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Blog
# The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
The brute-force way to solve this problem is literally to add up the first 15 positive perfect squares, from 1 to 225, inclusive. This is not necessarily completely out of bounds, given that we only have to sum up 15 numbers, all of which we should know already, and several of which are small. However, we should look for a shortcut using the formula.
Unfortunately, there is an unknown constant in the formula, but by using a small test number, we can solve for this constant. You can certainly pick n = 1, since it is a positive integer:
12 = 13/3 + c12 + 1/6
1 = 1/3 + c + 1/6
1/2 = c
If you feel uncomfortable picking n = 1, you can pick n = 2 and come to the same result almost as quickly.
Now, we plug n = 15 into the formula and solve:
12 + 22 + … + 152 = 153/3 + 152/2 + 15/6
= 15×15×15/3 + 15×15/2 + 15/6
= 15×15×5 + 15×15/2 + 5/2
= 225×5 + 225/2 + 5/2
= 1,125 + 230/2
= 1,125 + 115
= 1,240
The correct answer is (C).
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### 2023–2024 MBA Essay Tips
Click here for the 2022–2023 MBA Essay Tips
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## Ms. Brianne
### Target 1
###### Lesson Type:
New
Number Operation
:
Fractional Numbers
Visually distinguish equal parts of the same whole.
###### 1:
(This is an informal introduction to fractions).
###### 2:
Distinguish between equal and unequal parts.
###### 3:
Describe if parts are the same or if one is bigger or smaller.
1st
###### Vocabulary:
Equal, Unequal, half, third, fourth
Activities:
Students
• created shapes with playdoh and cut them into equal parts
• named the type of parts they created (halves, thirds, fourths)
• sorted pictures of shapes by whether they were shown in equal or unequal parts
## Absent Students:
### Target 2
:
###### 1:
Looking at a picture where information is chaotic or spread out, count to gather data to answer questions.
Kindergarden
###### Vocabulary:
Sort, Common, Belong
Activities:
Students
• sorted the same set of animals in 5 different ways (i.e., color, size, animal, number of legs, habitat, diet)
• identified the picture that did not belong when shown a set of 4 pictures
• used directions in a dichotomous key to determine the name of each monster
### Home Exploration
Have your child sort toys by various characteristics, such as color, type, patterns, texture, etc. encourage students to explore different categories/arrangements.
None
1. Can toy animals be sorted by where the animal would live/be (land, sea, air)?
2. Can toys be sorted by size (big, medium, small)?
3. Which categories include the most toys?
4. Which categories include the least amount of toys?
:
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# Randarray: Excel Formulae Explained
## Key Takeaway:
• RANDARRAY Formula generates random numbers or values in Excel, which can be useful in a variety of applications. It is a versatile and powerful tool for data analysis and modeling.
• Using the RANDARRAY function, you can generate random numbers within a specified range, generate unique values in a list, and generate non-repeating random numbers. This can be particularly useful for creating randomized quizzes or surveys, or for generating data sets for testing purposes.
• The RANDARRAY function is easy to use and can be customized to fit your specific needs. By specifying rows and columns, minimum and maximum values, and other parameters, you can create complex data sets that are tailored to your needs.
Have you ever wondered how Excel generates random numbers from formulae? Look no further! In this article, you will learn how the RANDARRAY formulae works and how you can use it to your advantage. Get ready to unlock the power of Excel!
## Understanding RANDARRAY Formula in Excel
Are you an Excel enthusiast? I sure am! I’m always on the lookout for fresh and exciting formulas. Recently, I discovered RANDARRAY! Let me take you on a journey to understand this formula. First, the basics: what it does and how it works. Then, we’ll go deeper and I’ll show you, step-by-step, how to use RANDARRAY in Excel. Ready to take your Excel skills up a notch? Let’s go!
### Introduction to RANDARRAY Formula
Introducing RANDARRAY – a new, exciting feature in Microsoft Excel! This formula creates an array of random numbers between 0 and 1. Every time it’s used, unique results are produced.
RANDARRAY is great for data analysis and deal tracking. It generates thousands of random scenarios to help you make decisions. It also offers built-in functions to modify outputs.
There are two ways to execute RANDARRAY – directly entering it into a worksheet cell or integrating with other functions like SUM or AVERAGE.
If you’re looking to experiment with data sets or gaming data, RANDARRAY is perfect. It offers an infinite number set to pull potentials from, and the parameters can be changed to suit your needs.
RANDARRAY was invented by James Harris Simons and Julian Simon in 1955. It was modeled after UNIVAC, a computer program used during WWII, but created with simpler math.
Let’s take the next step and get to know RANDARRAY better!
### Explaining how RANDARRAY Formula Works
RANDARRAY is a new formula in Excel 365. It can help you generate multiple series of random numbers with ease. When you don’t set the rows and columns, the default is 1 for both. This means you get one random number if you just use RANDARRAY.
Using this formula, you can quickly create samples or simulate data for analysis. You can also use it to compare multiple datasets.
It’s important to know that each time you recalculate the worksheet with RANDARRAY, it generates new sets of random numbers. It’s great for creating samples or picking random items from a list.
By understanding and using this feature in Excel 365, you can save time and get more accurate results. Don’t miss out on the benefits of this formula!
## Benefits of Using RANDARRAY Formula in Excel
I, an enthusiast of Excel, am always seeking time-saving formulas that can boost my productivity. One of these formulas is the RANDARRAY formula. So, I will explain some of its advantages that I have encountered. Firstly, I’ll explain how to use the RANDARRAY formula to generate random numbers in Excel. Secondly, I’ll show how it can be used to make unique values in a dataset. Lastly, I’ll discuss its fascinating use case: generating random numbers that do not repeat. Get ready for randomness in your Excel spreadsheets!
### Generating Random Numbers using RANDARRAY Formula
Generating random numbers with RANDARRAY is an efficient way to get your desired results. It was recently added to Excel, making it easier than ever.
1. With RANDARRAY, you specify the number of rows and columns you need and get random integers and/or decimals. No need for extra functions or manual calculations.
2. You can also set a range for the numbers. This helps get random numbers in the range specific to your data.
3. Press F9 after entering the function to repeat the values. Every time F9 is pressed, new sets of unique values appear.
RANDARRAY formula saves time and effort. Just define the parameters once and let Excel do the work! Unlock limitless possibilities today by generating random numbers effortlessly with RANDARRAY!
Now that you know how easy it is to produce sets of random numbers with RANDARRAY, let’s move ahead to our next topic: Generating Unique Values using RANDARRAY Formula.
### Generating Unique Values using RANDARRAY Formula
The RANDARRAY formula is the perfect tool for generating unique random numbers in Excel. It’s a dynamic array function introduced in Microsoft Excel 365 that gives you a way to make an array of random integers without repeats.
Here are four benefits of the formula:
• Simple to use – This formula is easy to work with. You can define the number and range of random values you need.
• No duplicates – The output won’t have any duplicates, which helps with data analysis and surveys since duplicate entries can mess up results.
• Multiple ranges – You can specify multiple ranges, so you can quickly generate random numbers in different ranges in the same worksheet.
• Dynamic generation – RANDARRAY creates a dynamic set of numbers that updates when the worksheet changes, so you don’t have to worry about static sets that repeat.
Using RANDARRAY is easy. To generate an array with five unique values from one to ten, type ‘=RANDARRAY(5,1,1,10,True)’ into any cell. Press enter and the cell will fill with five unique integer values.
You can also combine RANDARRAY with INDEX or SMALL formulas to get specific values from sorted or unsorted ranges according to their positions.
Generating unique random numbers in Excel is easier than ever. In short, this formula simplifies producing sets of non-repeating random integer series.
### Generating Non-repeating Random Numbers
Generating non-repeating random numbers in Excel? RANDARRAY formula can help! Here are five facts you should know:
• It returns an array of randomly generated decimal numbers between 0 and 1.
• Specify the size of the array by indicating the number of rows and columns in the formula.
• Use a combination of functions such as ROUND and MOD to convert the decimals into integers within a specified range.
• RANDARRAY ensures the random numbers generated are unique each time.
• Unlike older methods, RANDARRAY allows for non-repeating values without requiring extra steps.
When you need to generate a set of unique random values in Excel, RANDARRAY is the way to go. No extra steps to remove duplicates! With one formula, you can generate an array of unique decimal or integer values.
Need to shuffle a list or pick random names out of a hat? Avoid duplicates with RANDARRAY. It ensures each value is unique.
This tutorial will guide you through using the RANDARRAY formula step-by-step. Follow the instructions carefully and effectively utilize its features and benefits to generate sets of randomized data without any duplication.
## Utilizing RANDARRAY Formula: A Step-by-Step Guide
I’m an Excel fan! I’m really interested in the range of formulae it offers. RANDARRAY is one of them. It creates a range of random numbers in Excel. Here’s a step-by-step guide to using RANDARRAY:
1. First, learn how to pick a range with the tool.
2. Second, find out how to set rows and columns with RANDARRAY formula.
3. Last but not least, we’ll cover how to set min and max values with this powerful tool.
### How to Select Range using RANDARRAY
To use RANDARRAY in Excel, you need to:
1. Enter the formula =RANDARRAY(row_num [optional], column_num [optional]).
You can omit either parameter if you don’t need it.
2. Highlight the cells you want the result of the formula in.
3. Press Ctrl-Shift-Enter or Command-Return (for Mac) to make the formula an array formula.
This ensures each cell of the range gets a unique random number.
4. Use MIN or MAX to analyze and summarize the values.
Before selecting a range with RANDARRAY, decide how many rows and columns you need. This method is great for creating a concise yet randomized data set without having to input each value individually.
If you need dynamic ranges with fluctuating dimensions, try nesting INDEX and ROW formulas instead of RANDARRAY. The next part of the article – Specifying Rows and Columns with RANDARRAY Formula – will explain how to adjust the range size by calculating its dimensions from other existing values in your worksheet.
### Specifying Rows and Columns with RANDARRAY Formula
Follow these four steps to use RANDARRAY Formula to specify rows and columns:
1. Choose the starting cell.
2. Enter =RANDARRAY(rows, columns).
3. Put the desired number of rows in place of “rows” (e.g., 5).
4. Put the desired number of columns in place of “columns” (e.g., 3).
Excel will generate random numbers in the chosen cells. You can come back and adjust the values by changing either the rows or columns parameter.
Every time you recalculate it or make changes on your worksheet, new random numbers will be created automatically.
RANDARRAY Formula is useful when working with datasets. Always remember to update your work.
It may seem like a simple task, but it is quite powerful when used correctly. You can generate as many random data points as you need easily.
Let’s move on to the next topic: Specifying Minimum and Maximum Values with RANDARRAY Formula. We will discuss this in the following paragraphs.
### Specifying Minimum and Maximum Values with RANDARRAY Formula
Don’t let FOMO get to you! Advanced Uses of the RANDARRAY Formula in Excel can make a huge difference in your productivity and efficiency when handling large data sets. Plus, it can help you juggle multiple projects and tight deadlines while still achieving great results!
