscilloscope I saw at the Exploratorium was doing: Visualization of the Weirstrass Elliptic Function as a Sum of Terms John Baez used this in his AMS blog Visual Insight.	{"url":"https://www.davidchudzicki.com/posts/","timestamp":"2024-11-04T20:25:32Z","content_type":"application/xhtml+xml","content_length":"326820","record_id":"<urn:uuid:21fb292f-e3a3-46e7-af09-13778258d29f>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00307.warc.gz"}
Towards the theory of reheating after inflation Reheating after inflation occurs due to particle production by the oscillating inflaton field. In this paper we briefly describe the perturbative approach to reheating, and then concentrate on effects beyond the perturbation theory. They are related to the stage of parametric resonance, which we call preheating. It may occur in an expanding universe if the initial amplitude of oscillations of the inflaton field is large enough. We investigate a simple model of a massive inflaton field φ coupled to another scalar field χ with the interaction term g^2φ^2χ^2. Parametric resonance in this model is very broad. It occurs in a very unusual stochastic manner, which is quite different from parametric resonance in the case when the expansion of the universe is neglected. Quantum fields interacting with the oscillating inflaton field experience a series of kicks which, because of the rapid expansion of the universe, occur with phases uncorrelated to each other. Despite the stochastic nature of the process, it leads to exponential growth of fluctuations of the field χ. We call this process stochastic resonance. We develop the theory of preheating taking into account the expansion of the universe and back reaction of produced particles, including the effects of rescattering. This investigation extends our previous study of reheating after inflation. We show that the contribution of the produced particles to the effective potential V(φ) is proportional not to φ^2, as is usually the case, but to \|φ\|. The process of preheating can be divided into several distinct stages. In the first stage the back reaction of created particles is not important. In the second stage back reaction increases the frequency of oscillations of the inflaton field, which makes the process even more efficient than before. Then the effects related to scattering of χ particles on the oscillating inflaton field terminate the resonance. We calculate the number density of particles n[χ] produced during preheating and their quantum fluctuations <χ^2> with all back reaction effects taken into account. This allows us to find the range of masses and coupling constants for which one can have efficient preheating. In particular, under certain conditions this process may produce particles with a mass much greater than the mass of the inflaton field. Physical Review D Pub Date: September 1997 □ 98.80.Cq; □ Particle-theory and field-theory models of the early Universe; □ High Energy Physics - Phenomenology; □ Astrophysics; □ General Relativity and Quantum Cosmology; □ High Energy Physics - Theory 41 pages, revtex, 12 figures. Some improvements and additions are made. This version is scheduled for publication in Phys. Rev. on Sep. 15	{"url":"https://ui.adsabs.harvard.edu/abs/1997PhRvD..56.3258K","timestamp":"2024-11-02T14:46:00Z","content_type":"text/html","content_length":"45638","record_id":"<urn:uuid:baed3d6a-3ab0-4691-b992-6d47e6edde98>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00638.warc.gz"}
Filter disturbances using univariate ARIMA or ARIMAX model Y = filter(Mdl,Z) returns the numeric array of one or more response series Y resulting from filtering the numeric array of one or more underlying disturbance series Z through the fully specified, univariate ARIMA model Mdl. Z is associated with the model innovations process that drives the specified ARIMA model. [Y,E,V] = filter(Mdl,Z) also returns numeric arrays of model innovations E and, when Mdl represents a composite conditional mean and variance model, conditional variances V, resulting from filtering the disturbance paths Z through the model Mdl. Tbl2 = filter(Mdl,Tbl1) returns the table or timetable Tbl2 containing the results from filtering the paths of disturbances in the input table or timetable Tbl1 through Mdl. The disturbance variable in Tbl1 is associated with the model innovations process through Mdl. (since R2023b) filter selects the variable Mdl.SeriesName, or the sole variable in Tbl1, as the disturbance variable to filter through the model. To select a different variable in Tbl1 to filter through the model, use the DisturbanceVariable name-value argument. [___] = filter(___,Name,Value) specifies options using one or more name-value arguments in addition to any of the input argument combinations in previous syntaxes. filter returns the output argument combination for the corresponding input arguments. For example, filter(Mdl,Z,Z0=PS,X=Pred) filters the numeric vector of disturbances Z through the ARIMAX Mdl, and specifies the numeric vector of presample disturbance data PS to initialize the model and the exogenous predictor data X for the regression component. Filter Vector of Disturbances Through Model Compute the impulse response function (IRF) of an ARMA model by filtering a vector of zeros, representing disturbances, through the model. Specify a mean zero ARMA(2,0,1) model. Mdl = arima(Constant=0,AR={0.5 -0.8},MA=-0.5, ... Simulate the first 20 responses of the IRF. Generate a disturbance series with a one-time, unit impulse, and then filter it. z = [1; zeros(19,1)]; y = filter(Mdl,z); y is a 20-by-1 response path resulting from filtering the disturbance path z through the model. y represents the IRF. The filter function requires one presample observation to initialize the model. By default, filter uses the unconditional mean of the process, which is 0. Normalize the IRF such that the first element is 1. Plot the impulse response function. title("Impulse Response") The impulse response assesses the dynamic behavior of a system to a one-time, unit impulse. Alternatively, you can use the impulse function to plot the IRF for an ARIMA process. Simulate and Filter Multiple Paths Filter a matrix of disturbance paths. Return the paths of responses and innovations, which drive the data-generating processes. Create a mean zero ARIMA(2,0,1) model. Mdl = arima(Constant=0,AR={0.5,-0.8},MA=-0.5, ... Generate 20 random, length 100 paths from the model. rng(1,"twister"); % For reproducibility [ySim,eSim,vSim] = simulate(Mdl,100,NumPaths=20); ySim, eSim, and vSim are 100-by-20 matrices of 20 simulated response, innovation, and conditional variance paths of length 100, respectively. Because Mdl does not have a conditional variance model, vSim is a matrix completely composed of the value of Mdl.Variance. Obtain disturbance paths by standardizing the simulated innovations. Filter the disturbance paths through the model. [yFil,eFil] = filter(Mdl,zSim); yFil and eFil are 100-by-20 matrices. The columns are independent paths generated from filtering corresponding disturbance paths in zSim through the model Mdl. Confirm that the outputs of simulate and filter are identical. sameE = norm(eSim - eFil) < eps sameY = norm(ySim - yFil) < eps The logical values 1 confirm the outputs are effectively identical. Filter Disturbance Path in Timetable Since R2023b Fit an ARIMA(1,1,1) model to the weekly average NYSE closing prices. Supply a timetable of data and specify the series for the fit. Then, filter randomly generated Gaussian noise paths through the estimated model to simulate responses and innovations. Load Data Load the US equity index data set Data_EquityIdx. load Data_EquityIdx T = height(DataTimeTable) The timetable DataTimeTable includes the time series variable NYSE, which contains daily NYSE composite closing prices from January 1990 through December 2001. Plot the daily NYSE price series. title("NYSE Daily Closing Prices: 1990 - 2001") Prepare Timetable for Estimation When you plan to supply a timetable, you must ensure it has all the following characteristics: • The selected response variable is numeric and does not contain any missing values. • The timestamps in the Time variable are regular, and they are ascending or descending. Remove all missing values from the timetable, relative to the NYSE price series. DTT = rmmissing(DataTimeTable,DataVariables="NYSE"); T_DTT = height(DTT) Because all sample times have observed NYSE prices, rmmissing does not remove any observations. Determine whether the sampling timestamps have a regular frequency and are sorted. areTimestampsRegular = isregular(DTT,"days") areTimestampsRegular = logical areTimestampsSorted = issorted(DTT.Time) areTimestampsSorted = logical areTimestampsRegular = 0 indicates that the timestamps of DTT are irregular. areTimestampsSorted = 1 indicates that the timestamps are sorted. Business day rules make daily macroeconomic measurements Remedy the time irregularity by computing the weekly average closing price series of all timetable variables. DTTW = convert2weekly(DTT,Aggregation="mean"); areTimestampsRegular = isregular(DTTW,"weeks") areTimestampsRegular = logical DTTW is regular. title("NYSE Daily Closing Prices: 1990 - 2001") Create Model Template for Estimation Suppose that an ARIMA(1,1,1) model is appropriate to model NYSE composite series during the sample period. Create an ARIMA(1,1,1) model template for estimation. Set the response series name to NYSE. Mdl = arima(1,1,1); Mdl.SeriesName = "NYSE"; Mdl is a partially specified arima model object. Fit Model to Data Fit an ARIMA(1,1,1) model to weekly average NYSE closing prices. Specify the entire series. EstMdl = estimate(Mdl,DTTW); ARIMA(1,1,1) Model (Gaussian Distribution): Value StandardError TStatistic PValue ________ _____________ __________ ___________ Constant 0.86386 0.46496 1.8579 0.06318 AR{1} -0.37582 0.22719 -1.6542 0.09809 MA{1} 0.47221 0.21741 2.172 0.029858 Variance 55.89 1.832 30.507 2.1199e-204 EstMdl is a fully specified, estimated arima model object. By default, estimate backcasts for the required Mdl.P = 2 presample responses. Filter Random Gaussian Disturbance Paths Generate 2 random, independent series of length T_DTTW from the standard Gaussian distribution. Store the matrix of series as one variable in DTTW. rng(1,"twister") % For reproducibility DTTW.Z = randn(T_DTTW,2); DTTW contains a new variable called Z containing a T_DTTW-by-2 matrix of two disturbance paths. Filter the paths of disturbances through the estimated ARIMA model. Specify the table variable name containing the disturbance paths. Tbl2 = filter(EstMdl,DTTW,DisturbanceVariable="Z"); Time NYSE NASDAQ Z NYSE_Response NYSE_Innovation NYSE_Variance ___________ ______ ______ _____________________ ________________ ___________________ ______________ 16-Nov-2001 577.11 1886.9 -1.8948 0.41292 358.78 433.57 -14.166 3.087 55.89 55.89 23-Nov-2001 583 1898.3 1.3583 0.27051 367.95 436.63 10.155 2.0223 55.89 55.89 30-Nov-2001 581.41 1925.8 -0.9118 1.1119 363.35 445.61 -6.8165 8.3125 55.89 55.89 07-Dec-2001 584.96 1998.1 -0.14964 -2.418 361.61 428.95 -1.1187 -18.077 55.89 55.89 14-Dec-2001 574.03 1981 -0.40114 0.98498 359.6 434.9 -2.9989 7.3636 55.89 55.89 21-Dec-2001 582.1 1967.9 -0.57758 0.0039243 355.48 437.04 -4.318 0.029338 55.89 55.89 28-Dec-2001 590.28 1967.2 2.0039 -0.92415 370.84 430.2 14.981 -6.9089 55.89 55.89 04-Jan-2002 589.8 1950.4 -0.50964 -0.43856 369.19 427.09 -3.8101 -3.2787 55.89 55.89 Tbl2 is a 627-by-6 timetable containing all variables in DTTW, and the two filtered response paths NYSE_Response, innovation paths NYSE_Innovation, and constant variance paths NYSE_Variance (Mdl.Variance = 55.89). Supply Presample Responses Assess the dynamic behavior of a system to a persistent change in a variable by plotting a step response. Supply presample responses to initialize the model. Specify a mean zero ARIMA(2,0,1) process. Mdl = arima(Constant=0,AR={0.5 -0.8},MA=-0.5, ... Simulate the first 20 responses to a sequence of unit disturbances. Generate a disturbance series of ones, and then filter it. Set all presample observations equal to zero. Z = ones(20,1); Y = filter(Mdl,Z,Y0=zeros(Mdl.P,1)); Y = Y/Y(1); The last step normalizes the step response function to ensure that the first element is 1. Plot the step response function. title("Step Response") Simulate Responses from ARIMAX Model Create models for the response and predictor series. Set an ARIMAX(2,1,3) model to the response MdlY, and an AR(1) model to the MdlX. MdlY = arima(AR={0.1 0.2},D=1,MA={-0.1 0.1 0.05}, ... MdlX = arima(AR=0.5,Constant=0,Variance=0.1); Simulate a length 100 predictor series x and a series of iid normal disturbances z having mean zero and variance 1. z = randn(100,1); x = simulate(MdlX,100); Filter the disturbances z using MdlY to produce the response series y. Plot y. y = filter(MdlY,z,X=x); Filter Disturbances Through Composite Conditional Mean and Variance Model Create the composite AR(1)/GARCH(1,1) model $\begin{array}{l}{y}_{t}=1+0.5{y}_{t-1}+{\epsilon }_{t}\\ {\epsilon }_{t}={\sigma }_{t}{z}_{t}\\ {\sigma }_{t}^{2}=0.2+0.1{\sigma }_{t-1}^{2}+0.05{\epsilon }_{t-1}^{2}\\ {z}_{t}\sim N\left(0,1\ Create the composite model. CVMdl = garch(Constant=0.2,GARCH=0.1,ARCH=0.05); Mdl = arima(Constant=1,AR=0.5,Variance=CVMdl) Mdl = arima with properties: Description: "ARIMA(1,0,0) Model (Gaussian Distribution)" SeriesName: "Y" Distribution: Name = "Gaussian" P: 1 D: 0 Q: 0 Constant: 1 AR: {0.5} at lag [1] SAR: {} MA: {} SMA: {} Seasonality: 0 Beta: [1×0] Variance: [GARCH(1,1) Model] Mdl is an arima object. The property Mdl.Variance contains a garch object that represents the conditional variance model. Generate a random series of 100 standard Gaussian of disturbances. rng(1,"twister") % For reproducibility z = randn(100,1); Filter the disturbances through the model. Return and plot the simulated conditional variances. [y,e,v] = filter(Mdl,z); Input Arguments Z — Disturbance series paths z[t] numeric column vector \| numeric matrix Underlying disturbance paths z[t], specified as a numobs-by-1 numeric column vector or numobs-by-numpaths numeric matrix. numObs is the length of the time series (sample size). numpaths is the number of separate, independent disturbance paths. z[t] drives the innovation process ε[t]. For a variance process σ[t]^2, the innovation process is ${\epsilon }_{t}={\sigma }_{t}{z}_{t}.$ Each row corresponds to a sampling time. The last row contains the latest set of disturbances. Each column corresponds to a separate, independent path of disturbances. filter assumes that disturbances across any row occur simultaneously. Z is the continuation of the presample disturbances Z0. Data Types: double Tbl1 — Time series data table \| timetable Since R2023b Time series data containing the observed disturbance variable z[t], associated with the model innovations process ε[t], and, optionally, predictor variables x[t], specified as a table or timetable with numvars variables and numobs rows. You can optionally select the disturbance variable or numpreds predictor variables by using the DisturbanceVariable or PredictorVariables name-value arguments, For a variance process σ[t]^2, the innovation process is ${\epsilon }_{t}={\sigma }_{t}{z}_{t}.