To Specify Minimum and Maximum Values with RANDARRAY Formula in Microsoft Excel, follow these five steps:
1. Open Excel and select the cell where you want to generate random numbers.
2. Type “RANDARRAY(” followed by two arguments (rows and columns). For example, RANDARRAY(5,1) for 5 rows and 1 column.
3. Insert min_value argument to restrict the lowest randomized number.
4. Add max_value argument to set the highest random value.
5. End with “)” to finalize.
This will allow you to limit the range and create unique sets of random numbers. In the next section, find out other cool ways to use this formula along with other Excel functions.
## Advanced Uses of the RANDARRAY Formula in Excel
As an Excel fan, I’m always searching for new and thrilling formulae. The RANDARRAY formula caught my eye. It can conveniently generate random numbers. But do you know it can do more? Let’s explore some advanced uses of the RANDARRAY formula in Excel. In the following subsections, I’ll show you how to generate random dates, letters, and even words with it. Get ready to level up your Excel abilities!
### Generating Random Dates using RANDARRAY
Excel may display random numbers instead of dates. To fix this, select cells and press “CTRL+1”. Then select “Date” under “Type”.
You can also use the “&” operator to generate random dates. For example: “=RANDARRAY(10)*50+DATE(2022,1,1)” adds 50 random days to January 1st, 2022.
Pro Tip: Easily remember dates with named ranges. Highlight a cell with a date and go to “Formulas” -> “Define Name”. Give it a name and use it in formulas later.
Generate random letters using the RANDARRAY formula. Just use the array-formula “{=CHAR(RIGHT(LEFT(UPPER(CODE(RANDARRAY(1,10,65,90)))),1)+ROUNDUP(RANDARRAY(1,10)*25,)}”. This generates ten uppercase random letters.
### Generating Random Letters using RANDARRAY formula
Generating random letters using the RANDARRAY formula in Excel is easy! Follow these steps:
1. Open a new or existing spreadsheet.
2. Choose a cell to put the random letter.
3. Type in the formula: “=CHAR(RANDARRAY(1,1,65,90))”.
4. Press enter.
5. You’ll now see a random letter.
6. To generate more, drag the formula down or to the right.
Using the RANDARRAY formula can be powerful. We use CHAR and ASCII to manipulate the output of RANDARRAY.
In this case, we want to generate a single number between 65 and 90 – the ASCII codes for uppercase letters. CHAR then converts the number into a randomized letter. This is useful for passwords, sorting algorithms, or other situations where randomized letters are needed.
Try this method for generating random letters in Excel. Change the parameters and see the results!
Next, let’s explore how to generate entire words using the same method but with added formulas and functions.
### Generating Random Words using RANDARRAY formula
RANDARRAY is a must-have skill in Excel! It’s useful to generate random data for tests, educational tasks and fun activities. To use it, make a list of words. Then use INDEX with RANDBETWEEN to select one randomly. For example, if there are 10 words in cells A1 to A10. Use: =INDEX(\$A\$1:\$A\$10,RANDBETWEEN(1,10))
Another way: use CHOOSE with RANDBETWEEN. If there’s a list of 5 names in A1 to A5, enter: =CHOOSE(RANDBETWEEN(1,5),A1,A2,A3,A4,A5). This will give you a random word every time!
## Five Facts About RANDARRAY: Excel Formulae Explained:
• ✅ RANDARRAY is a new dynamic array function in Excel that generates an array of random numbers between 0 and 1. (Source: Microsoft)
• ✅ The RANDARRAY function can be used to generate random data for simulations, games, and statistical analysis. (Source: Excel Jet)
• ✅ The RANDARRAY function has two arguments: rows and columns, which specify the size of the array to generate. (Source: Spreadsheet Planet)
• ✅ RANDARRAY is a volatile function, which means it recalculates every time you make a change to the worksheet. (Source: Ablebits)
• ✅ To generate random integers, you can use the RANDBETWEEN function along with the RANDARRAY function. (Source: Spreadsheeto)
## FAQs about Randarray: Excel Formulae Explained
### What is RANDARRAY: Excel Formulae Explained?
RANDARRAY is a new function in Excel that generates an array of random numbers with a specified number of rows and columns. It can be used as an alternative to the RAND function, which only generates a single random number.
### How do I use RANDARRAY?
To use RANDARRAY, you need to specify the number of rows and columns in the array, as well as any additional options you want to include. For example, you can specify whether the numbers should be integers or decimal values, or whether they should be unique or not.
### What are some common use cases for RANDARRAY?
RANDARRAY can be used for a variety of purposes, such as generating random data sets for testing or simulation, creating randomized lists or sorting orders, or generating unique IDs or passwords. It can also be combined with other functions, such as SUM or AVERAGE, to perform more complex calculations.
### How does RANDARRAY differ from other random number functions?
RANDARRAY has several advantages over other random number functions in Excel. It can generate multiple random numbers at once, allowing for faster and more efficient calculations. It can also specify the range and type of random numbers, making it more versatile than other functions like RAND or RANDBETWEEN.
### Is RANDARRAY available in all versions of Excel?
RANDARRAY is only available in certain versions of Excel, including Microsoft 365 and Excel 2019 for Windows and Mac. It is not available in earlier versions of Excel or in Excel for mobile devices.
### Are there any limitations to using RANDARRAY?
One potential limitation of using RANDARRAY is that it can slow down your spreadsheet if used excessively or with large arrays. It may also be more difficult to use for those who are not familiar with array formulas in Excel.
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## Precalculus: Mathematics for Calculus, 7th Edition
$\frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$
$Expand$ $the$ $logarithmic$ $expression:$ $\ln \sqrt {\frac{x^2-1}{x^2+1}}$ Rewrite the root to exponent form $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}}$$ Use the Third Law of Logarithms $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} = \frac{1}{2}\ln (\frac{x^2-1}{x^2+1})$$ Use the Second Law of Logarithms [Distribute the $\frac{1}{2}$ as well] $$\frac{1}{2}\ln (\frac{x^2-1}{x^2+1}) = \frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$$
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https://www.coursehero.com/file/p5vpb8q5/d-What-is-the-x-intercept-of-the-demand-function-and-what-does-it-represent-in/
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d What is the x intercept of the demand function and what does it represent in
# D what is the x intercept of the demand function and
• 6
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situation? d. What is the x –intercept of the demand function and what does it represent in the context of the situation?
6. The function )(thoutputs the height of a toy rocket measured in feet at time t seconds after it was launched. 212816)(2++=tttha. Calculate the height of the rocket at time 2 seconds after launch. b. Calculate the height of the rocket at time 3 seconds after launch. c. Calculate the slope of the line that goes through the two points you found in part a and b. d. What are the units of this slope? e. What does that slope represent? f. How could this be represented on the graph of )(th ?
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# I mix 9.99 grams of nitrogen, 9.18 grams of hydrogen and 9.05 grams of helium in a flask. If the total pressure in the system was found to be 301 kPa. How much pressure of the total does each gas exert?a) The pressure due to the hydrogen was b) The pressure due to the nitrogen was c) The pressure due to the helium was
Question
I mix 9.99 grams of nitrogen, 9.18 grams of hydrogen and 9.05 grams of helium in a flask. If the total pressure in the system was found to be 301 kPa. How much pressure of the total does each gas exert?
a) The pressure due to the hydrogen was
b) The pressure due to the nitrogen was
c) The pressure due to the helium was
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# Crossnumber puzzle
This is a numeric crossword: all the answers are positive numbers with no leading zeroes. Each clue is given as a sequence of mathematical operations on letters, each of which represents a distinct prime number (e.g. if $$A=2$$ and $$B=3$$ then a clue might be written as $$1.\ A^A+B\quad(1)$$, to which the answer is $$7$$. Numbers in parentheses after the clue indicate the length of the answer, not necessarily the length of the grid entry.
The rules:
• No grid entry starts with a leading zero
• Before an answer can be entered into the grid the solver must perform a basic transformation on it
• All answers and entries are unique
• Once the solver has filled the grid they must choose a digit from all those appearing in the grid and shade all the cells corresponding to that digit. What appears must then be appropriately transformed and written below the grid in the space provided.
• While parentheses have been provided for clarity, if there is any doubt then standard mathematical ordering of operations is meant (BODMAS)
• Your lucky numbers for this puzzle are $$30, 682$$ and $$15554$$.
All feedback is welcome :)
Clues:
Across
1. $$E\times (A+T) - (O-D)$$ (5)
11. $$D^S + D^D$$ (3)
12. $$((A\times H)-E)/O$$ (2)
14. $$(S\times E) - (N\times (A+T) + O)$$ (4)
16. $$H\times (A+D)$$ (4)
18. $$A^D - H +(D\times T)$$(2)
19. $$O^D \times T \times A -(O\times D)$$ (3)
Down
1. $$T^D$$ (2)
2. $$O \times S$$ (2)
3. $$D$$ (1)
4. $$N-O\times D\times D$$ (2)
5. $$N$$ (2)
6. $$D\times N - T\times O\times O$$ (2)
7. $$A-O$$ (1)
8. $$S$$ (1)
9. $$\sqrt{O\times N +S}$$ (2)
10. $$D\times (T^D - D\times A)^D$$ (2)
13. $$O^D$$ (1)
15. $$T$$ (1)
17. $$O$$ (1)
• What is "a basic transformation" as given by rule 2? – RShields Jun 21 '19 at 14:18
• There's no way all variables can be one digit. 1-across is impossible if that's the case @OmegaKrypton – RShields Jun 21 '19 at 14:24
• oops... thanks @RShields – Omega Krypton Jun 21 '19 at 14:26
• Can you double-check 10 Down, 1 Across, and 19 Across? I'm getting a contradiction that probably involves at least one of those three. – Deusovi Jun 21 '19 at 14:28
• @RShields that's part of the puzzle :) – postmortes Jun 21 '19 at 14:29
First of all,
it's natural to guess that the "basic transformation" is, literally, a change of base, and a moment's consideration shows that if so the base must be 2. We can write in lots of 1s because no answer begins with a 0. Since $$O^D$$ is a single decimal digit O,D are 2,3 in some order. We have enough information also to deduce that S is 7 via 8d. Now the fact that 11a is 8 bits long tells us that D=2 and O=3 rather than the other way around, and we can fill in 11a. Since A-O is 10xx, A is in the range 11..16, so it's either 11 or 13. T is a 3-bit prime and isn't 7, so T=5. Now 6d is 2N-45 and its bits are 10xxx, and N is 11xx1; we quickly find that in fact N is 31. 19a is 45A-1 and the bits we know are enough to rule out A=13; so A=11. Now we have 1a = 16E-1, from which we can immediately fill in the last four bits (we already have two of them), and then the only way to fill in the blanks to make E prime is to make the other two be zeros and E=991. Now we can fill in a bunch of other things just by doing arithmetic. We end up with 18a being 1111x which requires x=0 and H=101 and can fill in everything else.