$ Each row is an observation, and measurements in each row occur simultaneously. The selected disturbance variable is a single path (numobs-by-1 vector) or multiple paths (numobs-by-numpaths matrix) of numobs observations of disturbance data. Each path (column) of the selected disturbance variable is independent of the other paths, but path j of all presample and in-sample variables correspond, for j = 1,…,numpaths. Each selected predictor variable is a numobs-by-1 numeric vector representing one path. The filter function includes all predictor variables in the model when it filters each disturbance path. Variables in Tbl1 represent the continuation of corresponding variables in Presample. If Tbl1 is a timetable, it must represent a sample with a regular datetime time step (see isregular), and the datetime vector Tbl1.Time must be strictly ascending or descending. If Tbl1 is a table, the last row contains the latest observation. Name-Value Arguments Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter. Before R2021a, use commas to separate each name and value, and enclose Name in quotes. Example: filter(Mdl,Z,Z0=PS,X=Pred) specifies the numeric vector of presample disturbance data PS to initialize the model and the exogenous predictor data X for the regression component. Y0 — Presample response data y[t] numeric column vector \| numeric matrix Presample response data y[t] to initialize the model, specified as a numpreobs-by-1 numeric column vector or a numpreobs-by-numprepaths numeric matrix. Use Y0 only when you supply the numeric array of disturbance data Z. numpreobs is the number of presample observations. numprepaths is the number of presample response paths. Each row is a presample observation (sampling time), and measurements in each row occur simultaneously. The last row contains the latest presample observation. numpreobs must be at least Mdl.P to initialize the AR model component. If numpreobs > Mdl.P, filter uses the latest required observations only. Columns of Y0 are separate, independent presample paths. The following conditions apply: • If Y0 is a column vector, it represents a single response path. filter applies it to each output path. • If Y0 is a matrix, each column represents a presample response path. filter applies Y0(:,j) to initialize path j. numprepaths must be at least numpaths. If numprepaths > numpaths, filter uses the first size(Z,2) columns only. By default, filter sets any necessary presample responses to one of the following values: • The unconditional mean of the model when Mdl represents a stationary AR process without a regression component • Zero when Mdl represents a nonstationary process or when it contains a regression component Data Types: double Z0 — Presample disturbance data z[t] numeric column vector \| numeric matrix Presample disturbance data z[t] providing initial values for the input disturbance series Z, specified as a numpreobs-by-1 numeric column vector or a numpreobs-by-numprepaths numeric matrix. Use Z0 only when you supply the numeric array of disturbance data Z. Each row is a presample observation (sampling time), and measurements in each row occur simultaneously. The last row contains the latest presample observation. numpreobs must be at least Mdl.Q to initialize the MA model component. If Mdl.Variance is a conditional variance model (for example, a garch model object), filter can require more rows than Mdl.Q. If numpreobs is larger than required, filter uses the latest required observations only. Columns of Z0 are separate, independent presample paths. The following conditions apply: • If Z0 is a column vector, it represents a single disturbance path. filter applies it to each output path. • If Z0 is a matrix, each column represents a presample disturbance path. filter applies Z0(:,j) to initialize path j. numprepaths must be at least numpaths. If numprepaths > numpaths, filter uses the first size(Z,2) columns only. By default, filter sets the necessary presample disturbances to zero. Data Types: double V0 — Presample conditional variance data σ[t]^2 positive numeric column vector \| positive numeric matrix Presample conditional variance data σ[t]^2 used to initialize the conditional variance model, specified as a numpreobs-by-1 positive numeric column vector or a numpreobs-by-numprepaths positive numeric matrix. If the conditional variance Mdl.Variance is constant, filter ignores V0. Use V0 only when you supply the numeric array of disturbance data Z. Each row is a presample observation (sampling time), and measurements in each row occur simultaneously. The last row contains the latest presample observation. numpreobs must be at least Mdl.Q to initialize the conditional variance model in Mdl.Variance. For details, see the filter function of conditional variance models. If numpreobs is larger than required, filter uses the latest required observations only. Columns of V0 are separate, independent presample paths. The following conditions apply: • If V0 is a column vector, it represents a single path of conditional variances. filter applies it to each output path. • If V0 is a matrix, each column represents a presample path of conditional variances. filter applies V0(:,j) to initialize path j. numprepaths must be at least numpaths. If numprepaths > numpaths, filter uses the first size(Z,2) columns only. By default, filter sets all necessary presample conditional variances to the unconditional variance of the conditional variance process. Data Types: double Presample — Presample data table \| timetable Since R2023b Presample data containing paths of response y[t], disturbance z[t], or conditional variance σ[t]^2 series to initialize the model, specified as a table or timetable, the same type as Tbl1, with numprevars variables and numpreobs rows. Use Presample only when you supply a table or timetable of data Tbl1. Each selected variable is a single path (numpreobs-by-1 vector) or multiple paths (numpreobs-by-numprepaths matrix) of numpreobs observations representing the presample of the response, disturbance, or conditional variance series for DisturbanceVariable, the selected disturbance variable in Tbl1. Each row is a presample observation, and measurements in each row occur simultaneously. numpreobs must be one of the following values: • At least Mdl.P when Presample provides only presample responses • At least Mdl.Q when Presample provides only presample disturbances or conditional variances • At least max([Mdl.P Mdl.Q]) otherwise When Mdl.Variance is a conditional variance model, filter can require more than the minimum required number of presample values. If you supply more rows than necessary, filter uses the latest required number of observations only. If Presample is a timetable, all the following conditions must be true: • Presample must represent a sample with a regular datetime time step (see isregular). • The inputs Tbl1 and Presample must be consistent in time such that Presample immediately precedes Tbl1 with respect to the sampling frequency and order. • The datetime vector of sample timestamps Presample.Time must be ascending or descending. If Presample is a table, the last row contains the latest presample observation. By default, filter sets the following values: • For necessary presample responses: □ The unconditional mean of the model when Mdl represents a stationary AR process without a regression component □ Zero when Mdl represents a nonstationary process or when it contains a regression component. • For necessary presample disturbances, zero. • For necessary presample conditional variances, the unconditional variance of the conditional variance model n Mdl.Variance. If you specify the Presample, you must specify the presample response, disturbance, or conditional variance name by using the PresampleResponseVariable, PresampleDisturbanceVariable, or PresampleVarianceVariable name-value argument. PresampleDisturbanceVariable — Disturbance variable z[t] to select from Presample string scalar \| character vector \| integer \| logical vector Since R2023b Disturbance variable z[t] to select from Presample containing presample disturbance data, specified as one of the following data types: • String scalar or character vector containing a variable name in Presample.Properties.VariableNames • Variable index (positive integer) to select from Presample.Properties.VariableNames • A logical vector, where PresampleDisturbanceVariable(j) = true selects variable j from Presample.Properties.VariableNames The selected variable must be a numeric matrix and cannot contain missing values (NaNs). If you specify presample disturbance data by using the Presample name-value argument, you must specify PresampleDisturbanceVariable. Example: PresampleDisturbanceVariable="StockRateDist0" Example: PresampleDisturbanceVariable=[false false true false] or PresampleDisturbanceVariable=3 selects the third table variable as the presample disturbance variable. Data Types: double \| logical \| char \| cell \| string X — Exogenous predictor data numeric matrix Exogenous predictor data for the regression component in the model, specified as a numeric matrix with numpreds columns. numpreds is the number of predictor variables (numel(Mdl.Beta)). Use X only when you supply the numeric array of disturbance data Z. X must have at least numobs rows. The last row contains the latest predictor data. If X has more than numobs rows, filter uses only the latest numobs rows. Each row of X corresponds to each period in Z (period for which filter filters errors; the period after the presample). filter does not use the regression component in the presample period. Columns of X are separate predictor variables. filter applies X to each filtered path; that is, X represents one path of observed predictors. By default, filter excludes the regression component, regardless of its presence in Mdl. Data Types: double • NaN values in Z, X, Y0, Z0, and V0 indicate missing values. filter removes missing values from specified data by list-wise deletion. □ For the presample, filter horizontally concatenates the possibly jagged arrays Y0, Z0, and V0 with respect to the last rows, and then it removes any row of the concatenated matrix containing at least one NaN. □ For in-sample data, filter horizontally concatenates the possibly jagged arrays Z and X, and then it removes any row of the concatenated matrix containing at least one NaN. This type of data reduction reduces the effective sample size and can create an irregular time series. • For numeric data inputs, filter assumes that you synchronize the presample data such that the latest observations occur simultaneously. • filter issues an error when any table or timetable input contains missing values. Output Arguments Y — Simulated response paths y[t] numeric column vector \| numeric matrix Simulated response paths y[t], returned as a length numobs column vector or a numobs-by-numpaths numeric matrix. filter returns Y only when you supply the input Z. For each t = 1, …, numobs, the simulated response at time t Y(t,:) corresponds to the filtered disturbance at time t Z(t,:) and response path j Y(:,j) corresponds to the filtered disturbance path j Z Y represents the continuation of the presample response paths in Y0. E — Simulated paths of model innovations ε[t] numeric column vector \| numeric matrix Simulated paths of model innovations ε[t], returned as a length numobs column vector or a numobs-by-numpaths numeric matrix. filter returns E only when you supply the input Z. The dimensions of Y and E correspond. Columns of E are scaled disturbance paths (innovations) such that, for a particular path ${\epsilon }_{t}={\sigma }_{t}{z}_{t}.$ V — Conditional variance paths σ[t]^2 numeric column vector \| numeric matrix Conditional variance paths σ[t]^2, returned as a length numobs column vector or numobs-by-numpaths numeric matrix. filter returns V only when you supply the input Z. The dimensions of Y and V If Z is a matrix, then the columns of V are the filtered conditional variance paths corresponding to the columns of Z. Columns of V are conditional variance paths of corresponding paths of innovations ε[t] (E) such that, for a particular path ${\epsilon }_{t}={\sigma }_{t}{z}_{t}.$ V represents the continuation of the presample conditional variance paths in V0. Tbl2 — Simulated response y[t], innovation ε[t], and conditional variance σ[t]^2 paths table \| timetable Since R2023b Simulated response y[t], innovation ε[t], and conditional variance σ[t]^2 paths, returned as a table or timetable, the same data type as Tbl1. filter returns Tbl2 only when you supply the input Tbl1. Tbl2 contains the following variables: • The simulated response paths, which are in a numobs-by-numpaths numeric matrix, with rows representing observations and columns representing independent paths, each corresponding to the input observations and paths of the disturbance variable in Tbl1. filter names the simulated response variable in Tbl2 responseName_Response, where responseName is Mdl.SeriesName. For example, if Mdl.SeriesName is StockReturns, Tbl2 contains a variable for the corresponding simulated response paths with the name StockReturns_Response. • The simulated innovation paths, which are in a numobs-by-numpaths numeric matrix, with rows representing observations and columns representing independent paths, each corresponding to the input observations and paths of the disturbance variable in Tbl1. filter names the simulated innovation variable in Tbl2 responseName_Innovation, where responseName is Mdl.SeriesName. For example, if Mdl.SeriesName is StockReturns, Tbl2 contains a variable for the corresponding simulated innovation paths with the name StockReturns_Innovation. • The simulated conditional variances paths, which are in a numobs-by-numpaths numeric matrix, with rows representing observations and columns representing independent paths, each corresponding to the input observations and paths of the disturbance variable in Tbl1. filter names the simulated conditional variance variable in Tbl2 responseName_Variance, where responseName is Mdl.SeriesName. For example, if Mdl.SeriesName is StockReturns, Tbl2 contains a variable for the corresponding simulated conditional variance paths with the name StockReturns_Variance. • All variables Tbl1. If Tbl1 is a timetable, row times of Tbl1 and Tbl2 are equal. Alternative Functionality filter generalizes simulate; both functions filter a series of disturbances to produce output responses, innovations, and conditional variances. However, simulate autogenerates a series of mean zero, unit variance, independent and identically distributed (iid) disturbances according to the distribution in Mdl. In contrast, filter enables you to directly specify custom disturbances. [1] Box, George E. P., Gwilym M. Jenkins, and Gregory C. Reinsel. Time Series Analysis: Forecasting and Control. 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1994. [2] Enders, Walter. Applied Econometric Time Series. Hoboken, NJ: John Wiley & Sons, Inc., 1995. [3] Hamilton, James D. Time Series Analysis. Princeton, NJ: Princeton University Press, 1994. Version History Introduced in R2012b R2023b: filter accepts input data in tables and timetables, and returns results in tables and timetables In addition to accepting input data (in-sample and presample) in numeric arrays, filter accepts input data in tables or regular timetables. When you supply data in a table or timetable, the following conditions apply: • filter chooses the default in-sample disturbance series and predictor data on which to operate, but you can use the specified optional name-value argument to select a different series. • If you specify optional presample data to initialize the model, you must also specify the presample response, disturbance, or conditional variance series name. • filter returns results in a table or timetable. Name-value arguments to support tabular workflows include: • DisturbanceVariable specifies the name of the disturbance series to select from the input data to filter through the model. • Presample specifies the input table or timetable of presample response, disturbance, and conditional variance data. • PresampleResponseVariable specifies the name of the response series to select from Presample. • PresampleDisturbanceVariable specifies the name of the disturbance series to select from Presample. • PresampleVarianceVariable specifies the name of the conditional variance series to select from Presample. • PredictorVariables specifies the names of the predictor series to select from the input data for a model regression component.	{"url":"https://au.mathworks.com/help/econ/arima.filter.html","timestamp":"2024-11-04T21:32:34Z","content_type":"text/html","content_length":"194608","record_id":"<urn:uuid:d45ed24e-ad67-4632-9b82-331a75c0772f>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00399.warc.gz"}