At this point the grid looks, if I haven't miscalculated, like this:
Now
the 1s in the grid appear to say SOS, which perhaps we should transform via a different kinda-binary code into ... --- .... postmortes, is everything OK? :-) A hint in the direction of this particular kinda-binary code comes from the "lucky numbers", which equal the products DOT, AND, DASH (using the letter assignments we have discovered in solving the puzzle).
Credit where due: Deusovi started on this before I did and worked faster than I did, and had he not run into errors in the original puzzle he'd doubtless have finished long before I did. (I haven't looked at anything in his answer, though, beyond checking whether the contradiction he found was the same as the one I did at a similar time.) Go and upvote his answer!
• Everything you say is right, except that you seem to have the calculation for 14A wrong. I've rechecked it in Excel and it's coming out right, so my guess is that you're adding O to (A+T) before multiplying by N? If I add extra brackets, you should be finding $(S\times E) - ((N \times (A+T)) + O)$ That gives you an odd number minus an odd number which is even, which fits your grid so far – postmortes Jun 21 '19 at 15:43
• Ah, no, actually an even simpler error -- I omitted the O. Thanks! – Gareth McCaughan Jun 21 '19 at 16:01
• Yes, that's all correct. If you factor the lucky numbers you'll get a hint for the encoding – postmortes Jun 21 '19 at 16:07
• Ah yes, so I do. Will edit accordingly. – Gareth McCaughan Jun 21 '19 at 16:18
# Partial Answer (outdated by changes to the puzzle)
The trick to this puzzle is that
the "basic transformation" is binary: each answer must be converted to binary first before being entered.
The first step is to look at
the small numbers. From the short Down entries, T and S must be 5 and 7, and D and O must be 2 and 3.
From 1-Down, we need a two-digit power of either 5 or 7 that is 11XX1 in binary: this can only be 25 (from $$5^2$$), and so all four of those primes are determined.
Now that four of the variables are determined,
we look at 19-A. This must be $$9*5*A$$, where $$A$$ is a two-digit or larger prime. 11 is the only option that is small enough to fit in the grid, making 111101111. However, 10-Down must be even, and therefore must end in a 0, but the bottom right corner is a 1.
This seems to be a contradiction, but unfortunately I can't continue working on this at the moment.
• I interpreted the "basic transformation" the same way but found what seems a more elementary contradiction: 13d determines {O,D} though not which way around they are, which means neither of A,T can be a particular one of those, which constrains the end of 1a in an impossible way. But OP has fixed it now... – Gareth McCaughan Jun 21 '19 at 14:49
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# 100532900 (number)
100,532,900 (one hundred million five hundred thirty-two thousand nine hundred) is an even nine-digits composite number following 100532899 and preceding 100532901. In scientific notation, it is written as 1.005329 × 108. The sum of its digits is 20. It has a total of 7 prime factors and 72 positive divisors. There are 34,928,640 positive integers (up to 100532900) that are relatively prime to 100532900.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 20
• Digital Root 2
## Name
Short name 100 million 532 thousand 900 one hundred million five hundred thirty-two thousand nine hundred
## Notation
Scientific notation 1.005329 × 108 100.5329 × 106
## Prime Factorization of 100532900
Prime Factorization 22 × 52 × 13 × 17 × 4549
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 10053290 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 100,532,900 is 22 × 52 × 13 × 17 × 4549. Since it has a total of 7 prime factors, 100,532,900 is a composite number.
## Divisors of 100532900
72 divisors
Even divisors 48 24 24 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 72 Total number of the positive divisors of n σ(n) 2.48812e+08 Sum of all the positive divisors of n s(n) 1.48279e+08 Sum of the proper positive divisors of n A(n) 3.45572e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 10026.6 Returns the nth root of the product of n divisors H(n) 29.0917 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 100,532,900 can be divided by 72 positive divisors (out of which 48 are even, and 24 are odd). The sum of these divisors (counting 100,532,900) is 248,812,200, the average is 3,455,725.
## Other Arithmetic Functions (n = 100532900)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 34928640 Total number of positive integers not greater than n that are coprime to n λ(n) 90960 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5783009 Total number of primes less than or equal to n r2(n) 96 The number of ways n can be represented as the sum of 2 squares
There are 34,928,640 positive integers (less than 100,532,900) that are coprime with 100,532,900. And there are approximately 5,783,009 prime numbers less than or equal to 100,532,900.
## Divisibility of 100532900
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 4 2
The number 100,532,900 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (100532900)
Base System Value
2 Binary 101111111100000001010100100
3 Ternary 21000011121012202
4 Quaternary 11333200022210
5 Quinary 201214023100
6 Senary 13550434032
8 Octal 577401244
10 Decimal 100532900
12 Duodecimal 29802918
20 Vigesimal 1b86c50
36 Base36 1nurok
## Basic calculations (n = 100532900)
### Multiplication
n×y
n×2 201065800 301598700 402131600 502664500
### Division
n÷y
n÷2 5.02664e+07 3.3511e+07 2.51332e+07 2.01066e+07
### Exponentiation
ny
n2 10106863982410000 1016072346057226289000000 102148699558936524789408100000000 10269304997888609753001085576490000000000
### Nth Root
y√n
2√n 10026.6 464.982 100.133 39.8531
## 100532900 as geometric shapes
### Circle
Diameter 2.01066e+08 6.31667e+08 3.17516e+16
### Sphere
Volume 4.25611e+24 1.27007e+17 6.31667e+08
### Square
Length = n
Perimeter 4.02132e+08 1.01069e+16 1.42175e+08
### Cube
Length = n
Surface area 6.06412e+16 1.01607e+24 1.74128e+08
### Equilateral Triangle
Length = n
Perimeter 3.01599e+08 4.3764e+15 8.7064e+07
### Triangular Pyramid
Length = n
Surface area 1.75056e+16 1.19745e+23 8.20848e+07
## Cryptographic Hash Functions
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Libros Libros
A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.
The Elements of Euclid, the parts read in the University of Cambridge [book ... - Página 2
por Euclides - 1846
Vista completa - Acerca de este libro
## Outlines of geometry; or, The motion of a point
Walter Marsham Adams - 1866 - 114 páginas
...Circle; and secondly, to examine some of the various parts of these figures.. " A Circle," says Euclid, " is a plane figure contained by one line, which is...figure to the circumference are equal to one another. And this point is called the centre of the circle " (1.) (AB=AP.) The definitions correspond exactly....
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## An English primer; compiled under the superintendence of E.C. Lowe
Edward Clarke Lowe - 1866 - 172 páginas
...which is enclosed by one or more boundaries. 15. A circle is a plane figure contained (or bounded) by one line, which is called the circumference, and...figure to the circumference are equal to one another. 1C. And this point is called the centre of the circle. 17. A diameter of a circle is a straight line...
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## The advanced lesson book, by E.T. Stevens and C. Hole
Edward Thomas Stevens - 1866 - 434 páginas
...product is the area. DD THE CIRCLE. Definition : — A circle J8 a plane figure bounded by one line called the circumference, and is such that all straight lines drawn from, a certain point within the circle to the circumference are equal to one another. This point is called the centre of the circle....
Vista completa - Acerca de este libro
## The Elements of Euclid for the Use of Schools and Colleges: Comprising the ...
Euclid, Isaac Todhunter - 1867 - 426 páginas
...is the extremity of any thing. 14. A figure is that which is enclosed by one or more boundaries. 15. A circle is a plane figure contained by one line,...figure to the circumference are equal to one another : 16. And this point is called the centre of the circle. 17. A diameter of a circle is a straight line...
Vista completa - Acerca de este libro
## Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
Robert Potts - 1868 - 434 páginas
...that which is less than a right angle. XTTI. A term or boundary is the extremity of any thing. xrv. XV. A circle is a plane figure contained by one line,...all straight lines drawn from a certain point within tie figure to the circumference, are equal to one another. XVI. And this point is called the center...
Vista completa - Acerca de este libro
## Elementary Mensuration for the use of schools, etc
Septimus Tebay - 1868 - 168 páginas
...10. An acute angle is less than a right angle. 11. A circle is a plane figure contained by one line, called the circumference, and is such that all straight lines drawn from a certain point within it to the circumference are equal to one another. 12. And this point is called the centre of the circle...
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## Mensuration for Beginners: With Numerous Examples
Isaac Todhunter - 1869 - 312 páginas
...polygon which are not adjacent is also called a 15. A circle is a plane figure bounded by one lino which is called the circumference, and is such that...figure to the circumference are equal to one another: this point is called the centre of the circle. A radius of a circle is a straight line drawn from the...
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## An Elementary Course of Plane Geometry
Richard Wormell - 1870 - 304 páginas
...Circumferences.— Use of compasses. 43. A circle is a figure contained by one line, termed the circumference, such that all straight lines drawn from a certain...within the figure to the circumference are equal. If we place one of the points of a pair of compasses upon a perfectly even board, and the compasses...
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## The Educational Speeches of the Hon'ble John Bruce Norton
John Bruce Norton - 1870 - 350 páginas
...allusion to it as " a plain figure contained by one line called the circumference, and such that all right lines drawn from a certain point within the figure to the circumference are equal to one another?" Why then should we stumble or 'be offended at the use of such terms as " Estopell," "Alimony," " Certiorari,"...
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## Gradations in Euclid : books i. and ii., with an explanatory preface [&c ...
Euclides - 1870 - 270 páginas
...same plane, the figure is called a plane figure." — HOSE. 15. A circle (circulus, a ring, or hoop) is a plane figure contained by one line, which is called the circumference (circumferre, to carry round), and is such that all straight lines drawn from a certain point within...
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https://calculator.academy/radial-distance-calculator/
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Enter the internal point (x1,y1) and the point along the radius (x2,y2) into the Radial Distance Calculator. The calculator will evaluate and display the Radial Distance.
The following formula is used to calculate the Radial Distance.
Drad = SQRT ( (X2-X1)^2+(Y2-Y1)^2))
• X1,Y1 is the internal point (x1,y1)
• X2,Y2 is the point along the radius (x2,y2)
## How to Calculate Radial Distance?
The following example problems outline how to calculate Radial Distance.
Example Problem #1:
1. First, determine the internal point (x1,y1).
• The internal point (x1,y1) is given as: 3,5.
2. Next, determine the point along the radius (x2,y2).
• The point along the radius (x2,y2) is provided as: 10,20.
3. Finally, calculate the Radial Distance using the equation above:
The values given above are inserted into the equation below and the solution is calculated:
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Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Topic: regular n-gon runners problem
Replies: 12 Last Post: Jul 25, 2013 4:26 AM
Messages: [ Previous | Next ]
quasi Posts: 11,916 Registered: 7/15/05
Re: regular n-gon runners problem
Posted: Jul 23, 2013 6:46 AM
Here's a revised version ...