Transfer Function Models Transfer function models Transfer function models describe the relationship between the inputs and outputs of a system using a ratio of polynomials. The model order is equal to the order of the denominator polynomial. The roots of the denominator polynomial are referred to as the model poles. The roots of the numerator polynomial are referred to as the model zeros. The parameters of a transfer function model are its poles, zeros, and transport delays. In continuous time, a transfer function model has the following form: Here, Y(s), U(s), and E(s) represent the Laplace transforms of the output, input, and noise, respectively. num(s) and den(s) represent the numerator and denominator polynomials that define the relationship between the input and the output. For more information, see What Are Transfer Function Models? System Identification Identify models of dynamic systems from measured data Create Transfer Function Model idtf Transfer function model with identifiable parameters tfest Estimate transfer function model pem Prediction error minimization for refining linear and nonlinear models spectrumest Estimate transfer function model for power spectrum data (Since R2022b) Model Initialization and Structure Parameters Extract or Set Model Parameters tfdata Access transfer function data getpvec Obtain model parameters and associated uncertainty data setpvec Modify values of model parameters getpar Obtain attributes such as values and bounds of linear model parameters setpar Set attributes such as values and bounds of linear model parameters addMinPhase Add minimum phase to frequency response magnitude (Since R2022b) Set Transfer Function Model Options Featured Examples	{"url":"https://uk.mathworks.com/help/ident/transfer-function-models.html?s_tid=CRUX_lftnav","timestamp":"2024-11-05T13:17:58Z","content_type":"text/html","content_length":"83288","record_id":"<urn:uuid:2a9ddf2c-801c-4b2c-83db-b94f0082d56c>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00706.warc.gz"}
Abrupt categories induced by categories with star-morphisms - Math Research of Victor PortonAbrupt categories induced by categories with star-morphisms In this blog post I introduced the notion of category with star-morphisms, a generalization of categories which have aroused in my research. Each star category gives rise to a category (abrupt category, see a remark below why I call it “abrupt”), as described below. Below for simplicity I assume that the set $latex {M}&fg=000000$ and the set of our indexed families of functions are disjoint. The general case (when they are not necessarily disjoint) may be easily elaborated by the reader. • Objects are indexed (by $latex {\mathrm{arity}m}&fg=000000$ for some $latex {m \in M}&fg=000000$) families of objects of the category $latex {C}&fg=000000$ and an (arbitrarily choosen) object $latex {\mathrm{None}}&fg=000000$ not in this set • There are the following disjoint sets of morphims: 1. indexed (by $latex {\mathrm{arity} m}&fg=000000$ for some $latex {m \in M}&fg=000000$) families of morphisms of $latex {C}&fg=000000$ 2. elements of $latex {M}&fg=000000$ 3. the identity morphism $latex {\mathrm{id}_{\mathrm{None}}}&fg=000000$ on $latex {\mathrm{None}}&fg=000000$ • Source and destination of morphims are defined by the formulas: □ $latex {\mathrm{Src}f = \lambda i \in \mathrm{dom}f : \mathrm{Src}f_i}&fg=000000$ □ $latex {\mathrm{Dst}f = \lambda i \in \mathrm{dom}f : \mathrm{Dst}f_i}&fg=000000$ □ $latex {\mathrm{Src}m =\mathrm{None}}&fg=000000$ □ $latex {\mathrm{Dst}m =\mathrm{Obj}_m}&fg=000000$. • Compositions of morphisms are defined by the formulas: □ $latex {g \circ f = \lambda i \in \mathrm{dom}f : g_i \circ f_i}&fg=000000$ □ $latex {f \circ m =\mathrm{StarProd} \left( m ; f \right)}&fg=000000$ □ $latex {m \circ \mathrm{id}_{\mathrm{None}} = m}&fg=000000$ • Identity morphisms for an object $latex {X}&fg=000000$ are: □ $latex {\lambda i \in X : \mathrm{id}_{X_i}}&fg=000000$ if $latex {X \neq \mathrm{None}}&fg=000000$ □ $latex {\mathrm{id}_{\mathrm{None}}}&fg=000000$ if $latex {X =\mathrm{None}}&fg=000000$ We need to prove it is really a category. Proof We need to prove: 1. Composition is associative 2. Composition with identities complies with the identity law. 1. $latex {\left( h \circ g \right) \circ f = \lambda i \in \mathrm{dom} f : \left( h_i \circ g_i \right) \circ f_i = \lambda i \in \mathrm{dom} f : h_i \circ \left( g_i \circ f_i \right) = h \circ \left( g \circ f \right)}&fg=000000$; $latex g \circ \left( f \circ m \right) = \mathrm{StarComp} \left( \mathrm{StarComp} \left( m ; f \right) ; g \right) = \\ \mathrm{StarComp} \left( m ; \lambda i \in \mathrm{arity} m : g_i \circ f_i \right) = \mathrm{StarComp} \left( m ; g \circ f \right) = \left( g \circ f \right) \circ m &fg=000000$; $latex {f \ circ \left( m \circ \mathrm{id}_{\mathrm{None}} \right) = f \circ m = \left( f \circ m \right) \circ \mathrm{id}_{\mathrm{None}}}&fg=000000$. 2. $latex {m \circ \mathrm{id}_{\mathrm{None}} = m}&fg=000000$; $latex {\mathrm{id}_{\mathrm{Dst} m} \circ m = \mathrm{StarComp} \left( m ; \lambda i \in \mathrm{arity} m : \mathrm{id}_{\mathrm{Obj} _m i} \right) = m}&fg=000000$. Remark I call the above defined category abrupt category because (excluding identity morphisms) it allows composition with an $latex m\in M$ only on the left (not on the right) so that the morphism $latex m$ is “abrupt” on the right.	{"url":"https://math.portonvictor.org/2012/06/13/tower-categories/","timestamp":"2024-11-10T21:38:02Z","content_type":"text/html","content_length":"101077","record_id":"<urn:uuid:eac256bc-b277-42c2-87e2-0fd4d2c7a735>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00559.warc.gz"}
Equivalent Expressions 6th Grade Alabama Course of Study Standards: 16 Generate equivalent algebraic expressions using the properties of operations, including inverse, identity, commutative, associative, and distributive. Arizona Academic Standards: 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x. Common Core State Standards: Math.6.EE.3 or 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. Georgia Standards of Excellence (GSE): 6.PAR.6.5 Apply the properties of operations to identify and generate equivalent expressions. North Carolina - Standard Course of Study: 6.EE.3 Apply the properties of operations to generate equivalent expressions without exponents. New York State Next Generation Learning Standards: 6.EE.3 Apply the properties of operations to generate equivalent expressions. e.g., Apply the distributive property to the expression 3(2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. Tennessee Academic Standards: 6.EE.A.3 Apply the properties of operations (including, but not limited to, commutative, associative, and distributive properties) to generate equivalent expressions. The distributive property is prominent here. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. Wisconsin Academic Standards: 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. Alabama Course of Study Standards: 17 Determine whether two expressions are equivalent and justify the reasoning. Arizona Academic Standards: 6.EE.A.4 Identify when two expressions are equivalent. For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. Common Core State Standards: Math.6.EE.4 or 6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. Georgia Standards of Excellence (GSE): 6.PAR.7.1 Solve one-step equations and inequalities involving variables when values for the variables are given. Determine whether an equation and inequality involving a variable is true or false for a given value of the variable. North Carolina - Standard Course of Study: 6.EE.4 Identify when two expressions are equivalent and justify with mathematical reasoning. New York State Next Generation Learning Standards: 6.EE.4 Identify when two expressions are equivalent. e.g., The expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y represents. Tennessee Academic Standards: 6.EE.A.4 Identify when expressions are equivalent (i.e., when the expressions name the same number regardless of which value is substituted into them). For example, the expression 5b + 3b is equivalent to (5 +3) b, which is equivalent to 8b. Wisconsin Academic Standards: 6.EE.A.4 Identify when two expressions are equivalent (e.g., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. Pennsylvania Core Standards: CC.2.2.6.B.1 Apply and extend previous understandings of arithmetic to algebraic expressions Pennsylvania Core Standards: M06.B-E.1.1.5 Apply the properties of operations to generate equivalent expressions. Florida - Benchmarks for Excellent Student Thinking: MA.6.AR.1.4 Apply the properties of operations to generate equivalent algebraic expressions with integer coefficients. Georgia Standards of Excellence (GSE): 6.PAR.6.5 Apply the properties of operations to identify and generate equivalent expressions.	{"url":"https://www.learningfarm.com/web/practicePassThrough.cfm?TopicID=1305","timestamp":"2024-11-13T22:18:04Z","content_type":"application/xhtml+xml","content_length":"49897","record_id":"<urn:uuid:e6dcef4c-ce58-426e-8faa-5d55b7abc61b>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00740.warc.gz"}
Third-grade math Curriculum Third-grade math Curriculum is pretty cool! You get to help young learners dive into some exciting stuff. We’re talking about numbers and operations, like adding and subtracting big numbers, getting into multiplication and division, and even dipping our toes into fractions. Then there’s geometry – you know, shapes and angles. In 3rd Grade, we tackle measurements and data, too, like telling time and dealing with lengths and stuff. Of course, we love to teach all this by mixing in some fun math games and activities. It’s all about making math enjoyable for our kiddos, right? Let’s take a look at what to cover in this crucial year! Key Math Concepts and Skills: 1. Number and Operations: • Understanding place value up to thousands. • Addition and subtraction of multi-digit numbers. • Multiplication facts and strategies. • Division facts and basic division concepts. • Rounding to the nearest ten and hundred. • Recognizing and generating equivalent fractions. • Comparing and ordering fractions with like denominators. • Adding and subtracting fractions with like denominators. • Measurement and data interpretation. 2. Third Grade Math Curriculum Geometry: • Identifying and classifying two-dimensional shapes (e.g., polygons, quadrilaterals, triangles). • Understanding lines, angles, and perpendicular lines. • Measuring and drawing angles. • Identifying and understanding the properties of three-dimensional shapes (e.g., cubes, spheres, cylinders). 3. Measurement and Data: • Telling time to the nearest minute. • Measuring and comparing lengths using standard units. • Understanding the concepts of area and perimeter. • Collecting and interpreting data using tables and graphs. • Solving problems involving time, money, and measurement. 4. Fractions: • Recognizing and identifying unit fractions. • Exploring fractions on a number line. • Comparing and ordering fractions with different denominators. • Adding and subtracting fractions with different denominators. • Understanding the concept of equivalent fractions. 5. Patterns and Algebra: • Understanding multiplication as repeated addition. • Recognizing, extending, and creating patterns. • Solving problems involving multiplication and division. • Using multiplication and division to solve word problems. • Introduction to basic concepts of multiplication and division. 6. Problem Solving: • Developing problem-solving strategies. • Solving real-world mathematical problems. • Communicating mathematical thinking and reasoning. • Applying critical thinking skills to mathematical situations. Third-Grade Math Curriculum Teaching Strategies: Games, activities and so much more. • Hands-on activities and manipulatives to reinforce concepts. • Group work and collaborative problem-solving. • Math games and puzzles to make learning fun. • Real-world application of math concepts. • Differentiated instruction to meet the needs of all learners. • Regular formative assessments to monitor student progress. Assessment and Evaluation: • Regular quizzes and tests to gauge understanding. • Homework assignments to practice and reinforce skills. • Performance tasks and projects to assess application of concepts. • Teacher observations and conferences with students. • Use of rubrics and grading criteria to provide feedback. Third-grade math is an exciting journey of discovery for students. Teachers play a vital role in fostering a strong mathematical foundation by providing a balanced and engaging curriculum that encourages problem-solving, critical thinking, and a love for math. It is essential to support all students in their mathematical growth by differentiating instruction and addressing individual needs while ensuring they have a solid understanding of foundational math concepts. Let’s teach!	{"url":"https://funtoteach.com/2023/09/third-grade-math-curriculum/","timestamp":"2024-11-13T08:58:07Z","content_type":"text/html","content_length":"111240","record_id":"<urn:uuid:a9610e6e-a686-4933-aa71-4d8e31319cc9>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00718.warc.gz"}
In a weakly supervised multi-label classification (WSML) task, labels are given as a form of partial label, which means only a small amount of categories is annotated per image. This setting reflects the recently released large-scale multi-label datasets, e.g. JFT-300M or InstagramNet-1B, which provide only partial label. Thus, it is becoming increasingly important to develop learning strategies with partial labels. Target with Assume Negative Let us define an input $x \in X$ and a target $y \in Y$ where $X$ and $Y$ compose a dataset $D$. In a weakly supervised multi-label learning for image classification task, $X$ is an image set and $Y = \{0,1,u\}^K$ where $u$ is an annotation of `unknown', i.e. unobserved label, and $K$ is the number of categories. For the target $y$, let $S^{p}=\{i\|y_i=1\}$, $S^{n}=\{i\|y_i=0\}$, and $S^{u}=\{i\| y_i=u\}$. In a partial label setting, small amount of labels are known, thus $\|S^{p}\| + \|S^{n}\| < K$. We start our method with Assume Negative (AN) where all the unknown labels are regarded as negative. We call this modified target as $y^{AN}$, $y^{AN}_i = \begin{cases} 1, & i \in S^{p}\\ 0, & i \in S^{n} \cup S^{u} , \end{cases}$ and the set of all $y^{AN}$ as $Y^{AN}$. $\{y_i^{AN} \| i \in S^{p}\}$ and $\{y_i^{AN} \| i \in S^{n}\}$ are the set where each element is true positive and true negative, respectively. $\{y_i^{AN} \| i \in S^{u}\}$ contains both true negative and false negative. % and $\{y_i^{AN} \| i \in S^{u}\}$ are true negative and false negative, respectively. % Therefore, $y_i^{AN} = 0$ would be either true negative or false negative. % Note that in WSML with AN target, there are no labels with false positive. The naive way of training the model $f$ with the dataset $D^{\prime} = (X, Y^{AN})$ is to minimize the loss function $L$, $L = \frac{1}{\|D^{\prime}\|} \sum_{y^{AN} \in D^{\prime}} \frac{1}{K} \sum_{i=1}^{K} \mathrm{BCELoss} (f(x)_i, y_i^{AN}) ,$ where $f(\cdot) \in [0,1]^{K}$ and $\mathrm{BCELoss}(\cdot, \cdot)$ is the binary cross entropy loss between the function output and the target. We call this naive method as Naive AN. Memorization effect in WSML We observe that a memorization effect occurs in WSML when the model is trained with the dataset with AN target. To confirm this, we make the following experimental setting. We convert Pascal VOC 2012 dataset into partial label one by randomly remaining only one positive label for each image and regard other labels as unknown (dataset $D$). These unknown labels are then assumed as negative (dataset $D^{\prime}$). We train ResNet-50 model with $D^{\prime}$ using the loss function $L$ in the equation above. We look at the trend of loss value corresponding to each label $y_i^{AN}$ in a training dataset while the model is trained. A single example for true negative label and false negative label is shown in the above figure. For a true negative label, the corresponding loss value keeps decreasing as the number of iteration increases (blue line). Meanwhile, the loss of a false negative label slightly increases in the initial learning phase, and then reaches the highest in the middle phase followed by decreasing to reach near $0$ at the end (red line). This implies that the model starts to memorize the wrong label from the middle phase. In this section, we propose novel methods for WSML motivated from the ideas of noisy multi-class learning which ignores the large loss during training the model. Remind that in WSML with AN target, the model starts memorizing the false negative label in the middle of the training with having a large loss at that time. While we can only observe that the label in the set $\{y_i^{AN}\| i \in S^{u} \}$ is negative and cannot explicitly discriminate whether it is false or true, we are able to implicitly distinguish between them. It is because the loss from false negative is likely to be larger than the loss from true negative before memorization starts. Therefore, we manipulate the label in the set $\{y_i^{AN}\| i \in S^{u}\}$ that corresponds to the large loss value during the training process to prevent the model from memorizing false negative labels. We do not manipulate the known true labels, i.e. $\{y_i^{AN}\| i \in S^{p}\cup S^{n}\}$, since they are all clean labels. Instead of using the original loss function, we further introduce the weight term $\lambda_i$ in the loss function, $L = \frac{1}{\|D^{\prime}\|} \sum_{(x, y^{AN}) \in D^{\prime}} \frac{1}{K} \sum_{i=1}^{K} l_i \times \lambda_i .$ We define $l_i = \mathrm{BCELoss} \, (f(x)_i, y_i^{AN})$ where arguments of function $l_i$, that are $f(x)$ and $y^{AN}$, are omitted for convenience. The term $\lambda_i$ is defined as a function, $ \lambda_i=\lambda(f(x)_i, y_i^{AN})$, where arguments are also omitted for convenience. $\lambda_i$ is the weighted value for how much the loss $l_i$ should be considered in the loss function. Intuitively, $\lambda_i$ should be small when $i \in S^{u}$ and the loss $l_i$ has high value in the middle of the training, that is, to ignore that loss since it is likely to be the loss from a false negative sample. We set $\lambda_i=1$ when $i \in S^{p}\cup S^{n}$ since the label $y_i^{AN}$ from these indices is a clean label. We present three different schemes of offering the weight $\ lambda_i$ for $i\in S^{u}$. The schematic description is shown below. Large Loss Rejection. This is to gradually increase the rejection rate during the training process. We set the function $\lambda_i$ as $\lambda_i = \begin{cases} 0, & i\in S^{u} \mathrm{and} l_i > R(t) \\ 1, & \mathrm{otherwise} , \end{cases}$ where $t$ is the number of current epochs in the training process and $R(t)$ is the loss value that has $[(t-1) \cdot \Delta_{rel}]\%$ largest value in the loss set $\{ l_i \| (x, y^{AN}) \in D^{\ prime}, i\in S^{u}\}$. $\Delta_{rel}$ is a hyperparameter that determines the speed of increase of rejection rate. Defining $\lambda_i$ as above makes rejecting large loss samples in the loss function. We do not reject any loss values at the first epoch, $t=1$, since the model learns clean patterns in the initial phase. In practice, we use mini-batch in each iteration instead of full batch $D^{\prime}$ for composing the loss set. We call this method as LL-R. We also propose LL-Ct and LL-Cp which refer to large loss correction (temporary) and large loss correction (permanent), respectively. The readers can find these variants in detail in the paper. The figure above shows the qualitative result of LL-R. The arrow indicates the change of categories with positive labels during training and GT indicates actual ground truth positive labels for a training image. We see that although not all ground truth positive labels are given, our proposed method progressively corrects the category of unannotated GT as positive. We also observe in the first three columns that a category that has been corrected once continues to be corrected in subsequent epochs, even though we perform correction temporarily for each epoch. This conveys that LL-R successfully keeps the model from memorizing false negatives. We also report the failure case of our method on the rightmost side where the model confuses the car as truck which is a similar category and misunderstands the absent category person as present. The quantitative comparison and more analysis of our method can be found in the paper.	{"url":"https://www.eml-munich.de/publication/Large-Loss-Matters-in-Weakly-Supervised-Multi-Label-Classification","timestamp":"2024-11-08T14:43:12Z","content_type":"text/html","content_length":"167139","record_id":"<urn:uuid:c72b8907-7ef6-42de-966d-1099f3e41d18>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00764.warc.gz"}
No Cost Refinance Calculator - Certified Calculator No Cost Refinance Calculator Introduction: Refinancing your mortgage can be a smart financial move, especially when it comes to a no-cost refinance. To help you determine the potential savings and benefits of a no-cost refinance, we’ve created the No Cost Refinance Calculator. Formula: The calculator uses the following formula to estimate your monthly payment and total payment over the loan term: 1. Calculate the monthly interest rate by dividing the annual interest rate by 1200 (to convert from percentage to decimal). 2. Determine the total number of payments by multiplying the loan term (in years) by 12 (to get the number of months). 3. Use the formula for calculating a fixed monthly payment on a loan to find the monthly payment. 4. Calculate the total payment over the loan term by multiplying the monthly payment by the total number of payments. How to Use: 1. Enter the loan amount, which is the amount you wish to refinance. 2. Provide the annual interest rate as a percentage. 3. Specify the loan term in years. 4. Click the “Calculate” button to get your results. Example: Suppose you have a loan amount of $200,000, an interest rate of 3.5%, and a loan term of 15 years. After clicking “Calculate,” the tool will provide your estimated monthly payment and total payment over the loan term. 1. What is a no-cost refinance? A no-cost refinance is a type of mortgage refinance where the lender covers the closing costs. In exchange, you might have a slightly higher interest rate. 2. How does the no-cost refinance calculator work? The calculator estimates your monthly payment and total payment by using your loan amount, interest rate, and loan term. 3. Is the result accurate for my specific situation? The calculator provides estimates and should be used as a starting point. For precise figures, consult with a mortgage professional. 4. What’s the benefit of a no-cost refinance? You can reduce your out-of-pocket expenses by not paying closing costs upfront, potentially saving you money. 5. Can I refinance without any fees at all? While a no-cost refinance eliminates upfront fees, you may still encounter some costs, such as appraisal fees or title insurance. 6. Should I refinance my mortgage? Deciding to refinance depends on your current interest rate, your financial goals, and how long you plan to stay in your home. 7. How does the loan term affect my payments? Shorter loan terms typically result in higher monthly payments but lower overall interest costs. 8. What if I want to pay extra towards my mortgage? You can reduce the total interest paid and the loan term by making extra principal payments. 9. Can I refinance multiple times? Yes, but the decision should align with your financial goals and the costs involved. 10. Are there other factors to consider when refinancing? Yes, factors like credit score, home equity, and market conditions can also influence the refinance process. Conclusion: The No Cost Refinance Calculator is a useful tool to estimate your potential savings when considering a no-cost refinance for your mortgage. Remember that this calculator provides estimates, and it’s important to consult with a mortgage professional for accurate information tailored to your unique situation. Refinancing can be a financially savvy move, and this calculator helps you make an informed decision about your mortgage. Leave a Comment	{"url":"https://certifiedcalculator.com/no-cost-refinance-calculator/","timestamp":"2024-11-09T22:35:12Z","content_type":"text/html","content_length":"55526","record_id":"<urn:uuid:5dce118e-f8af-4f42-865d-09c6fec93d19>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00616.warc.gz"}
Rational Choice and Strategic Conflict : The Subjectivistic Approach to Game and Decision Theory Rational Choice and Strategic Conflict : The Subjectivistic Approach to Game and Decision Theory Gabriel Frahm 'This book is refreshing, innovative and important for several reasons. Perhaps most importantly, it attempts to reconcile game theory with one-person decision theory by viewing a game as a collection of one-person decision problems. As natural as this approach may seem, it is hard to find game theory books that really implement this view. This book is a wonderful exception, in which the transition between decision theory and game theory is both smooth and natural. It shows that decision theory and game theory can go—and, in fact, must go—hand in hand. The careful exposition, the many illustrative examples, the critical assessment of traditional game theory concepts, and the enlightening comparison with the subjectivistic approach advocated in this book, make it a pleasure to read and a must have for anyone interested in the foundations of decision theory and game theory.'Andrés Perea (Maastricht University)'Gabriel Frahm's relatively nontechnical book is a bold synthesis of decision theory and game theory from a Bayesian or subjectivist perspective. It distinguishes between decisions, or one-person games, and games with two or more players, but Frahm argues that this distinction is not always necessary—the two kinds of games can be analyzed within a common theoretical framework. He models the dynamics of choice in several different settings (e.g., information may be complete or incomplete as well as perfect or imperfect), including one in which players look ahead and make farsighted calculations on which they base their choices. His book contains many provocative examples that illustrate the advantages of a unified theory of rational decision-making.'Steven J. Brams (New York University) De Gruyter Oldenbourg Год издания: Полный текст книги доступен студентам и сотрудникам МФТИ через Личный кабинет https://profile.mipt.ru/services/. После авторизации пройдите по ссылке «Books.mipt.ru Электронная библиотека МФТИ» Если Вы считаете нужным сообщить об опечатке, ошибке или о другой проблеме, Вы можете это сделать.	{"url":"https://books.mipt.ru/book/305211","timestamp":"2024-11-06T07:41:09Z","content_type":"text/html","content_length":"13379","record_id":"<urn:uuid:e290c328-a220-425f-8ff7-a53c0f3cc957>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00406.warc.gz"}