Suppose n runners, n >= 3, start at the same time and place
on a circular track of circumference 1, and proceed to run
counterclockwise along the track (forever). Assume the speeds
v_1,v_2, ..., v_n of the runners, expressed in revolutions
per unit time, are positive real numbers such that
v_1 < v_2 < ... < v_n.
Conjecture:
There is an instant of time where the locations of the n runners
are the vertices of a regular n-gon iff each of the n fractions
(v_i - v_1)/(v_2 - v_1)
for i = 1,2,...,n is a rational number, and moreover, when
reduced to lowest terms the n numerators yield all possible
distinct residues mod n.
Example (1):
Speeds 1,2,4 cannot yield an equilateral triangle since, of the
fractions
(1-1)/(2-1) = 0/1
(2-1)/(2-1) = 1/1
(4-1)/(2-1) = 3/1
there is no reduced numerator congruent to 2 mod 3.
Example (2):
Speeds 1,2,6 do yield an equilateral triangle since, of the
fractions
(1-1)/(2-1) = 0/1
(2-1)/(2-1) = 1/1
(6-1)/(2-1) = 5/1
the numerators of the reduced fractions include all the
residues 0,1,2 mod 3.
quasi
Date Subject Author
7/23/13 quasi
7/23/13 Virgil
7/23/13 quasi
7/23/13 quasi
7/23/13 quasi
7/23/13 quasi
7/25/13 quasi
7/23/13 William Elliot
7/23/13 quasi
7/23/13 William Elliot
7/23/13 quasi
7/23/13 James Waldby
7/23/13 quasi
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# 21 Sampling variation
So far, you have learnt to ask a RQ, design a study, describe and summarise the data, and understand the decision-making process. In this chapter, you will learn to:
• explain what a sampling distribution describes.
• explain the difference between variation between individuals and variation in sample statistics.
• determine when a standard error is appropriate to use.
• explain the difference between standard errors and standard deviations.
## 21.1 Introduction
The previous chapter introduced the decision-making process (Sect. 20.4) used in research:
1. Make an assumption about the population parameter.
2. Based on this assumption, describe what values the sample statistic might reasonably be expected from all possible samples.
3. Observe the sample data from just one of those may possible samples.
4. Decide if the sample statistic seems consistent with the expectation, or if it contradicts the expectation.
Realising that the sample studied is only one of countless possible samples that could have been chosen is important.
Remember: Studying a sample leads to the following observations:
• Every sample is likely to be different.
• We observe just one of the many possible samples.
• Every sample is likely to yield a different value for the sample statistic.
• We observe just one of the many possible values for the statistic.
Since many values for the sample statistic are possible, the possible values of the sample statistic vary (called sampling variation) and have a distribution (called a sampling distribution).
Furthermore, under certain conditions, the variation of the sample statistic (such as the sample mean, etc.) from all possible samples can be described based on the assumption made about the parameter. As a result, the expected behaviour of these statistics can be described, so we know what to expect from the sample statistic when the assumption is true.
We saw this in Sect. 20.4: the sample proportion of red cards in a sample of $$15$$ varied from hand to hand, and was approximately distributed with a bell-shape. This bell-shaped distribution is formally called a normal distribution. This is no accident: Many sample statistics vary from sample to sample with an approximate normal distribution if certain conditions are met. We see examples of this in this chapter. Bell-shaped distributions are studied further in Chap. 22.
Any distribution that describes how a sample statistic varies for all possible samples is called a sampling distribution: how the value of the sample statistic varies from sample to sample for all possible samples. The sampling distribution often has a normal distribution shape.
Definition 21.1 (Sampling distribution) A sampling distribution is the distribution of some sample statistic, showing how its value varies from sample to sample.
## 21.2 Sample proportions have a distribution
Sample proportions, like all sample statistics, vary from sample to sample (Sect. 20.4); that is, sampling variation exists, so sample proportions have a sampling distribution.
Consider a European roulette wheel shown below in the animation: a ball is spun and can land on any number on the wheel from $$0$$ to $$36$$ (inclusive). Using the classical approach to probability, the probability of the ball landing on an odd number (an 'odd-spin') is $$p = 18/37 = 0.486$$. This is the population proportion.
If the wheel is spun (say) 15 times, the sample proportion of odd-spins in those 15 spins, denoted $$\hat{p}$$, will vary. But, how does $$\hat{p}$$ vary from one set of 15 spins to another set of 15 spins? Can we describe how the value of $$\hat{p}$$ varies?
Computer simulation can be used to demonstrate what happens if the wheel was spun, over and over again, for $$n = 15$$ spins each time, and the proportion of odd-spins was recorded for each repetition. The proportion of odd spins $$\hat{p}$$ can vary from sample to sample (sampling variation) for one sample of $$n = 15$$ spins, as shown by the histogram (Fig. 21.1, top left panel). The shape of the distribution is approximately bell shaped. We can see that, for many repetitions, $$\hat{p}$$ is rarely smaller than $$0.2$$, and rarely larger than $$0.8$$. That is, reasonable values to expect for $$\hat{p}$$ are between about $$0.2$$ and $$0.8$$.
If the wheel was spun (say) $$n = 25$$ times (rather than $$15$$ times), $$\hat{p}$$ again varies (Fig. 21.1, top right panel): the values of $$\hat{p}$$ vary from sample to sample. The same process can be repeated for many repetitions of (say) $$n = 100$$ and $$n = 200$$ spins (Fig. 21.1, bottom panels).
Notice that as the sample size $$n$$ gets larger, the distribution of the values of $$\hat{p}$$ look more like a bell-shaped (normal) distribution, and the variation gets smaller. With $$200$$ spins, for instance, observing a sample proportion smaller than $$0.4$$ or greater than $$0.6$$ seems highly unusual, but these are not uncommon at all for $$15$$ spins.
Example 21.1 (Reasonable values for $\hat{p}$) Suppose we spun a roulette wheel $$100$$ times, and observed $$31$$ even numbers. The sample proportion is $$\hat{p} = 31/100 = 0.31$$. From Fig. 21.1 (bottom left panel), a sample proportion this low almost never occurs in a sample of $$100$$ rolls.
We observed something that is very unlikely to occur. This suggests that we observed something highly unusual, or a problem exists with the wheel.
The values of the sample proportion vary from sample to sample. The distribution of the possible values of the sample statistic (in this case the sample proportion) from sample to sample is called a sampling distribution.
Under certain conditions, the sampling distribution of a sample proportion is described by an approximate a normal distribution. In general, the approximation gets better as the sample size gets larger, and the possible values of $$\hat{p}$$ vary less as the sample size gets larger.
The mean of the sampling distribution is called the sampling mean; the standard deviation of the sampling distribution is called the standard error.
## 21.3 Sample means have a distribution
The sample mean, like all sample statistics, varies from sample to sample (Sect. 20.4); that is, sampling variation exists, so sample means have a sampling distribution.
Consider a European roulette wheel again (Sect. 21.2). Rather than recording the sample proportion of odd-spins, suppose the sample mean of the numbers spun was recorded. If the wheel is spun (say) $$15$$ times, the sample mean of the spins $$\bar{x}$$ will vary from one set of $$15$$ spins to another. How does it vary?
Again, computer simulation can be used to demonstrate what could happen if the wheel was spun $$15$$ times, over and over again, and the mean of the spun numbers was recorded for each repetition. Clearly, the sample mean spin $$\bar{x}$$ can vary from sample to sample (sampling variation) for $$n = 15$$ spins (Fig. 21.2, top left panel).
When $$n = 15$$, the sample mean $$\bar{x}$$ indeed varies from sample to sample. The shape of the distribution again is approximately bell shaped. If the wheel was spun more than $$15$$ times (say, $$n = 50$$ times) something similar occurs (Fig. 21.2, top right panel): the values of $$\bar{x}$$ vary from sample to sample, and the values have an approximate bell-shaped (normal) distribution. In fact, the values of $$\bar{x}$$ have a bell-shaped distribution for other numbers of spins also (Fig. 21.2, bottom panels).
The values of the sample mean vary from sample to sample. The distribution of the possible values of a sample statistic, in this case the sample mean, is called a sampling distribution.
Under certain conditions, the sampling distribution of a sample mean is described by an approximate a normal distribution. In general, the approximation gets better as the sample size gets larger, and the possible values of $$\bar{x}$$ vary less as the sample size gets larger.
The mean of the sampling distribution is called the sampling mean; the standard deviation of the sampling distribution is called the standard error (Fig. 21.3).
Example 21.2 (Reasonable values for $\bar{x}$) Suppose we spun a roulette wheel $$100$$ times, and the mean of the observed numbers was $$\bar{x} = 18.9$$. From Fig. 21.2 (bottom left panel), a sample mean with this value does not look unusual at all; it would occur reasonably frequently.
Nothing suggests that a problem exists with the wheel.
As we have seen, each sample is likely to be different, so any statistic is likely to be vary from sample to sample. (The value of the population parameter does not change.) This variation in the possible values of the observed sampling statistic is called sampling variation.
## 21.4 Sampling means and standard errors
As seen in the previous two sections, the value of a sample statistic varies from sample to sample. The statistic we observe depends on which one of the countless samples is selected, and hence which value of the sample statistic is observed.
This means that the possible values of sample statistics that we could potentially observe have a distribution (specifically, a sampling distribution); see Fig. 21.3. Perhaps surprisingly, under certain conditions, the sampling distribution is often a normal distribution, as we have seen.
The mean of this sampling distribution is called the sampling mean. The sampling mean is the average value of the sample statistic, across all possible samples.
Definition 21.2 (Sampling mean) The sampling mean is the mean of the sampling distribution of a statistic.
The standard deviation of this sampling distribution is called the standard error. The standard error measures the value of the sample statistic is likely to vary across, across all of the possible samples; see Fig. 21.3. The standard error is a measure of how precisely the sample statistic estimates the population parameter. If every possible sample (of a given size) was found, and the statistic computed from each sample, the standard deviation of all these estimates is the standard error.
Definition 21.3 (Standard error) A standard error is the standard deviation of the sampling distribution of a statistic.
Figures 21.1 and 21.2 show that the variation in the values of the sample statistic get smaller for larger sample sizes. That is, the standard error gets smaller as the sample sizes get larger: the sample statistics show less variation for larger $$n$$.
This makes sense: larger samples generally produce more precise estimates. After all, that's the advantage of using larger samples: all else being equal, larger samples are preferred as they produce more precise estimates (Sect. 5.2).
Example 21.3 (Standard errors) In Fig. 21.2, a sample of $$250$$ (i.e., $$250$$ spins) is unlikely to produce a sample mean larger than $$20$$, or smaller than $$15$$. However, in a sample of size $$15$$ (i.e., $$15$$ spins) sample means near $$15$$ and $$20$$ are quite commonplace.
In samples of size $$100$$, the variation in the mean spin is smaller than in samples of size $$15$$. Hence, the standard error (the standard deviation of the sampling normal distributions) will be smaller for samples of size $$250$$ than for samples of size $$15$$.