Calculus You Forgot (Or Never Learned): Derivatives Calculus You Forgot (Or Never Learned): Derivatives Intuitive ideas about the derivative Photo by Andrea Piacquadio from Pexels If you went to college, it is a fair bet that you took some sort of calculus class. If you went for engineering or science, you probably took a LOT of calculus classes. Unfortunately, a lot of math teachers like to show you how smart they are and they make simple ideas pretty hard to understand. Plus, in school you have a lot of things — both academic and non-academic — competing for your attention. So it isn’t surprising that a lot of people don’t have an intuitive grasp of some calculus ideas. Sometimes, I think some of the professors don’t either and that’s part of the problem. It is one thing to know the notes on each piano key and another thing to be able to play the piano. What is Calculus? So what is calculus? Easy answer: it is the mathematics of change. Regular math lets us answer questions like “Which rug is larger?” or “How many eggs do we need every day to feed a certain number of people?” Algebra answers questions like “If we earn 8% interest on a bank account and after a year we have $1122, how much money did we start with?” Geometry and trigonometry answer questions about shapes and angles. Calculus answers questions about change. Examples of questions you might answer with calculus are: If a pipe develops a 1mm leak that doubles in size every hour, how long does it take to empty a 300-gallon tank? Or, if a hockey stick snaps off the ice according to a math formula, what is the maximum acceleration it will sustain? If you want to deal with stock or option trades, write amazing real-world computer simulations in 3D, or build skyscrapers, you are probably going to need calculus at some point. Good thing it isn’t as hard as people say. What is a Derivative? A derivative is simply the rate of change of something. That’s all it is. A ball standing still or a hose with the water faucet off all have a formula that describes them and the derivative of that formula is zero. Because they are not changing. Sometimes, we really want to know about the actual rate of change. How fast is a rocket accelerating after its engine is on for 3 seconds? But sometimes we want to know the sign of the rate of change. For example, if we have a formula that approximates the cost of soybeans, we will note that the rate of change is positive when the price is going up, negative when it is going down, and zero at places where it changes from plus to minus or vice versa. Those zero points will be the minimum and maximum price of soybeans, at least according to our model. So this is a useful math trick. When I was a student, I used to think “How did Newton and Libnietz figure this s*** out from scratch? Crazy!” But now I see that it is actually really simple if you want to ask those kinds of questions that you would work out this kind of math. We are going to work it out, probably like they did. Start Simple Let’s say we have a function f(x)=4. That means that any value of x you plug in, the answer is 4. This could be a math model for a ball sitting 4 meters off the ground on the floor of a big building. At time 0, the position is 4 meters. At time 1,000 the position is… 4 meters. It isn’t moving. What’s the rate of change? Zero. The same goes for f(x)=20. Or f(x)=1000. No change based on x so therefore the rate of change — and, thus, the derivative is zero. Graphically, that looks like this: f(x)=4 is boring More Interesting Look at this graph: This is a bit more interesting. It is a line. You might remember that a line is y=mx+b where m is the slope and b is the y-intercept. In this case, b=0 and you can see that m=4, so the slope is 4. For a line, though, the slope is the rate of change. that is at every point on the line, a change in x of 1 will cause a change in y of 4 in the same direction. If you didn’t know the formula for this line, you could compute the slope by measuring the rise over the run. What this means is you pick some run — a pair of x values. Find the y values for those two x’s. The difference in x values is the run and the difference in y values is the rise. The rise divided by the run is the slope. So if we pick x=0 and x=1, we see that f(0)=0 and f(1)=4. So the run is 1–0 or 1. The rise is 4–0 or 4. The rise over the run is 4/1 = 4. Same result. But it doesn’t matter what numbers we pick. Since f(1)=4 and f(11)=44 we can see that for a run of 11–1=10 we have a rise of 44–4=40 and 40/10 is… 4. You can do that with any two numbers you care to name. It also would not matter if we added an offset because that won’t change the slope: If you do the same math, the +3 term will cancel out when you subtract and so the slope of this line is still 4. Since we can use any number we like as the difference in x — delta x, or we will call it d, we could write a formula: slope = ((f(x+d)-f(x))/d This is just what we did for x=0,1 or x=1,11, just written as a formula. No matter what d is, we get 4 for our example. So what if d were the smallest number we could have without being 0 (because, after all, we can’t divide by zero)? I don’t mean like 0.1, or 0.01, or even 0.001. I mean millions and millions of zeros followed by a 1. Such a small quantity won’t matter, but the answer will still come out to 4. So let’s plug into our formula that f(x)=4x+3: slope = (((4(x+d)+3)-(4x+3))/d = (4x+4d+3–4x-3)/d = 4d/d = 4 That gives us a lot of confidence that our original formula, ((f(x+d)-f(x))/d is correct. It makes sense and it gives us an answer we know to be correct. We know that the derivative of f(x)=10 is zero. If you think about it, it is really the formula where y=mx+b has values of m=0, b=10. Let’s try our formula there: = 0/d = 0 Keep in mind that f(x+d)=10 not 10+d. So our formula still works. Taking it to the Limit In math, there’s the idea of taking a limit as something approaches something else. So, for example, if I told you g(x)=5/x, then x can’t be zero. But it can approach zero and as it does the result will be bigger and bigger. So 5/1=5 but 5/.1=50 and 5/.001 = 5000. So we would say the limit of 5/x as x approaches zero is infinity. This can be a problem if you wind up with a result from the formula like 10/d. We have not seen that yet, but if we had, the answer to that is infinity because we are really taking the limit of the formula when d approaches zero. In most problems, the d part works itself out, but if we add or subtract d, it is so small we can ignore it. If we multiply by d, we can treat the result as zero. If we divide by d, treat the result as infinite. Next Step This all seems like a lot of work just to find the slope of a line. It is. But we can now take this formula and apply it to other functions and we have some confidence it will give us the rate of change at some value x, too. With a line, the rate of change never changes. But think about the function f(x)=x². If you were going to try to approximate this curve using straight lines, you could say, well, from 0 to 1 the result is 0 and 1, so we can draw line with slope 1 for that part. From 1 to 2 the result is 2 and 4. That’s a rise/run of 2. Then from 2 to 3, the result is 4 and 9. That’s a slope of 5. If you looked at the negative x values, it would be the same but negative (that is, from -1 to 0 is a slope of -1). Now if you did it from 0 to 0.1 and 0.1 to 0.2 you’d get a better approximation. If you did it from 0 to 0.01 and 0.01 to 0.02, it would be even better (but would take a long time). Our formula should be able to tell us the rate of change between any x and an infinitely small increase from x. A rocket ship’s position between now and the infinitely tiny next time period is its acceleration, for example. We can also tell some things by looking at the slope. If I told you I took the derivative of f(x) at -1 and the slope was negative you could tell me that the line is descending at that point (look at the graph; it is). If I told you the derivative was positive, you know the line is going upwards. But if I told you the slope was negative, went to zero (at x=0), and then started going positive you would know the curve hits bottom at x=0. What if I told you the slope was positive, zero, then negative (it isn’t, but pretend). That would mean you’d hit a peak, right? So if you have the math that describes, say, the price of some good based on input factors, you could use this to determine when the price is going to be lowest and when it is going to be highest. We won’t try this function but imagine trying to work with sin(x)+(1/3 sin(3x))+1/5: Lots of slope changes, peaks, and valleys. You can find them all using derivatives. Let’s go back to f(x)=x². Here is a plot that includes a tangent line at x=0.25. The slope of that line is the rate of change at that point, just like if we were approximating with a straight line that was infinitely tiny. That slope is also the derivative of the function at x=0.25. You can see the slope is positive at that point; the line is lower on the left than it is on the right. Red dot at x=0.25 Here’s the tangent line at x=-0.5: Red dot at x=-0.5 Here the slope is negative. We can find the slopes of these lines using our formula. So if x=0.25, the slope of the tangent line, and thus the derivative, is 2*0.25 or 0.5. At x=-0.5 we get -1. Just like with the line, adding a constant won’t matter. So if you plotted x²+10 or x²-33 you’d get the same answer. The derivative is 2x. Knowing this, it is easy to ask where is 2x=0? That happens at 0 and that’s where the minimum is. Is there any place the slope changes from positive to negative? No. So this curve has no peaks. In Real Life You know that multiplying 4 x 5 is the same as saying 5+5+5+5, right? But no one really does it that way. Derivatives are the same. All the common cases are worked out and you either remember it or you look it up in a table. A lot of practical calculus is trying to rearrange things to look like a combination of things you can look up in a table. There are also rules about how to combine things. So, for example, we know how to find the derivative of a square and a straight line, so what if we had: We can show that a constant multiplication also affects the derivative, so the derivative of 4x² is 4(2x) = 8x. We know the derivative of 10x is 10. And the -3 doesn’t matter because the derivative is zero. So f’(x) (a common way to write the derivative of a function) is 8x+10. The situation is more complex if you multiply things together or use trig functions, etc. But you can look up the rules for those very easily. If you aren’t sure that the derivative of 4x² is the same as 4 times the derivative of x², consider this: You can also convince yourself that if the derivative of, say, 10x is 10, 8x is 8, and 2x=2, then: f’(x)=g’(x)+h’(x) we want to know if this is true 10 = 8 + 2 yep, it is true! Again, you can look up these rules instead of working them out each time. For example, here is a good set of rules: https://en.wikibooks.org/wiki/Calculus/Tables_of_Derivatives Keep in mind that instead of writing f’(x), you may also see something like: d/dx x² That’s just another way of saying take the derivative of x². If you want some practice problems, this site will make some up and check your work: https://homepages.bluffton.edu/~nesterd/apps/derivs.html Of course, Wolfram Alpha is great and I used it to produce the graphs in this post. However, to get the full step-by-step results, you have to pay. A nice option that is free is from Microsoft. It will give you the answer, but also has a button that will reveal the work step by step: Microsoft Math Solver shows its work Wrap Up I hope this has given you a better feeling for why we want to do derivatives and that the underlying concepts are pretty easy. Of course, it is easy to learn how to play tennis. It is hard to become Venus Williams. Some math problems are going to be hard for one reason or another. That’s why the pros practice, practice, practice.	{"url":"https://www.cantorsparadise.org/calculus-you-forgot-or-never-learned-derivatives-5a69833c594c/","timestamp":"2024-11-12T03:53:06Z","content_type":"text/html","content_length":"44915","record_id":"<urn:uuid:44c120ec-076a-4beb-bb64-5fc306521034>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00510.warc.gz"}
1. n. [] The pressure, usually measured in pounds per square inch (psi), at the bottom of the hole. This pressure may be calculated in a static, fluid-filled wellbore with the equation: BHP = MW * Depth * 0.052 • BHP is the bottomhole pressure in pounds per square inch • MW is the mud weight in pounds per gallon • depth is the true vertical depth in feet • 0.052 is a conversion factor if these units of measure are used. For circulating wellbores, the BHP increases by the amount of fluid friction in the annulus. The BHP gradient should exceed the formation pressure gradient to avoid an influx of formation fluid into the wellbore. On the other hand, if BHP (including the added fluid friction pressure of a flowing fluid) is too high, a weak formation may fracture and cause a loss of wellbore fluids. The loss of fluid to one formation may be followed by the influx of fluid from another formation. Alternate Form: BHP See related terms: formation pressure	{"url":"https://glossary.slb.com/zh-cn/terms/b/bottomhole_pressure","timestamp":"2024-11-04T06:56:36Z","content_type":"text/html","content_length":"36522","record_id":"<urn:uuid:c180bae3-d71c-44ac-b557-c068831570a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00604.warc.gz"}
On the role of surface permeability for the control of flow around a circular cylinder The circular cylinder with porous materials coating (PMC) is studied in detail to reveal the sensitivity of surface permeability to the flow control and noise reduction. Two-dimensional simulations were firstly used to identify the critical values of permeability and thickness. Parametric study results show that, there is a critical permeability value which produces the minimum force fluctuation and maximum noise reduction. Additionally, the porous coating can work more efficient for noise reduction with larger thickness. The further three-dimensional simulation is employed to understand the underlying physical mechanisms of flow control. The results show that the spanwise vorticity is modified more than that of other directions and behaves more synergistically. The pressure field adjacent to the cylinder surface indicates that the adverse pressure gradient is changed to the favorable pressure gradient around the porous surface which contributes partly to the vortex shedding suppression. 1. Introduction Flow past circular cylinders has been extensively investigated in fluid dynamics associated with engineering application of aircraft landing gear system, pantograph of high-speed trains, heat exchangers etc. The cross-flow interaction with the cylinder induces unsteady loading on the surface which can cause serious structural and environmental problems, such as noise radiation, drag and Passive flow control is an efficient approach to mitigate these adverse effects [1]. Porous media has potential benefits for flow-control [2, 3] through introducing a fluid permeable medium between the fluid and solid and modify the boundary layer and wake characteristics. A number of researchers have studied flow pass and through porous bluff bodies to reveal the interesting flow control phenomenon. Some of them focus on the low Reynolds flow aiming at e.g. enhancing heat transfer [4-6]. The high Reynolds flow is associated to the applications such as drag reduction, vibration and noise reduction. For example, Bruneau and Mortazavi [7] found that porous wraps around rectangular cylinders with suitable choice of permeability coefficient can potentially reduce the hydrodynamic drag by up to 30 %. Zhao et al. [8] and Yu et al. [9] performed numerical simulations for a porous cylinder and found that the drag reduction can be obtained using porous treatment dependent greatly on materials property and Reynolds number. Experimental [10] and numerical studies [11, 12] have also shown that the porous coatings can stabilize the shear layer and the near wake region and subsequently cause significant reduction in lift fluctuations and noise. It is attributed partly to the existence of slip boundary condition at the porous-fluid interface and dissipation of energy by the porous media. More recently, PIV (Particle Image Velocimetry) measurements have revealed again that perforated cylinders can elongate the shear layer and prevent the Karman Vortex Street [13]. Through acoustic measurements, Geyer et al. [14] found that the use of soft porous covers for circular cylinders can lead to noticeable aerodynamic noise reduction. Syamir et al. [15] experimentally found that the porous cover can reduce the fluctuating force of bluff body significantly and delay the vortex shedding and increase the formation length. Liu et al. [16] and Yamamoto et al. [17] investigated porous-cover concept for noise reduction of bluff-body component of landing gear and showed the promising results. However, the sensitivity and critical values of some important parameters of porous materials coating, such as typically permeability and thickness, for circular cylinder flow control and potential noise reduction, were not identified clearly by the previous studies. Also, to better understand the underlying mechanisms, more systematically researches are necessary. For this reason, the present paper reports some important results of our study on the use of porous coatings for controlling the flow field of circular cylinder and reduction of the flow-generated noise. In particular, parametric study indicates the significance of the permeability constant for effective control of the unsteady aerodynamic features and far-field noise. The critical values of permeability and thickness are also studied systematically. The present study also contributes to reveal more underlying physical mechanisms, e.g. adverse pressure gradient medication. This paper is organized as follows. Section 2 presents the methodology used for the simulations. Section 3 is devoted to the results of parametric study. Section 4 gives the results of three-dimensional simulation for understanding the mechanisms of flow control. Results are presented for the unsteady aerodynamic forces, wake development and also the far-field noise. Section 5 concludes the critical values of porous coating parameter and the important control mechanisms. 