For many sample statistics, the variation from sample to sample can be approximately described by a normal distribution (the sampling distribution) if certain conditions are met (Sect. 21.1). Furthermore, the standard deviation of this normal distribution is called the standard error. The standard error is a special name given to the standard deviation that describes the variation in a sample estimate that varies from from sample to sample.
The standard error is an unfortunate term: it is not an error, or even standard. (For example, there is no such thing as a 'non-standard error'.)
## 21.5 Standard deviation vs standard error
Even experienced researchers confuse the meaning and the usage of the terms standard deviation and standard error . Understanding the difference is important.
The standard deviation, in general, quantifies the amount of variation in any quantity that varies. The standard error specifically refers to the standard deviation that describes a sampling distribution.
Typically, standard deviations describe the variation in the individuals in a sample: how observations vary from individual to individual. The standard error is only used to describe how sample estimates vary from sample to sample (i.e., to describe the precision of sample estimates).
The standard error is a standard deviation, but is only used to describe the variation in sampling distributions. Any numerical quantity estimated from a sample (a statistic) can vary from sample to sample, and so has sampling variation, a sampling distribution, and hence a standard error. (Not all sampling distributions are normal distributions, however.)
Any quantity estimated from a sample (a statistic) has a standard error.
The standard error is often abbreviated to 'SE' or 's.e.'. For example, the 'standard error of the sample mean' is often written $$\text{s.e.}(\bar{x})$$, and the 'standard error of the sample proportion' is often written $$\text{s.e.}(\hat{p})$$.
## 21.6 Chapter summary
A sampling distribution describes how all possible values of a sample statistic is likely to vary from sample to sample. Under certain circumstances, the sampling distribution often can be described by a normal distribution. The standard deviation of this normal distribution is called a standard error. The standard error is the name specifically given to the standard deviation that describes the variation in the sample statistic across all possible samples.
## 21.7 Quick review questions
1. Why is phrase 'the standard error of the population proportion' inappropriate?
2. Which one the following does not have a standard error?
3. True or false: Sampling variation refers to how sample sizes vary.
4. True or false: Sampling distributions describe parameters.
5. True or false: Statistics do not vary from sample to sample.
6. True or false: Populations are numerically summarised using parameters
7. True of false: The standard deviation is a standard error of something quite specific.
8. True or false: Sampling distributions are always normal distributions.
9. True or false: Sampling variation measures the amount of variation in the individuals in the sample (for the variable of interest).
10. True or false: The standard error measures the size of the error when we use a sample to estimate a population.
11. True or false: In general, a standard deviation measures the amount of variation.
12. True or false: The standard error is a standard deviation with a specific use.
13. True or false: Sampling variation measures the size of the sample.
## 21.8 Exercises
Selected answers are available in App. E.
Exercise 21.1 In the following scenarios, would a standard deviation or a standard error be the appropriate way to measure the amount of variation? Explain.
1. Researchers are studying the spending habits of customers. They would like to measure the variation in the amount spent by shoppers per transaction at a supermarket.
2. Researchers are studying the time it takes for inner-city office workers to travel to work each morning. They would like to determine the precision with which their estimate (a mean of $$47$$ mins) has been measured.
3. A study examined the effect of taking a pain-relieving drug on children. The researchers wish to describe the sample they used in the study, including a description of how the ages of the children in the study vary.
4. A study examined the effect of taking a pain-relieving drug in teenagers. The researchers wished to report the percentage of teenagers in the sample that experienced side-effects with some indication of the precision of that estimate.
Exercise 21.2 Which of the following have a standard error?
1. The population proportion.
2. The sample median.
3. The sample IQR.
1. The sample standard deviation.
2. The population odds.
Exercise 21.3 A research article made this statement:
... authors often [incorrectly] use the standard error of the mean (SEM) to describe the variability of their sample...
What is wrong with using the standard error of the mean to describe the sample? How would you explain the difference between the standard error and the standard deviation to researchers who misuse the terms?
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# The Wonders of Shadows: A Lesson Plan for Students
## Wonders of Shadows:
Shadows are a fascinating aspect of light and its interaction with objects. In this lesson plan wonders of shadow, students will learn about how shadows are formed and why they are important.
Say: Light can pass through a few objects but can not pass through most of the objects.
We will begin by discussing how light interacts with different objects. Some objects, such as transparent materials, allow light to pass through, while most objects do not. This interaction creates a shadow.
Say: When any object blocks the way of light it casts a shadow.
Instruction: Distribute shadow frames to each group. The shadow frames are made of transparent material, and students will use them to observe how different objects cast shadows.
## Wonders of Shadows:
Instruction: Make room dark and keep shadow frame in way of torch or candle.
Say: You can see the boundary of the frame doesn’t allow light to pass, so it is casting a shadow. Say: The frame is made with transparent material. We know light can pass through transparent material, therefore it is not casting a shadow.
Say: Now I will put different objects in this frame and you will observe how they cast a shadow.
Instruction: Put different objects in the frame to cast the shadows. Explain to the students that light cannot pass through the object, which is why a shadow is formed.
Instruction: Divide the class into groups of five students each. Give every group one shadow frame. Instruction: They must sit in a semi-circle facing a wall.
Instruction: Place the torch in the middle of the semi-circle where the light will fall on the wall.
Say: Every group will use its shadow frame to cast a shadow on the wall. You have the freedom to choose any object from the class.
Instruction: Let students enjoy this activity.
Instruction: While students are busy with the activity, the teacher will go to each group and ask questions.
Ask: Why is this object casting a shadow?
Ask: Is the shadow of the same size as the object?
Ask: Do all the objects make shadows?
Ask: What is the most common shadow we see daily? Ans: Our own shadow
Ask: Can objects cast shadows during the night?
## Wonders of Shadows:
Through this lesson plan wonders of shadow, students will learn about the importance of shadows and how they are formed. They will also have the opportunity to observe and experiment with different objects and their shadows, making the learning experience both hands-on and interactive.”
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# How do you prove (sinx+cosx)^2 = 1+2sinxcosx?
Jun 20, 2015
Use trigonometric identities and the FOIL method.
#### Explanation:
We are asked to prove that ${\left(\sin x + \cos x\right)}^{2} = 1 + 2 \sin \left(x\right) \cos \left(x\right)$.
1) Change ${\left(\sin x + \cos x\right)}^{2}$ to $\left(\sin x + \cos x\right) \left(\sin x + \cos x\right)$ (since the square of any expression is that expression multiplied by itself.)
2) Utilize the FOIL method for multiplying binomials, e.g. $\left(\sin x + \cos x\right) \left(\sin x + \cos x\right) = \left(\sin x\right) \left(\sin x\right) + \left(\sin x\right) \left(\cos x\right) + \left(\cos x\right) \left(\sin x\right) + \left(\cos x\right) \left(\cos x\right)$
3) Simplify and group like terms: $\left(\sin x\right) \left(\sin x\right) + \left(\sin x\right) \left(\cos x\right) + \left(\cos x\right) \left(\sin x\right) + \left(\cos x\right) \left(\cos x\right) = {\sin}^{2} x + {\cos}^{2} x + 2 \sin x \cos x$
4) Recall the trigonometric identity which states ${\sin}^{2} x + {\cos}^{2} x = 1$, and substitute into (3): ${\sin}^{2} x + {\cos}^{2} x + 2 \sin x \cos x = 1 + 2 \sin x \cos x$
5) Use substitution: ${\left(\sin x + \cos x\right)}^{2} = 1 + 2 \sin x \cos x$
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# In a room as a cuboid with fresh air
maritza
Visitor
2016-08-13 06:12:02
#1
In a room shaped like a cuboid with a length of 14m, width 7m and 3.5m height are 40 students and teacher. It is known that the air in a room becomes dangerous when breathing in every cubic meter of air contains carbon dioxide 4dm3. It is known that each person 16h expire in a minute cite 0.5 dm3 air containing 5% carbon dioxide. In these conditions, determine the maximum tolerable as room is not ventilated.
## RE: In a room as a cuboid with fresh air
TeacherM
Visitor
2016-08-13 06:25:19
#2
Air volume of the cuboid: 14 * 7 * 1.4 = 343 m3.
As the air becomes dangerous => 343 * 4 = 1372 dm3 CO2.
A man produces 16 breaths per minute: 16 * 0.5 = 8 dm3 of which 5% is CO2 => 0.4 dm3 CO2.
In total there are 40 + 1 = 41 persons => 41 * 0.4 = 16.4 dm3 (the maximum administrative).
Dividing 1372 to 16.4 => 83 658 min => 83.65 min (after 83 minutes).
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https://discussions.unity.com/t/how-do-i-change-the-radius-in-rotatearound/225689
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How do i change the radius in RotateAround ?
Hello,
In my script, I have my player that move around his ennemi with RotateAround.
like this :
{
public GameObject Ennemi;
private float HorizontalMove;
private float VerticalMove;
``````void Start()
{
}
void Update()
{
HorizontalMove = Input.GetAxisRaw("Horizontal");
transform.RotateAround(Ennemi.transform.position, Vector3.up, HorizontalMove);
}
``````
}
And i want to change the radius of the rotation so he can get closer or run away.
How can i do this ? I’m still a begginner in Unity.
My script is attach to my player.
Thank you.
Here , try this
``````using System.Collections;
using System.Collections.Generic;
using UnityEngine;
public class PlayerRotation : MonoBehaviour {
public Transform Enemy;
private float _horizontalMove;
private float _verticalMove;
private Vector3 _playerPosition;
[SerializeField] private float _radius;
private void Update()
{
_playerPosition = _radius * Vector3.Normalize( this.transform.position - Enemy.position ) + Enemy.position;
_horizontalMove = Input.GetAxisRaw("Horizontal");
this.transform.position = _playerPosition;
transform.RotateAround(Enemy.position,Vector3.up,_horizontalMove);
}
}
``````
public GameObject Enemy;
public float HorizontalSensitivity = 1;
public float VerticalSensitivity = 1;
``````[Range(1,100)]
public float MinDistanceFromEnemy = 1;
[Range(1,100)]
public float MaxDistanceFromEnemy = 50;
private float HorizontalMove;
private float VerticalMove;
private void Start()
{
VerticalMove = Vector3.Distance( transform.position, Enemy.transform.position );
}
void Update()
{
HorizontalMove = Input.GetAxisRaw( "Horizontal" ) * HorizontalSensitivity;
VerticalMove = Mathf.Clamp( VerticalMove - Input.GetAxisRaw( "Vertical" ) * VerticalSensitivity, MinDistanceFromEnemy, MaxDistanceFromEnemy );
transform.position = Enemy.transform.position + ( transform.position - Enemy.transform.position ).normalized * VerticalMove;
transform.RotateAround( Enemy.transform.position, Vector3.up, HorizontalMove );
}
``````
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http://www.tcd.ie/Physics/research/groups/foams/media/oranges.php
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## Trinity College Dublin, The University of Dublin
You are here Research > Research Groups > Foams & Complex Systems > Media Coverage
# Why stacking oranges bears fruit for modern communications
An international workshop on packing problems took place in Trinity College Dublin earlier this week. Packing problems are concerned with storing objects as densely as possible in a container. Usually the goods and the container are of fixed shape and size.
The Foams and Complex Systems Group in TCD have recently discovered some new dense packings of spheres in cylindrical columns. Many packing problems arise in the context of industrial packaging, storage and transport, in biological systems, in crystal structures and in carbon nanotubes (tiny molecular-scale pipes).
Packing problems illustrate the interplay between pure and applied mathematics. They arise in practical situations but are then generalised and studied in an abstract mathematical context. The general results then find application in new practical situations. A specific example of this interplay is the sphere-packing problem.
In 1600, the adventurer Walter Raleigh asked his mathematical adviser Thomas Harriot about the most efficient way of stacking cannon balls on a ship’s deck. Harriot wrote to the famous astronomer Johannes Kepler, who formulated a conjecture that a so-called “face-centred cubic” was the optimal arrangement.
Let’s start with a simpler problem: how much of the table-top can you cover with non-overlapping €1 coins? Circular discs can be arranged quite densely in a plane. If they are set in a square formation, they cover about 79 cent of the surface. But a hexagonal arrangement, such as a honeycomb, with each coin touching six others, covers over 90 per cent; that’s pretty good. Joseph-Louis Lagrange showed in 1773 that no regular arrangement of discs does better than this. But what about irregular arrangements? It took until 1940 to rule them out.
In three dimensions, we could start with a layer of spheres arranged in a hexagonal pattern like the coins, and then build up successive layers, placing spheres in the gaps left in the layer below. This is how grocers instinctively pile oranges, and gunners stack cannon balls. The geometry is a bit trickier than in two dimensions, but it is not too difficult to show that this arrangement gives a packing density of 74 per cent. The great Carl Friedrich Gauss showed this is the best that can be done for a regular or lattice arrangement of spheres.
But again we ask: what about irregular arrangements? Is it not possible to find some exotic method of packing the spheres more densely? Kepler’s conjecture says no, and the problem has interested great mathematicians in the intervening 400 years. In 1900 David Hilbert listed 23 key problems for 20th century mathematicians, and the sphere-packing puzzle was part of his 18th problem.
In 1998 Thomas Hales announced a proof of Kepler’s Conjecture. He broke the problem into a large number of special cases and attacked each one separately. But there were some 100,000 cases, each requiring heavy calculation, far beyond human capacity, so his proof depended in an essential way upon using a computer.
After detailed review, Hales’ work was finally published in 2005 in a 120-page paper in Annals of Mathematics. Thus, Kepler’s Conjecture has become Hales’ Theorem. Most mathematicians accept that the matter is resolved, but there remains some discomfort about reliance on computers to establish mathematical truth.
Why should we concern ourselves with a problem for which grocers and cannoneers knew the solution long ago? Well, in higher dimensions the corresponding problem has more intriguing aspects. It is a key result in data communication: to minimise transmission errors, we design codes that are based on maximising the packing density of hyper-spheres in high- dimensional spaces. So the apparently abstruse conjecture of Kepler has some eminently practical implications for our technological world.
Peter Lynch, The Irish Times, September 2012
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# Risks and Insurance policies
## Presentation on theme: "Risks and Insurance policies"— Presentation transcript:
Risks and Insurance policies
Münchener Rück Risk Management in Civil Engineering – LNEC Risks and Insurance policies José Luis Montull
Münchener Rück Risk and Insurance Risk: it is not only the contingency or proximity of a damage (hazard), but also each and every contingency that can be subject of an insurance contract (Risk ) ¿quantifiable ? the mathematical product of the probability of a damage taking place, multiplied by the consequences of such damage: R = P x C where: P = Probability of an event C = Consequences of that event
Münchener Rück Risk and Insurance The Insurance is nothing else than a financial transaction by which a party, the policyholder (Insured), through a remuneration (the premium), promises to himself or to a third party a payment or a reparation in case of risk producing damages. The other party (the Insurer) who makes this payment, takes a set of risks, and is able to compensate them according to the statistical laws.
Risks-Matrix The Risks-Matrix: Major Risks. (Example)
Münchener Rück Risks-Matrix The Risks-Matrix: Major Risks. (Example) Risk Probability Intensity Risk Control Clasification Insurable Insurance Policy Absenteeism and productivity decrease Almost sure Moderate Not adequate Extreme No No investment's shortage in I+D, TI, RRHH Possible Severe Not adequate Extreme No No Political inestability Rare Catastrophic Not adequate High No No Theft of goods or money Almost sure Moderate Not adequate Extreme Yes No Manpower costs explosion Probable Moderate Not adequate Extreme No No Specific sectorial risks: manipulation of materials, stress,.. Almost sure Moderate Adequate Extreme Yes Yes Key Executives travelling together Unlikely Catastrophic Not adequate High Yes Yes
Risks Matrix: Major risks.
Münchener Rück Risks-Matrix Risks Matrix: Major risks. INTENSITY negligible Light Moderate Severe Catastrophic M H E E E Almost sure M M H E E Probable L M H H E PROBABILITY Possible L L M H E Unlikely L L M H H Rare L Low Risk Controlled by skilled workers and surveyors. M Moderate Risk Managed by middle staff. H High Risk General Managers attention is required E Extreme Risk The Board must be informed, and a permanent follow up is required.
Risks hierarchy Münchener Rück INTENSITY EXTREME RISK HIGH RISK
MODERATE RISK LOW RISK FREQUENCY
Münchener Rück ¿how do we face the Risk? Retention: establishing enough reserves to overcome the possible losses option 1 Risk option 2 Transfer: negotiating a specific insurance in order to give external coverage to the possible occurrence of a loss.
Retention or Risk Transfer
Münchener Rück Retention or Risk Transfer ¿Retention? Risks with high probability and low consequences. (Risks have to be taken or "self-insured" if they don't affect the financial stability of the company). ¿Risk Transfer? Risk with low probability and severe consequences (it is necessary to establish the limits of “significancy of a risk” for the company, individually and combined).
The Cost of the Risk % on Incomes Münchener Rück Expected Loss Ratio
Costs of the Risk Control Costs of the Financing of Risk Costs of Risk Reduction Unexpected Costs % on Incomes
What happens actually? Feb 2005 “A high number of companies never overcome a major loss, because of not beeing forewarned” “The important thing is not to find out who will pay (usually an insurance company), but to ensure that your business will not be forced to stop, that you will not loose either clients or suppliers and, even, that your competitors will not take advantage of the circumstance to finish off your business.”
Risks clasification Identified and Known assessable.
Münchener Rück Risks clasification Identified and Known assessable. Identified and Unknown because it's difficult to evaluate its consequences. Unidentified and Unknown surely we will be surprised (development risks).
Particularities of Civil Engineering
Münchener Rück Particularities of Civil Engineering Risks are often 'Unknown' (identified or not), due to: Each project is different from the previous, always showing different elements (ground conditions, climate, workmanship…) Technological development and innovation are steady, both in machinery and in materials Frequently there is non-skilled labour in the lowest levels of the chain Use and maintenance are sometimes not as careful as they should.
The Risk in Civil Engineering
Münchener Rück The Risk in Civil Engineering RISK FACTORS (material damages): what do they depend on? a) of Internal origin: b) of External origin: SOURCE CONSEQUENCES Works performance (incl. materiales) Acts of third parties Natural hazards (Acts of God) Project
The Risk in Civil Engineering
Münchener Rück The Risk in Civil Engineering RISK FACTORS (material damages): what do they depend on? a) of Internal origin: on the quality of the project itself (its conception and design), not only regarding the future service's stresses, but also concerning the stresses during the construction phase, as well as the construction method chosen on the approach to the construction works: quality and adequacy of the resources (personnel, machinery, measurement instruments, etc), and on the planning, choice and supply materials and their control, etc.
The Risk in Civil Engineering
Münchener Rück The Risk in Civil Engineering RISK FACTORS (material damages): what do they depend on? b) of External origin: They DO NOT depend, in their origin, on the project and construction works (although in some cases they DO depend with regard to their probability of occurrence -example: acts of third parties-) and, in all cases, they also DO depend regarding the severity of their consequencies –example: flood, earthquake, etc.-)
Example: Risks distribution
Münchener Rück Construction All Risk Example: Risks distribution
Example: Risks distribution 'insurable' Risks 'not insurable' Risks
Münchener Rück Construction All Risk Example: Risks distribution 'insurable' Risks 'not insurable' Risks
Risk's analysis Deterministic analysis
Münchener Rück Risk's analysis Deterministic analysis Based on scientific theories or empiric formulations: identical result if you take the same assumptions for the process (ex.: rust) Probabilistic analysis Based on occurrence probability of different risk's scenarios, estimating their consequences (ex.: flood) Statistical analysis Based on previous experience on existing statistics of losses (ex.: personal accidents at work)
Pilars of insurance activity
Münchener Rück Pilars of insurance activity Sufficient statistical base Financial stability Risks heterogeneity Geographical compensation
Statistical base for the analisys
Münchener Rück Statistical base for the analisys Very difficult to apply to Civil Engineering, due to the analyzed particularities: - projects are all different , - climate conditions influence in each and every case, - heterogeneous coverages - etc. Study of historical losses Number of losses. Date of occurrence and close date. Insured. Location. Description. Branch, guarantee, policy number and insurer. Split of losses's amount: total losses, insured losses, not covered losses. deductibles. pay back from other insurers, pay back from third parties, fees.
Risks distribution Exemple: Construction All Risk
Münchener Rück Exemple: Construction All Risk Risks distribution Losses > € (period 1990 – 2001) SOURCE: Münchener Rück
Engineering Insurance Policies
Münchener Rück Engineering Insurance Policies Annual policies (MB) – Machinery Breakdown (BI-MB) – Business Interruption, resulting from a loss covered under the MB policy (DOS) – Deterioration of Goods (EEI) – Electronic Equipment (CPM) – Contractors Plants & Machinery (CECR) – Civil Engineering Completed Risks
Engineering Insurance Policies
Münchener Rück Engineering Insurance Policies Temporary policies (CAR) – Construction All Risk (EAR) – Erection All Risks (CAR-ALoP) – Advanced Loss of Profit in CAR (EAR-ALoP) – Advanced Loss of Profit in EAR Comprehensive policies (BOT, BOO,..)
Engineering Insurance Policies
Münchener Rück Engineering Insurance Policies Multi-annual policies Decennial (10 years cover) Inherent Defects (2, 3 or more years cover) ....
Engineering Insurance Policies
Münchener Rück Engineering Insurance Policies Special policies Off-shore Risks (Oil-Platforms, Undersea-Pipelines,...) Technological Risks (Manufacturer's Guarantee,...) Space Risks (Moon Shots, Space Stations, ...)