2. Numerical methodology This paper considers the case of a single circular cylinder with diameter of $D=$0.025 m, wrapped by a porous layer with thickness $h$, placed in the uniform air inflow (density $\rho =$1.225 kg/m^3, viscosity $u =$1.79×10^-5 kg/(m·s)) at $R{e}_{D}=$ 4.7×10^4, as shown in Fig. 1. Although the flow in this problem is intrinsically three-dimensional, as demonstrated in some prior research [18, 19], two-dimensional unsteady CFD can reveal some important aspects of flow-porous interaction, particularly the dynamics of vortex shedding and effects of unsteady aerodynamic forces. The related validation for two-dimensional simulation can be found in Ref. [20], where the numerical method has also been compared against the available data in other numerical and experimental investigations [21-24]. So, to balance the huge computing cost and parametric study aims, with some degree of accuracy, a number of two-dimensional simulations are implemented firstly. Subsequently, specific three-dimensional simulation is employed to reveal more underlying information to better understand the physical mechanisms of flow control by porous coating. The combination of two and three-dimensional simulation is a common strategy for investigating passive flow control [11]. The incompressible Navier-Stokes equations were solved using the finite volume CFD package of ANSYS-Fluent [25]. Large eddy simulation (LES) with smagorinsky subgrid model is employed to calculate the flow field. All current computations were performed with second-order accuracy. The computation domain and the boundary conditions are also shown in Fig. 1. The dimensionless time-step $\mathrm{\ Delta }t{U}_{\infty }/D$ for unsteady simulation is 0.05, which is corresponding to 110 time-steps per vortex shedding cycle for sufficient resolution of shedding process. The present unsteady time step produces the maximum acoustics frequency up to 10 kHz which is enough for the present problem. The Brinkman-Forchheimer extended Darcy model is used to describe the mass and momentum conservation in the isotropic and homogeneous porous medium, similar to Vafai [26] and Hsu and Cheng [27]. The continuity condition is employed to couple the interface flow between the fluids outside and inside porous media. Further details of the simulation method can be found in Ref. [12]. In this paper, the Darcy number $Da=K/{D}^{2}$ is used as non-dimensionalized permeability and the porosity ($\varphi$) is set to 0.97. The relationship of permeability to other parameters, such as porosity and microscopic structure of the porous medium, is not the focus of this paper. Ffowcs Williams-Hawkings (FW-H) equation with integral solution [28, 29] are used to predict the far-field noise. The acoustics integral surface is defined on the porous surface. Fig. 1Schematic of flow past the circular cylinder with porous materials coating 3. Sensitivity study of permeability and thickness 3.1. Unsteady force fluctuation and drag The effect of porous coating permeability and thickness on the effective control of the unsteady forces acting on the cylinder is presented in Fig. 2. The RMS (Root-Mean-Square) force coefficients and time-averaged drag coefficient are normalized with the value of rigid cylinder (baseline). In Fig. 2(a), the dimensionless porous coating thickness is fixed at normalized value of $h/R=$ 0.80, where $R=D/2$. Fig. 2(a) shows that both the lift and drag fluctuations and mean drag reach their minimum when the dimensionless permeability $Da$ is between $D{a}_{2}=$ 6.4×10^-3 and $D{a}_{3}=$ 1.8×10^-2. It is considered here the critical value to be approximately $D{a}_{critical}=\mathrm{}$6.4×10^-3, because that is close to Naito and Fukagata’s value of 2.0×10^-2 [11]. The permeability deviating from the critical value causes the degenerative flow control effects. The porous materials coating (PMC) with extremely high permeability behave similar to a pure fluid medium and therefore the unsteady forces and drag level should gradually approach the level of the baseline case. In the permeability range considered in this study, the lift and drag fluctuations of the treated cylinders are always lower than that of the baseline. When the permeability is proper e.g. larger than $D{a}_{1}=$3.5×10^-3 as seen in Fig. 2(a), the drag level seems to be reduced compared with baseline value. Fig. 2(b) presents the influences of the porous coating thickness ($h/R$). Based on the results in Fig. 2(a), the dimensionless permeability is set to $D{a}_{critical}=$ 6.4×10^-3. Results have shown that the unsteady forces on the cylinder can be significantly reduced by adding a porous coating on the cylinder. Results indicate that the mean drag force can be decreased if a relatively thicker porous layer is used (i.e. $h/R>$ 0.48). Seen from the results, even a very thin porous layer (i.e. $h/R=$ 0.05) can lead to an increase in the mean drag force, because of the change of surface condition and the porous intrinsic drag force. However, when the extreme thickness of porous layer reach zero, the drag level should be equal to that of baseline. The rapid drop observed during thickness changes from $h/R=$ 0.48 to 0.64, indicates significant changes of flow structure which was also investigated in prior investigations [12, 20]. Fig. 3 shows the power spectrum density (PSD) of the lift fluctuations for cylinders treated with porous materials with different permeability ($\mathrm{D}\mathrm{a}$) and thickness ($h/R$). The dimensionless frequency is $St=fD/{U}_{\infty }$. Fig. 3(a) shows that the lift PSD reduction depends strongly on the permeability of the porous material. In addition, the dominant vortex shedding frequency is reduced to about 0.1 when permeability of porous coating is $Da=$ 6.4×10^-3. Furthermore, results in Fig. 3(b) show that increasing the thickness of the porous layer leads to a significant decrease of the lift fluctuation energy and associated frequency, when the non-dimensional permeability is fixed at $Da=$ 6.4×10^-3. The PSD results at$\mathrm{}D{a}_{critical}$ and $h/R\ ge$ 0.64 show even energy attenuation of two orders. The recent work [30] of using plasma to control the tandem cylinder flow also reports similar dramatic peak value reduction of PSD up to two orders. Here, it is explained as the significant vortex shedding suppression in terms of vorticity energy dissipation and frequency reduction which can see in the later flow results. Fig. 2Unsteady forces fluctuation and drag for a cylinder with porous coating: a) Cylinder with h/R= 0.8 over a wide range of Darcy number, b) Effect of varying the coating thickness at Da= 6.4×10-3 Fig. 3Power spectral density of the lift coefficient of cylinders covered with porous material: a) The effect of Darcy number at constant h/R= 0.80 and b) the effect of porous coating thickness at Da = 6.4×10-3 3.2. Effects on flow behavior To better understand the aerodynamic results presented in Section 3.1, the flow field results are given in this section. The instantaneous spanwise vorticity fields of cylinders covered by porous layer of different $\mathrm{D}\mathrm{a}$ values are shown in Fig. 4. The solid and dashed lines represent the solid surface and the porous surface, respectively. The non-dimensional thickness ($h/R$ ) of porous layer is set to 0.8. In Fig. 4(a), it can be seen that when the permeability is quite small ($Da=$ 9.6×10^-5), the porous materials behave more like solid, and intense vortex shedding happens from the surface of the porous coating. Fig. 4(d) shows that, in contrast, when the permeability is quite large ($Da=$ 0.272), the porous materials behave more like pure fluid and intense vortex shedding appears around the inside solid cylinder. Fig. 4 shows that the vortex shedding formation and turbulence level in the near wake can be effectively controlled by employing a porous material cover with an optimum permeability value. It can be seen in Fig. 4(b) that the initial roll-up location can be moved downstream to position of several diameter (almost 8$D$) using an optimum permeability. According to the studies of Lam and Lin [31] on waved-surface cylinder, the elongated vortex formation length consequently can contribute to the pressure-drag reduction which demonstrates the result in previous Fig. 2. The porous coating is also found to decrease the vorticity intensity within the wake, which is consistent with the recently experimental reports of perforated cylinders [13]. The instantaneous spanwise vorticity field for cylinders with different thicknesses $h/R$ are shown in Fig. 5, where the permeability value is fixed at $Da=$ 6.4×10^-3. With the increase of porous cover thickness, the free shear layer extends further downstream and rolls up weakly. Significant changes can be observed between Fig. 5(b) and Fig. 5(c), when the porous coating thickness increases from $h/R=$ 0.48 to $h/R=$ 0.64. Fig. 4Instantaneous dimensionless vorticity contours 0 ≤ωzD/U∞≤ 16 for a) Da= 9.6×10-5, b) Da= 6.4×10-3, c) Da= 4.16×10-2, d) Da= 0.272. The non-dimensional thickness is 0.8 Fig. 5Instantaneous dimensionless vorticity contours ωzD/U∞ for a) h/R= 0.30, b) h/R= 0.48, c) h/R= 0.64, d) h/R= 0.80. The Darcy number is 6.4×10-3 3.3. Effects on acoustic The effect of the permeability and thickness of the porous treatment on the relative overall noise reduction, $\mathrm{\Delta }OASPL$, (i.e. noise from cylinders with porous treatment relative to the bare cylinder) are presented in Fig. 6. The acoustics measurement point is located at 80$D$ above the cylinder center. The influence of the dimensionless permeability $Da$ on noise reduction is shown in Fig. 6(a). Results show that in the case of thin porous covers, the level of noise reduction is almost independent of the permeability of the porous material. For cylinders covered with a thick porous coating e.g. $h/R\ge$ 0.64, however, the level of noise reduction can change significantly with varying permeability. The maximum noise reduction can be achieved at $D{a}_{critical}=$ 6.4×10^ -3, as also indicated by the previous unsteady force results in Section 3.1. The effect of porous coating thickness on noise reduction is shown in Fig. 6(b). It can be seen that for cases of $h/R<$ 0.48, the $\mathrm{\Delta }OASPL$ is quite small, while for cases with $h/R\ge$ 0.48, the porous treatment lead to a sharp $\mathrm{\Delta }OASPL$ increase, especially for critical permeability $Da{} Fig. 6The influence of porous coating a) permeability and b) thickness on ΔOASPL 4. Three-dimensional simulation for understanding the mechanisms For the three-dimensional simulation, the spanwise length $\pi D$ is chozen for better representing the three-dimensional flow features and balancing the computaitonal cost [32]. The properties of porous materials coating are $Da=$ 4.16×10^-2 and $h/R=$ 0.80 which are similar to the parameters used by Takeshi et al. [10] in experiment. The three-dimensional vorticity structure is identified by $Q$ criterion which is expressed by: $Q=\left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial x}\frac{\partial u}{\partial y}\right)+\left(\frac{\partial v}{\partial y}\frac{\partial w}{\partial z}- \frac{\partial w}{\partial y}\frac{\partial v}{\partial z}\right)+\left(\frac{\partial u}{\partial x}\frac{\partial w}{\partial z}-\frac{\partial w}{\partial x}\frac{\partial u}{\partial z}\right),$ where the maximum positive value of $Q$ corresponds to the vorticity core region and the negative value corresponds to the pure shear flow without vortex motion. Seen from the instantaneous three-dimensional vorticity structures in Fig. 7(a), the shear layer of the rigid cylinder just develops over the rear surface. In addition, it comprises complicate three-dimensional coherent structure and small-scale structure associated to the Kelvin-Helmholtz instability (KHI) [33]. In this case, the near wake disturbance is strong and ‘flaps’ behind the cylinder. More explains of this instability mode can be found in the work of Blevins [33]. Regarding the porous materials coating (PMC) cylinder, the smooth free shear layer over the porous surface detaches from two sides. The stable laminar shear layer develops further downstream and transition to the turbulent vorticity wake. Additionally, the three-dimensionality and KHI of free shear layer are suppressed significantly so that it behaves more synergetic along spanwise direction. The shedding vortex from the core cylinder below porous surface is actually very weak because the dissipation effects of porous materials. The far downstream wake of different cylinders has similar vortex characteristic such as appearance of streanwise vortex, necklace vortex and spanwise vortex. The complex flow structures here were not yet found by the previous two dimensional approaches. So, the three-dimensional study is the important supplement to reveal more information of flow modification by PMC. To identify the effects of flow control by PMC on vorticity modification at different directions, Fig. 8 presents the contours of vorticity of $X$, $Y$ and $Z$ directions respectively, in three columns. The top and bottom row represents the rigid cylinder and PMC cylinder respectively. Generally, the scale of flow structure in PMC case is always larger than that of rigid cylinder which implies possible suppression of small-scale vortex. It is more obvious for the modification of ${\omega }_{z}$ (${\omega }_{z}=\partial v/\partial x-\partial u/\partial y$) which means that shear effects on $X$-$Y$ plane is significantly controlled, just like the previous two-dimensional results shown in Fig. 4 and Fig. 5. Five cut-planes vorticity contours equal-spaced along the spanwise direction are shown in the followed Fig. 9. The left one represents the rigid cylinder and the right one represents the PMC cylinder respectively. It is found that the flow structure of rigid cylinder has remarkable spanwise difference. In contrast, for the PMC cylinder at different $Z$ planes, it shows similar flow structure. This result provides evidence again of the spanwise synergy effects of PMC on flow structure. Fig. 7Instantaneous iso-contours