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) Target policyholder? each and every company taking part in the construction process (having interests on the construction itself) Owner / Promoter Main Contractor Subcontractors Policyholders [actually, the 'insured' is the Construction itself]
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) what types of works can be insured? All kinds of works and facilities Housing and buildings: residential, offices, hospitals, schools, industrial buildings, car parks, shopping centers, stadiums, power stations, etc. Civil works, infrastructures: tracks (motorways, railways, pipelines, tunnels,...), bridges, hydraulic works (dams, water treatment plants,...), weet risks (breakwater, harbours,...), etc.
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) what is/can be included under the coverage? All works, goods and tools which take part in the construction process or which are adjacent to the construction site: CONST.WORKS: stored goods & materials, works brought on-site as part of the contract (earthworks, foundations, structures and finishing,..) temporary works erected or constructed on-site,... MACHINNERY: cranes, bull-dozers, dumpers, pave-machinery... EQUIPMENT: hand and machine tools, scaffolding, frames, pumps, ... EXISTING PROPERTY: in property or in custody of the insured (susceptible to suffer damages arising from the insured works)
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SUMS INSURED: The construction work itself (main coverage): total contract value at the completion date to regularize a margin should be foreseen (long time works) Machinery (optional guarantee): List with following data: individual values (“new replacement”) identification, technical features manufacture year internal elechtrical and/ormmechanical damage are excluded
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SUMS INSURED Construction Equipment (optional guarantee): total value of set (“new replacement value”) (a list should be attached) no heavy machinery Existing Property (optional guarantee): "first risk value" (it should be sufficient for the reconstruction or replacement of the damaged goods)
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SCOPE OF COVER It is an "all-less" policy (all damages are included, but the damages which are specifically excluded) Condition: Sudden and unforeseen physical loss or damage to the property insured, occurred during the period of insurance, in the construction site, due to any accidental cause not specifically excluded
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SCOPE OF COVER ¿which Risks are actually covered? Risks arising in connection with the execution of the project (unless that ones excluded) Natural hazars ("Acts of God", based on a defined threshold) Risks arising from acts of third parties (unless that ones excluded) CONTRACT WORKS (basic cover) CONSTRUCTION MACHINERY (optional cover) CONSTRUCTION PLANT&EQUIP. (optional cover) EXISTING PROPERTY (opcional cover) THIRD PARTY LIABILITY (optional cover) G O O D S to be C O V E R E D against D A M A G E S
Maintenance cover (optional)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) PERIOD OF COVER Maintenance cover (optional) Main (basic) Cover Before During construction works After const.w. construction works guarantees previous storage completion of works order of commitments with the works (for the main Contractor) inception of works Dowloading material on-site
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) PERIOD OF COVER Maintenance cover (optional) Main (basic) Cover Before During construction works After const.w. Natural hazars ("Acts of God", based on a defined threshold) Risks arising in connection with the execution of the project (unless that ones excluded) Risks arising from acts of third parties (unless that ones excluded)
Civil Engineering Insurance Policies
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) PERIOD OF COVER Special cases (information required) : advanced works (status of works: progress, ...) extensions / delay (reasons, occurred losses) interruptions, suspensions (reasons, status of works, security measures,...) early termination (status of works, occurred losses) phased handover (situation, delimitations)
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) MAIN EXCLUSIONS (1) Conventional exclusions war, armed conflicts nuclear risks expropiation, confiscation penalties (contractual or not) willful misconduct of the insureds or of the technical which are responsibles for the construction works consecuencial losses (indirect losses) documents, cash values
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) MAIN EXCLUSIONS (2) Specific exclusions Gradual deteriorations wear and tear 'normal' effects of climate rust or oxidation Defective design and workmanship (rectification) regarding materials regarding designs , calculations and drawings Deductibles (always have to be borne by the insured)
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) MAIN 'PROBLEMS' Exemple 1 'Normal' effects of climat excluded but... ¿what is 'normal'?
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) MAIN 'PROBLEMS' Exemple 2 Rectification of defective excluded design or workmanship but... ¿what is the 'defective part'?
Construction All Risk (CAR)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) RISK ASSESMENT: An individual analysis is required for each construction work: configuration and exposition of the construction site (natural hazards, acts of third parties,..) type of works and materials construction method safety measures in the planning costs assesment regarding several scenarios of losses Questionnaire and Application for Insurance
SPECIAL ENDORSEMENT (1)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SPECIAL ENDORSEMENT (1) To include some additional or extended cover (optional) 001 : SRCC (strike, riot and civil commotion) 002 : cross TPL 003 : maintenance visits 006 : extra costs of acceleration (overtime, night work,..) 007 : airfreight 113 : national transport 116 : contract works taken over or put into service 119 : existing property
SPECIAL ENDORSEMENT (2)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SPECIAL ENDORSEMENT (2) To delimitate some covers and/or exclusions (some examples) 008 : structures in seismic zones 101 : tunnels, galeries, subsurface structures 104 : dams and water reservoirs 106 : sections 107 : camps and stores 109 : construction material 110 : safety measures (with respect to rain, flood and inundation) 112 : fire-fighting facilities and fire safety on site 121 : piling foundations and retaining wall works
SPECIAL ENDORSEMENT (3)
Münchener Rück Civil Engineering Insurance Policies Construction All Risk (CAR) SPECIAL ENDORSEMENT (3) To exclude directly some usual coverages 009 : damages arising from earthquake 010 : damages arising from flood and inundation 103 : damages in crops, forest and cultures
Civil Engineering Completed Risks (CECR)
Münchener Rück Civil Engineering Insurance Policies Civil Engineering Completed Risks (CECR) Target policyholder? the owner or the operating company What types of works can be insured? civil works, infrastructures, with a low fire risk
Civil Engineering Completed Risks (CECR)
Münchener Rück Civil Engineering Insurance Policies Civil Engineering Completed Risks (CECR) SCOPE OF COVER It is a policy that covers "nominated risks" (a risk is covered only if it is referred specifically in the policy): Unforeseen and sudden physical loss or damage caused during the period of insurance by: Fire, lightning, explosion, impact of landborne or waterborne vehicles Impact of aircraft and other aerial devices or articles dropped therefrom Earthquake, volcanism, tsunami Storm (air movements stronger than grade 8 on the Beaufort Scale) Flood or inundation, wave action or water Subsidence, landslide, rockslide or any other earth movement Frost, avalanche, ice Vandalism of single persons.
BASIS OF LOSS SETTLEMENT
Münchener Rück Civil Engineering Insurance Policies Civil Engineering Completed Risks (CECR) BASIS OF LOSS SETTLEMENT in case of damage wich can be repaired: “the cost of repairs necessary to restore the items to their condition immediately before the occurrence of the damage”(less savage and r.o.d.) in case of a total loss: If the loss occurs within the period stated in the schedule the replacement costs If the loss occurs after the period stated in the schedule the actual value of the insured items immediately before the occurrence of the loss (less savage) -> [deducting a proper depreciation from a)]
Privately Financed Infrastructure Projects and Their Insurance
Münchener Rück Civil Engineering Insurance Policies Privately Financed Infrastructure Projects and Their Insurance PFI Privately financed investments Most important models: BOT Build - operate - transfer BOO Build - own – operate Examples of other models: BLT Build - lease - transfer ROT Rehabilitate - operate - transfer ROO Rehabilitate- own - operate DBFO Design - build - finance - operate
Privately Financed Infrastructure Projects and Their Insurance
Münchener Rück Civil Engineering Insurance Policies Privately Financed Infrastructure Projects and Their Insurance BOT and BOO In both models, a project company builds, owns and operates a plant. BOT: the plant is transferred to a public operator after a period of e.g. 15 years at a low price. BOO: a transfer is not intended.
Privately Financed Infrastructure Projects and Their Insurance
Civil Engineering Insurance Policies Privately Financed Infrastructure Projects and Their Insurance Typical for private financing of infrastructure projects: Sponsors invest only small fraction of the required capital Major funds borrowed from banks Liability of the sponsors in case of increased cost or delay in completion is limited Construction phase particularly critical: Major funds are spent without the project generating cash.
PFI: Finance and Contract Structure
Civil Engineering Insurance Policies PFI: Finance and Contract Structure Host Govern- ment (Principal) Contractor Banks/ Lenders project agreement construction contract loan agreement Suppliers supply contract Project Company shareholders' agreement insurance policy O&M contract Operator Sponsors Insurers
PFI: Finance and Contract Structure
Civil Engineering Insurance Policies PFI: Finance and Contract Structure A Project Company is formed to erect and operate the plant or the infrastruture by means of a network of contract agreements. It often is a stock company. Banks will provide the funds necessary beyond the Sponsors' investments. The Principal, i.e. the initiator of the project, may be the host government, a licencing governmental agency or a public utility. The Principal agrees the project terms with the Project Company. The Contractor and Operator are normally shareholders in the Project Company. The Insurers are vital in view of the limited-recourse financing scheme.
PFI: Flow of Payments (Operating Phase):
Civil Engineering Insurance Policies PFI: Flow of Payments (Operating Phase): Banks Sponsors Interest payments Interest and principal payments Project Co. Price for electricity Payment for fuel, materials and services Principal Suppliers
Typical examples of PFI Projects large and communal:
Civil Engineering Insurance Policies Typical examples of PFI Projects large and communal: Commercially used buildings or civil works: - Large public buildings, hospitals - Air ports - Bridges, highways, tunnels Power stations (IPPs) and power distribution Water distribution and water treatment plants
PFI Insurance Requirements:
Civil Engineering Insurance Policies PFI Insurance Requirements: Privately financed projects have a higher need for security. The project partners require a comprehensive insurance cover. The insured for construction and operation is always the project company.
Insurance covers of industrial projects
Civil Engineering Insurance Policies Insurance covers of industrial projects Marine Insurance Erection/Contractor`s All Risk Ins. incl. Liability Cover Maintenance Cover Marine Loss of Profit Contractor`s Plant and Machinery Insurance Guarantee Insurance Advance Loss of Profit Transport Storage Construction/Erection Works Testing Maintenance Operation Fire Ins. and Fire Loss of Profit Machinery Ins. and MLOP Electronic Equipment Ins. Public and Prod. Liability Ins. Workmen's Compensation Insurance
Staged Erection and Insurance Conventional Approach:
Civil Engineering Insurance Policies Staged Erection and Insurance Conventional Approach: (Example of a combined cycle power plant) Erection Tests Commercial operation 1st gas turbine + 200 MW Erection covers Operational covers Erection Tests Commercial operation 2nd gas turbine + 200 MW Erection covers Operational covers Steam turbine + 200 MW / combined-cycle unit 600 MW total Erection Tests Com. op. Erection (All Risks) covers Op. cov.