with Q-criterion value of 0.8. The colors represent the streamwise velocity value u/U∞ Fig. 8Instantaneous iso-contours of vorticity magnitude with normalized value of 2: ωxD/U∞, ωyD/U∞ and ωzD/U∞ for rigid cylinder and PMC cylinder respectively Fig. 9The vorticity structure at various planes along Z direction Fig. 10 shows the time history of lift coefficient and acoustics spectrum. The sound measurement position is above the cylinder center with distance of 80$D$ in the middle plane, computed through three-dimensional LES approach. In Fig. 10(a), the fluctuating magnitude and frequency can be reduced significantly by PMC and become more regular. The fast Fourier transform (FFT) of lift coefficient are also inserted as subplot of the figure which can demonstrate the conclusion from the view of frequency. Fig. 10(b) shows the acoustics spectrum changes caused by PMC. The tonal noise level is reduced by 15 dB. The previous two-dimensional approach over-predicted the noise reduction value around 2 dB which can be attributed to the assumption of fully correlated flow over the spanwise direction of cylinder in the 2D simulation [22]. In Fig. 10(b), the dominant frequency is moved to lower range which is similar to the conclusions of early publication using two-dimensional method [20]. However, through three-dimensional simulation, additionally, it is also found that the tonal bandwidth become narrow for PMC case compared with the rigid case. And the PMC seems to filter the ‘needling’ of sound spectrum, which appears in the results of rigid case. Because the employed method of signal process is identical, it is attributed to the acoustics modulation by flow control. The Kelvin-Helmholtz instability (KHI) produces the additional small-scale vortex that is considered to be the reason for these extra acoustics component (‘needling’). It can be concluded that the PMC can modify not only the large vorticity structure (coherent structure) but also the small vorticity structure (KHI). Fig. 10Time history of lift and acoustics spectrum at (0, –80D, 0) position Fig. 11The pressure coefficient contour around cylinder surface in the middle plane of Z direction (dashed line denotes the negative value) Until now, it is well known that the vortex shedding can be suppressed by the porous materials coating (PMC) [7, 9, 10, 12-14]. To reveal more essential mechanisms, Fig. 11 presents the pressure field around the cylinder. Adverse pressure gradient ($\partial p/\partial x>0$) normally appears around the rigid cylinder which is responsible for the boundary layer separation and subsequent vortex shedding. However, for the PMC cylinder, flow around the PMC surface is with favorable pressure gradient ($\partial p/\partial x<0$) which causes the vortex shedding suppression and flow stability. Even the adverse pressure gradient also appears under the porous surface, but the dissipation attenuates the intensity of shedding from the inside solid cylinder significantly which therefore, has little influences on the whole flow field. 5. Conclusions Numerical investigations on aerodynamics and acoustic performance of a single circular cylinder treated with porous materials coatings at $Re=$ 4.7×10^4 has revealed that the permeability plays an important role and can determine the effectiveness of the porous treatment for controlling the vortex shedding, unsteady forces and the far-field noise. The non-dimensional permeability about $Da=$ 6.4×10^-3 is found to achieve the maximum noise reduction and deviation from this critical value causes the degradation of control ability. Under the critical permeability, the free shear layer is elongated significantly. The porous coating thickness $h/R$ for effective noise reduction is suggested to be larger than 0.3 at least. In addition, the permeability behaves more sensitively under the thicker porous coatings. Three-dimensional simulation reveals that the spanwise modification is more obvious rather than other directions by PMC and the flow structure is more synergetic along spanwise direction. The favorable pressure gradient form around the porous surface is partly the underlying physical mechanisms of vortex shedding suppression. • Gad-el-Hak M. Flow Control: Passive, Active, and Reactive Flow Management. Cambridge University Press, New York, 2007. • Heenan A., Morrison J. 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Unsteady flowfield around tandem cylinders as prototype component interaction in airframe noise. AIAA Journal, Vol. 45, Issue 8, 2007, p. 1930-1941. • Casalino D., Jacob M. Prediction of aerodynamic sound from circular rods via spanwise statistical modelling. Journal of Sound and Vibration, Vol. 262, Issue 4, 2003, p. 815-844. • Liu H., Azarpeyvand M., Wei J., Qu Z. Tandem cylinder aerodynamic sound control using porous coating. Journal of Sound and Vibration, Vol. 334, 2015, p. 190-201. • Revell J., Prydz R., Hays A. Experimental Study of Airframe Noise vs. Drag Relationship for Circular Cylinders. Lockheed Report 28074. Final Report for NASA Contract NAS1-14403, 1977. • Cox J. S., Brentner K. S., Rumsey C. L. Computation of vortex shedding and radiated sound for a circular cylinder: subcritical to transcritical Reynolds numbers. Theoretical and Computational Fluid Dynamics, Vol. 12, Issue 4, 1998, p. 233-253. • Norberg C. Fluctuating lift on a circular cylinder: review and new measurements. Journal of Fluids and Structures, Vol. 17, Issue 1, 2003, p. 57-96. • Orselli R. M., Meneghini J. R., Saltara F. Two and three-dimensional simulation of sound generated by flow around a circular cylinder. 15th AIAA/CEAS Aeroacoustics Conference, AIAA Paper 2009-3270, 2009. • Ansys-Fluent 14.0 User’s Manual. Fluent Inc., USA, 2006. • Vafai K. Convective flow and heat transfer in variable-porosity media. Journal of Fluid Mechanics, Vol. 147, Issue 1, 1984, p. 233-259. • Hsu C., Cheng P. Thermal dispersion in a porous medium. International Journal of Heat and Mass Transfer, Vol. 33, Issue 8, 1990, p. 1587-1597. • Ffowcs-Williams J. F., Hawkings D. L. Sound generation by turbulence and surfaces in arbitrary motion. Philosophical Transactions of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 264, 1151, p. 321-342. • Brentner K. S. B., Farassat F. 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About this article Flow induced structural vibrations flow control porous materials coating noise and vibration reduction numerical simulation This work was supported by the National Natural Science Foundation of China (Grant No. 51506179) and the Fundamental Research Funds for the Central Universities (Grant No. 3102016ZY018). The helpful discussion with Dr. Mahdi Azarpeyvand of University of Bristol is greatly acknowledged. Copyright © 2016 JVE International Ltd. This is an open access article distributed under the Creative Commons Attribution License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.	{"url":"https://www.extrica.com/article/17281","timestamp":"2024-11-07T16:43:23Z","content_type":"text/html","content_length":"159902","record_id":"<urn:uuid:7ebef533-0e54-49d7-a2cf-51551f31656a>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00119.warc.gz"}
Comments on Computational Complexity: If P=NP then we HAVE an alg for SAT.Another (trivial) assumption + algorithm is the fo...You're right; it works if we add also the P=NP...If I understand your assumption correctly, the pol...My mistake! it's equivalent to NOT x OR y :-)...... More on the first point: - if the algorithm (...> So it seems that the only way to fail would b...I don't understand the claim about the finite ...In factorization you can use the polynomial time p...Why do you say that "X -> Y" would be...I don't understand the "(all of which are...Typo: ... Exists P -> polytime(P) .... should ...I think that Q1 is equivalent to proving P=NP: P=...Huh! No! I bet P!=NP and furthermore the proof wil...Are you user2925716? tag:blogger.com,1999:blog-3722233.post3593116095045748693..comments2024-11-13T15:38:29.005-06:00Lance Fortnowhttp://www.blogger.com/profile/ 06752030912874378610noreply@blogger.comBlogger14125tag:blogger.com,1999:blog-3722233.post-25774643377971321762018-11-02T16:50:55.815-05:002018-11-02T16:50:55.815-05:00Another (trivial) assumption + algorithm is the following:<br /><br />* Assume that if P=NP we can build a polynomial time algorithm for SAT which is provably correct (in PA or ZFC)<br /><br />Let M1,M2,... be a standard TM enumeration;<br />P1,P2,... a standard PA proof enumeration;<br /><br />. On input x (a boolean formula):<br />. for i = 1 to \|x\| do<br />... for j = 1 to i do<br />..... if Pi is a valid proof that Mj solves SAT in polynomial time<br />....... then simulate Mj(x) and accept/reject according to it<br />. If no proof is found solve SAT(x) using an exhaustive searchMarzio De Biasihttps:// www.blogger.com/profile/18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-38810687150805269072018-11-02T14:13:46.162-05:002018-11-02T14:13:46.162-05:00You're right; it works if we add also the P=NP assumption.Marzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-74388374840773616422018-11-01T12:56:11.937-05:002018-11-01T12:56:11.937-05:00If I understand your assumption correctly, the poly boundedness of a proof system for tautologies only implies NP=CoNP (There is an oracle A such that P\neq NP, but NP=CoNP with respect to A, hence it seems that the poly boundedness does not imply P=NP if we assume relativizable proofs.) Erfan Khanikinoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-52475042134788397492018-11-01T11:23:07.672-05:002018-11-01T11:23:07.672-05:00My mistake! it's equivalent to NOT x OR y :-)<br /><br />It's equivalent to prove<br /><br />P <> NP OR [polytime (A) and A=SAT]<br />Marzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-18453694619027109332018-10-31T12:38:40.705-05:002018-10-31T12:38:40.705-05:00... More on the first point:<br /><br />- if the algorithm (that behaves well on small instances) outputs a satisfying assignment then you can check it (and "patch" it if is not valid)<br />- if it outputs "unsatisfiable" then you must "trust" it (otherwise you'll patch it infinitely many times)<br /><br />So the explicit algorithm can output "unsatisfiable" on some satisfiable instances (finitely many if P=NP).<br /><br />Marzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-84038926811924920962018-10-31T12:29:31.222-05:002018-10-31T12:29:31.222-05:00> So it seems that the only way to fail would be to claim an instance is unsatisfiable when in fact it is<br /><br />Yes.<br /><br />> ... Or would that count as non-constructive?<br /><br />Yes. It would be constructive if you explicitly write those hard-coded instances (or, alternatively, you could give the explicit x_0 such that for all instances >= x_0 the algorithm solves it correctly "without patches") Marzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-81910580377626549272018-10-30T20:41:08.242-05:002018-10-30T20:41:08.242-05:00I don't understand the claim about the finite number of exceptions:<br />- Any algorithm deciding SAT could also be used to output a satisfying assignment when the instance is satisfiable. So it seems that the only way to fail would be to claim an instance is unsatisfiable when in fact it is.<br /><br />- If the algorithm has a finite number of instances where it is wrong, couldn't those be hard-coded into the algorithm? Or would that count as non-constructive?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-11068788460748682862018-10-30T16:44:26.579-05:002018-10-30T16:44:26.579-05:00In factorization you can use the polynomial time prime algorithm to detect the bad answers and correct them searching exaustively for their factors (and if factoring is in P you need to do this only on a finite number of & quot;bad algorithms" that behaved well on smaller numbers). In a similar fashion <br /><br />*under the assumption (much? stronger than P=NP) that Frege (or Extended Frege) systems are polynomially bounded proof systems*<br /><br />then the following algorithm solves SAT correctly and runs in polynomial time:<br /><br />Given a formula x, find the smallest i < log log \|x\| such that M_i outputs a satisfying assignment on all satisfiable formulas y of length \|y\| < log log \|x\| or a valid Frege unsatisfiability proof on all unsatisfiable formulas (in both cases running at most for \|y\|^\|M_i\| steps). Then run M_i on x for at most \|x\|^\|M_i\| steps; check (in polynomial time) the satisfying assignment or the correctness of then Frege unsatisfiability proof; if the assigment or the proof are not correct (or there is no M_i that satisfies the above condition) then check exhaustively (exponentially) whether x is in SAT or not.Marzio De Biasihttps:// www.blogger.com/profile/18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-63646817862288718312018-10-29T18:44:31.645-05:002018-10-29T18:44:31.645-05:00Why do you say that "X -> Y" would be equivalent to "NOT Y OR X"?Jakitohttps://www.blogger.com/profile/ 08235089048981338795noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-28561451399395732492018-10-29T18:37:03.405-05:002018-10-29T18:37:03.405-05:00I don't understand the "(all of which are NOT in SAT)" part. Whenever the program fails, it claims that the given formula is not satisfiable. (Otherwise it provides a valid witness, and hence doesn't fail.) I don't understand why you claim that such a formula would be NOT in SAT. Isn't it the other way around?Jakitohttps://www.blogger.com/profile/ 08235089048981338795noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-7370808234988765072018-10-29T12:13:06.916-05:002018-10-29T12:13:06.916-05:00Typo:<br /><br />... Exists P -> polytime (P) .... should be<br /><br />... Exists P . polytime(P) ... (. read as "such that")<br /><br /><br />Marzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-26988501651382174152018-10-29T08:36:12.336-05:002018-10-29T08:36:12.336-05:00I think that Q1 is equivalent to proving P= NP:<br /><br />P=NP is equivalent to Exists P -> polytime(P) AND forall x . P(x) = SAT(x)<br /><br />So what we are trying to prove is that for a particular concrete A :<br /><br />Exists P -> polytime(P) AND forall x . P(x) = SAT(x) -> polytime(A) AND forall x . A(x) = SAT(x)<br /><br />Which is equivalent to prove:<br /><br />NOT [ polytime(A) AND forall x . A(x) = SAT(x) ] OR Exists P -> polytime(P) AND forall x . P(x) = SAT(x)<br /><br />but we want [ polytime(A) AND forall x . A(x) = SAT(x) ] = True (Q1) so we must prove P=NPMarzio De Biasihttps://www.blogger.com/profile/ 18441670787376943932noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-695827498761785962018-10-29T06:16:42.910-05:002018-10-29T06:16:42.910-05:00Huh! No!<br />I bet P!=NP and furthermore the proof will be trivial. :-)<br /><br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-27570345916068731322018-10-29T04:38:03.622-05:002018-10-29T04:38:03.622-05:00Are you	{"url":"https://blog.computationalcomplexity.org/feeds/3593116095045748693/comments/default","timestamp":"2024-11-14T15:00:13Z","content_type":"application/atom+xml","content_length":"29937","record_id":"<urn:uuid:69cedd53-d6a6-4add-88ba-f940f1dcc4e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00636.warc.gz"}
Convert Natural Exponential Equation In Logarithmic Form Worksheets [PDF]: Algebra 2 Math How Will This Worksheet on "Convert Natural Exponential Equation in Logarithmic Form" Benefit Your Student's Learning? • Converting natural exponential equations to logarithmic form helps students see how exponents and logarithms relate, specifically with the natural base $e$. • This approach offers another way to solve equations involving $e$, useful in various math problems. • Learning this conversion is crucial for advanced math topics like calculus and differential equations. • It provides additional techniques for solving complex math problems, making students more versatile. • Properly converting between exponential and logarithmic forms helps students solve math problems accurately and quickly. How to Convert Natural Exponential Equation in Logarithmic Form? • Recognize that the equation involves the natural base $e$, written in the form $e^x = y$. Here, $e$ is the base, $x$ is the exponent, and $y$ is the result. • Know that the natural logarithm ($\ln$) is the inverse of the natural exponential function. This means the natural logarithm undoes the exponentiation by $e$. • Convert the equation by using the natural logarithm. The result of the exponential equation becomes the argument of the natural logarithm. • Express the equation as $\ln(y) = x$, indicating that the natural logarithm of $y$ is equal to $x$. This shows the inverse relationship clearly. Q. Convert the exponential equation in logarithmic form.$ewline$$e^4 \approx 54.598$	{"url":"https://www.bytelearn.com/math-algebra-2/worksheet/convert-natural-exponential-equation-in-logarithmic-form","timestamp":"2024-11-10T04:38:20Z","content_type":"text/html","content_length":"117679","record_id":"<urn:uuid:a357db96-07a7-4326-9e71-b85b88519465>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00636.warc.gz"}
Talks & Conferences \| Mysite top of page 1. 7th International Conference on Economics and Statistics (EcoSta 2024), Beijing, China, 2024 Gave an online talk in the EcoSta 2024 conference organized at Beijing Normal University, China (Certificate). 2. Bangalore Probability Seminar, Bangalore, India, 2024 Gave an online talk in the Bangalore Probability Seminar (2024) organized at TIFR Centre for Applicable Mathematics. 3. The Mathematics of Data Workshop on Optimization and Discrete Structures, IMS, Singapore, 2024 4. 38th Annual Conference of Ramanujan Mathematical Society, Guwahati, India, 2023 Gave a talk in the 38th annual conference of RMS organized by the Indian Institute of Technology (IIT) Guwahati (Certificate). 5. CFE-CMStatistics 2023, Berlin, Germany 2023 Gave an online talk in the CFE-CMStatistics conference 2023, Berlin, Germany (Certificate). 6. IMS Young Mathematical Scientists Forum - Statistics and Data Science, Singapore, 2023 Was a member of the organizing committee, and gave a talk at the IMS Young Mathematical Scientists Forum, Statistics and Data Science in Singapore, during November, 2023 (IMS-2023). 7. EcoSta Conference, Tokyo, Japan, 2023 Gave a talk at the EcoSta 2023 Conference organized at Waseda University, Tokyo, Japan (Certificate). 8. Indian Statistical Institute, 2023 Gave talks in the Stat-Math Unit and the Applied Stat Unit, ISI Kolkata, during the summer of 2023. 9. AMNS-2023 Conference, Pokhara, Nepal, 2023 Gave an invited talk in the Third International Conference on Applications of Mathematics to Nonlinear Sciences (AMNS-2023). 10. McGill University, Department of Mathematics and Statistics, 2023 Gave a talk in the department of Mathematics and Statistics, McGill University, Canada. 11. IISA Conference, IISC Bengaluru, 2022 12. Indian Statistical Institute, 2022 Gave talks in the Stat-Math Unit and the Interdisciplinary Statistical Research Unit, ISI Kolkata, during the summer of 2022. 13. SASI Probability Theory and Related Areas Workshop, NYU Abu Dhabi, 2022 Gave a talk in the 2022 SASI workshop organized at NYU, Abu Dhabi. 14. Applied Probability Seminar Series, Columbia University, 2021. Gave a talk in the Applied Probability Seminar Series, Columbia University. 15. Statistics Department Seminar, Stanford University, 2021. Gave a talk in the Statistics department seminar, Stanford University. 16. Tel Aviv University Statistics Seminar, 2020. Gave a talk in the Statistics seminar, Department of Statistics & Operations Research, Tel Aviv University. 17. Joint Statistical Meeting (JSM), 2020 Gave a talk in the Joint Statistical Meeting (2020). 18. International Indian Statistical Association, 2020 Gave a talk in the International Indian Statistical Association (IISA) student paper competition after being selected among the top five papers in the Probability/Theory/Methodology category. Presented the paper Phase Transitions of the Maximum Likelihood Estimates in the p-Spin Curie-Weiss Model in this competition. Received the best student paper award in this competition. 19. Stat-Math Unit, Indian Statistical Institute, 2019 Gave a talk in the Stat-Math Unit in ISI, Kolkata (2019) on applications of dependent combinatorial data in statistics. 20. Interdisciplinary Statistical Research Unit, Indian Statistical Institute, 2019 Gave a talk in the Interdisciplinary Statistical Research Unit in ISI, Kolkata (2019) on high-dimensional central limit theorems. 21. CombinaTexas, 2019 Attended the CombinaTexas Conference (2019) in Texas A&M University. Gave a talk in this conference. 22. 17th Northeast Probability Seminar, 2018 Attended the 17th Northeast Probability Seminar (2018) in New York University. Gave a talk in this conference. 23. Frontier Probability Days, 2018 Attended the Frontier Probability Days Conference (2018) in the Oregon State University. Gave a talk in this conference. 24. Second Warsaw Summer School in Probability, 2017 Attended the Second Warsaw Summer School in Probability (2017) in the University of Warsaw. Gave a talk in this program. 25. P.C. Mahalanobis Gold Medal Presentation, 2016 Gave a talk in the P.C. Mahalanobis Gold Medal Presentation (2016) in the Indian Statistical Institute. Won the award. 26. D. Basu Memorial Lecture, 2014 Gave a talk in the D. Basu Memorial Lecture (2014) in the Indian Statistical Institute. 1. Intertwining between Probability, Analysis and Statistical Physics, 2024 2. Institute of Mathematical Statistics, Asia Pacific Rim Meeting, Melbourne, 2024 Organized and chaired a session in the IMS APRM 2024 conference held at Melbourne, Australia. 3. Talking Across Fields, 2020 Attended the Talking Across Fields conference, 2020 in Stanford University. 4. International Workshop on Inference on Graphical Models, 2019 5. Columbia-Princeton Probability Day, 2019 6. Summer School on Random Matrices, 2018 Attended the Summer School on Random Matrices, 2018 in the University of Michigan, Ann Arbor. 7. Graph Limits, Groups and Stochastic Processes Summer School, 2017 bottom of page	{"url":"https://www.somabha.com/talks-conferences","timestamp":"2024-11-03T23:15:45Z","content_type":"text/html","content_length":"706350","record_id":"<urn:uuid:44683200-203f-4fcf-a749-acea57377046>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00617.warc.gz"}
Problem Set Week 04 Before First Class Before Second Class Q107. Our neighborhood association has a ten member board. Each year it plans to add four members. Write the difference equations that describe the size of the board (S) each year. Q108. You are a small non-profit. Your sole funder says that each year it will double what you have as your balance at the end of the year. Each year you project spending 20,000 for programs. Ignore interest. Write difference equations describing your balance (B). What special situations can you imagine we might get into? What, for example, happens if B[0]=$32,000? What happens if it is 50,000? 40,000? Q109. Each year the feral cat population grows by 3%. Let C[n] be the number of cats n years from now. Assume there are presently 350. Write a difference equation that describes the cat population from year to year. Q110. Each year the feral cat population grows by 3%. Let C[n] be the number of cats n years from now. Assume there are presently 350. Suppose that each year we catch and euthanize or place in homes 20 cats. Write the equations for this situation. Q111. Let's say we have a 2 year graduate program. The first year class is growing at a rapid rate 5% per year. Between the first and second years, 25% of the students change their minds or get jobs and leave the program. Among the second years, 10% leave before graduation. The program currently has 20 first year and 12 second year. Write difference equations to describe population in future Before Lab Q136. Write out the difference equation that represents the following scenario and the first five terms of the corresponding sequence given the stated starting value. 1. Membership in a club goes up by 4 people each year. At year one it has 21 members. 2. A community's population increases by 4% each year. At year one it is 350. 3. A swimming pool, currently containing 100,000 gallons of water, is leaking at the rate of 2% per day but is being filled at the rate of 1,000 gallons per day. 4. A retirement account which stands at $120,000 earns 3% interest annually. The owner needs to withdraw $1500 per month to pay for eldercare. For each of these, graph P[n] vs. time. For each of these, graph P[n+1] vs. P[n] In Class/Lab page revision: 2, last edited: 28 Aug 2013 22:57	{"url":"http://djjr-courses.wikidot.com/ppol225:problemset-week04","timestamp":"2024-11-08T01:15:57Z","content_type":"application/xhtml+xml","content_length":"25990","record_id":"<urn:uuid:a1ecf06b-f355-454b-8494-f8db61e560a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00653.warc.gz"}
50 Nodal Analysis Multiple Choice Questions (MCQs) with Answers This article lists 50 Nodal Analysis MCQs for engineering students. All the Nodal Analysis Questions & Answers given below include a hint and a link wherever possible to the relevant topic. This is helpful for users who are preparing for their exams, interviews, or professionals who would like to brush up on the fundamentals related to Nodal Analysis. An electric circuit in which the voltage (potential difference) between "nodes" (points where elements or branches connect) in terms of the branch currents is nodal analysis. It is calculated using Ohms's law, KCL, and KVL. They are classified as reference and non-reference node type analysis. The datum node is another name for the reference node. Chassis ground and Earth ground are two types of Reference nodes. The nodal analysis of a circuit is calculated in 4 main steps. Identify primary nodes & consider them as reference nodes. So, this node is treated like the Ground. The second step is to all the main nodes excluding the reference node and writes nodal equations. This equation can be attained by applying KCL first & after that Ohm’s law. The 3^rd step is to identify the main nodes & choose one of them as a reference node. So, this node is treated like the Ground. This analysis benefits in terms of including the minimum number of equations required. 1). A nodal analysis is defined to determine ___ parameter between nodes? 2). Nodal Analysis uses which Kirchhoff's circuit based laws for analysing a circuit ? 3). At which point a Nodal analysis occurs? 4). How a circuit can be resolved when admittance representations are absent for elements? 5). In what kind of circuits can nodal analysis be applied? 6). How many simultaneous equations are required with ‘n’ nodes in circuit. 7). In which of the following analysis, reference and non-reference nodes are parts? 8). Datum Node is other name for? 9). Which of the following are reference nodes? 10). Identify the first step of nodal analysis 11). Identify the second step of nodal analysis. 12). Identify the 3rd step of nodal analysis. 13). What is Ohm’s law equation? 14). Nodal analysis/node voltage method is also called as? 15). For formulating circuit equations Nodal method was uses also in....? 16). Nodal method has _____ disadvantage? 17). Nodal Analysis can be said as sum of ____ laws? 18). How many nodes does a super rnode analysis has? 19). The definite node voltage at a node in a circuit is known as ? 20). ____ is called as reference node? Nodal Analysis MCQs for Students 21). How many nodal equations to be solved for a given circuit? 22). A 11 nodes in network in nodal analysis requires ____ number of equations to represent the circuit? 23). Nodal analysis is applicable for which type of networks? 24). A supernode requires application of which laws? 25). Which type of analysis is preferred in power systems network between nodal and mesh analysis? 26). Can nodal analysis be applied for circuit containing _____? 27). Which of the following are sources for Nodal circuits? 28). A reference node in nodal analysis is classified as? 29). Electrical sources are classified as? 30). Which of the following are dependent sources? 31). Which of the following are dependent sources? 32). Which of the following are the characteristics of amplifiers? 33). Current gain is represented as? 34). Which of the following are the examples of linear dependent sources? 35).Which of the following is the equation of VCVS Linear dependent source? 36). Which of the following is the CCVS equation of Linear dependent source? 37). Which of the following is the CCCS equation of Linear dependent source? 38). Which of the following is the VCCS equation of Linear dependent source? 39). What is the value obtained from no of nodal equations and non-reference nodes? 40). Identify the type of reference node? 41). Identify the type of reference node? Nodal Analysis MCQs for Exams 42). Current is measured in terms of? 43). Resistance is measured in terms of? 44). Which of the following is the advantage of nodal analysis? 45). Which of the following law falls short for calculating an electric circuit? 46). Which of the following law is used for calculating the impedance factor? 47). Kirchoffs laws are invalid at ____ frequency? 48). I1+ I2+I3…. =0 is which law? 50). ___ is reciprocal of impedance?	{"url":"https://www.watelectronics.com/mcq/nodal-analysis/","timestamp":"2024-11-08T01:14:56Z","content_type":"text/html","content_length":"202211","record_id":"<urn:uuid:adeed8df-0088-41d1-9858-4628a2b91d14>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00244.warc.gz"}
The 25 students in your class will present.your teacher randomly selects a student to give the first presentation. Find the probability that you are NOT selected first? \| Socratic The 25 students in your class will present.your teacher randomly selects a student to give the first presentation. Find the probability that you are NOT selected first? 1 Answer See a solution process below: For you to NOT be the first presenter one of the other 24 students must be selected. The chances of one of these other 24 students being randomly selected is: #24/25 = 96/100 = 96%# Impact of this question 1781 views around the world	{"url":"https://socratic.org/questions/the-25-students-in-your-class-will-present-your-teacher-randomly-selects-a-stude","timestamp":"2024-11-06T10:22:38Z","content_type":"text/html","content_length":"32744","record_id":"<urn:uuid:4082cd2b-5ea3-43e5-96b5-feed14d65823>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00259.warc.gz"}
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