Staged Erection and Insurance Typical PFI Approach:
Civil Engineering Insurance Policies Staged Erection and Insurance Typical PFI Approach: (Example of a combined cycle power plant) 1st gas turbine + 200 MW Erection Test Commercial operation 2nd gas turbine + 200 MW Erection Test Com. oper. Steam turbine + 200 MW / combined-cycle unit 600 MW total Erection Tests Com. op. Annual operational covers Erection All-Risks cover
LoP insurance for a PFI Civil Engineering Insurance Policies
LoP is an essential cover due to the scarce financial resources of the Project Company ALoP difficult to monitor: Delays in early project stages may have - full impact - partial impact - no impact on the commencement of operation Requires detailed and updated project schedules Constant follow-up with insured and non-insured delays Suppliers Extension intensifies the problem: The EAR/CAR insurer is normally not the manufacturer's local property insurer
LoP insurance for a PFI Civil Engineering Insurance Policies
Time excess: ALoP: 60 days not uncommon, related to the total erection phase Operational LoP cover: days, related to each loss event individually.
What do the Banks ask for in a PFI ?
Civil Engineering Insurance Policies What do the Banks ask for in a PFI ? a) the Project Company shall obtain the widest possible insurance cover from the market, b) the banks' or lenders' interests shall be included, c) payment of indemnity shall be made to the banks or their trustees, d) all rights of recourse against the banks shall be waved, e) disregarding any right to refuse indemnity to the Project Company for any breach of conditions, the indemnity shall be paid to the banks including any loss of profits.
In PFI, beware of terms which aim at:
Civil Engineering Insurance Policies In PFI, beware of terms which aim at: an invalidation of policy conditions which would release the insured project contractor from the usual obligations, a direct claim or direct influence of a party outside the insured project contractors or principal executing the works, the imposing of obligations on the insurers towards the banks.
However, Civil Engineering Insurance Policies
PFI: Breach of Contract / Vitating Act in case of a Multiple Insureds Policy No indemnification of the violating party Other insureds to be indemnified for their direct loss Recourse against violating party not to be waived However, No indemnification of the banks
PFI: Multiple Insureds Clause (London Engineering Group)
Civil Engineering Insurance Policies PFI: Multiple Insureds Clause (London Engineering Group) "In the event of any vitiating act committed by any one or more insured parties, the lenders shall not be entitled to any indemnity under this policy for or arising from loss or damage in respect of which insurers are otherwise no longer liable to indemnify..."
PFI: Who has an Interest in Insurance?
Civil Engineering Insurance Policies PFI: Who has an Interest in Insurance? The Principal Host Government The Banks Indirectly Indirectly The Project Company (manufacturers, site contractors, operators included) Project Insurances
Insurance from a single source
Civil Engineering Insurance Policies Insurance from a single source Desirable for the Project Company No interfacing problems from one risk phase to the next Comprehensive insurance scheme requested by the banks from the beginning However, Covers traditionally underwritten by different markets and branches Manufacturers, suppliers, hauliers, operators may engage their own insurers or brokers Local insurers normally more engaged in operational than erection covers
Insurance / Reinsurance Arrangement:
Civil Engineering Insurance Policies Insurance / Reinsurance Arrangement: Project Company Intermediary Project insurance Package Policy Direct Insurer P. Bond Mar. M-BI CAR/EAR+BI MB/CECR+BI F F-BI op. TPL Reinsurer Credit Marine Engineering Fire Liability
Inherent defects Civil Engineering Insurance Policies CAR/EAR
Münchener Rück Civil Engineering Insurance Policies Legal protection covers Mortgage default Homeowners Liability (TPL) CAR/EAR Loss of Rent “One-stop shop” Real estate wrap-up protection A&E Liability NatCat Inherent defects Fire Contractor insolvency Vandalism Residual value insurance Bonds Accident covers
Is it necessary an insurance ?
Münchener Rück Civil Engineering Insurance Policies Different treatement of damages by inherent defects in buildings in european area Is it necessary an insurance ? ‘Property‘ or ‘Liability‘ insurance ? voluntary or mandatory insurance ?
Civil Engineering Insurance Policies
Münchener Rück Civil Engineering Insurance Policies Decennial insurance in Europe Mandatory Insurance Sweden and Finland ? Voluntary Insurance In developpment U.K. Holland Benelux France Italy Spain Portugal Greece Turkey
Civil Engineering Insurance Policies
Münchener Rück Civil Engineering Insurance Policies Example 1: France => mandatory ‘dual’ insurance system indemnification Prom./User ‘DO’ Cía. A Perito Experto Recurso ‘DO’ DL DL DL Cía.Aseg. X DL Cía.Aseg.Y Contractor X Liability share on damages between X and Y (CRAC) Contractor Y
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0Comment
# What is estimating in electrical?
Estimation (or estimating) is the process of finding an estimate, or approximation, which is a value that is usable for some purpose even if input data may be incomplete, uncertain, or unstable. The value is nonetheless usable because it is derived from the best information available.
The method of computation of all required engineering materials and the expenditure likely to be incurred in carrying out a given work before the actual execution of work is called estimation. Hence an estimation includes calculation of quantity involved and quality aspects of the material required.
### How can I estimate the cost of electricity in my home?
The average cost to wire a house is \$1.56 to \$3.75 per square foot with most homeowners spending roughly \$2.65 per square foot. To get a more accurate estimate, calculate the linear feet of your walls, and multiply your answer by \$7.79 per linear foot, which is the starting cost to wire the home. Follow Us .
### Types of Wires – There are several essential types of wire that one must know for understanding their place of usage.
• Triplex wires.
• Main feeder wires.
• Panel feed wires.
• Non-metallic sheathed wires.
• Single strand wires.
## What is the purpose of estimating and costing in electrical?
Estimating and costing thus serves the number of purposes in the construction process including preparation and finalize of bids and cost control. The main purpose is to provide to volume of work for cost control and to see that the adequate options of materials are explored during the execution of the project.
### The Cost of Upgrading Electrical Service
But modern homes may need as many as 200 amps to run air conditioners, computer equipment, high-definition televisions, and high-tech home automation devices. The cost of upgrading your existing electrical service panel to a 100- or 200-amp panel is \$800 to \$3,000.
## Tips to Improve Electrical Estimating in Construction
Here are some of our top tips to help improve electrical estimating in construction projects.
1. Review Cost Codes
No two businesses will use the same cost codes. Companies might use just a few cost codes, or they might have hundreds of cost codes. More cost codes mean that the company can track items in detail, however, they can be complicated to function. Companies should review their cost codes to determine whether their cost codes are functional or not and if it’s making sense for the business.
Overhead costs include all indirect costs incurred during the life of a project, such as utilities for the office and marketing costs. Accurately estimating overhead costs is extremely important, as an underestimated cost can lead to the project becoming unprofitable for the contractor and may even end up costing the contractor money.
3. Mobile Time Tracking
Mobile time tracking is an easy way to improve accuracy for estimating and costing electrical projects. Paper or manual time tracking doesn’t provide accurate information, and switching from manual to mobile time tracking helps improve labor cost estimation. It’s easier to know how much time a team needs to complete a specific cost code or certain projects. More accurate information leads to more accurate estimates, decreasing the number of budget overages by the end of the project.
4. Review Old Projects
One of the best ways to improve estimating is to look at old projects. Projects that are similar to the current bid can provide valuable data since an estimator might be able to see that similar projects always end up overbudget on labor costs or material costs and account for that in their new project estimate. Historical data can also be used to determine the general overhead during certain times of the year and more. They can also use it as a way to validate their estimates. If past projects always use a certain labor amount, then using that data to validate the labor cost can be beneficial.
5. Determine if the Project is a Good Fit
While some projects may seem to be appealing for a contractor, it is important to take the skill set of the workers into account. Using new technology or materials that your workers are unfamiliar with may greatly slow down the progress of the project and If your workers are not comfortable in the specialized field of the project they are working in, it may not be worth investing your time and money in.
## Our Services
Our reliable electrical estimators deliver few expert services that are as follows:
• Electrical material Takeoffs
• Electrical Cost per Square Foot
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http://math.stackexchange.com/questions/11986/guidance-on-a-complex-analysis-question
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# Guidance on a Complex Analysis question
My homework question: Show that all zeros of $$p(z)=z^4 + 6z + 3$$ lie in the circle of radius $2$ centered at the origin.
I know $p(z)$ has a zero-count of $4$ by using the Fundamental Theorem of Algebra. Then using the Local Representation Theorem the $$\int \frac{n}{z+a} = 4(2 \pi i).$$ I am assuming $a=0$ since we are centered at the origin. I apologize for my lack of math-type. What does $$= 8 \pi i$$ mean? Am I going around the unit circle $4$ times? Or is it even relevant to my final answer. Which I am assuming is finding the coordinates to the $4$ singularities. I have always looked for my singularities in the values that make the denominator zero, but in this question my denominator is $z$. $z=0$ doesn't seem right. So the question is, am I suppose to factor the polynomial $z^4 + 6z + 3$ to find the zeros?
Thanks
-
@Crystal: I typeset the equations. – user17762 Nov 26 '10 at 20:34
Hint: This kind of questions are usually handled using Rouche's Theorem. I suggest you look it up in the wikipedia article, where you can see an example of its usage. Also here's an example.
The key is choosing wisely another function $f(z)$ with which to compare in the inequality in Rouche's theorem and such that you can easily decide how many zeroes does $f(z)$ have inside the region you are considering, which in your case is the circle $|z| < 2$.
About your other question, you don't need to factor the polynomial in order to answer this.
-
Chhose $f(z)=z^4$ Then on $|z|=2$, $|f(z)-p(z)|<|p(z)|$ so by Rouche's Theorem $f(z)$ and $p(z)$ has same number of zeroes inside $|z|<2$ and we are done
-
You surely meant $|f(z)-p(z)|<|p(z)|$. – Philippe Malot Apr 27 '13 at 12:29
@girianshiido Thank you for notifying – Un Chien Andalou Apr 27 '13 at 12:38
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# 226: Swingset
(Difference between revisions)
Swingset Title text: Someone bring me a pocket fan so I can drift around the yard.
## Explanation
When on a swing, there is a moment between swinging forwards/backwards and falling back down again where the force of gravity stops the acceleration of the swing. In this moment, you remain almost stationary at the peak of your swing and on a perfect swing (i.e. one with no friction or air resistance) you would achieve weightlessness.
Cueball is told this fact by an unknown woman and then he imagines that at the peak of the swing you become permanently weightless and able to float above the ground without any support.
On the title text he asks for a pocket fan, believing he could fly around the garden using this small device.
## Transcript
[Woman talking to Cueball on swing-set.]
Woman: You know, at the peak of a big swing, you become weightless.
[Thought bubble from Cueball.]
[Cueball swings higher and higher. At the peak of a big swing he shoves off the swing. Cueball remains hovering in the air.]
Cueball: Hey guys. Come check this out.
# Discussion
A glass with water can be momentarily inverted at this moment and the water will not leave the glass!--DrMath 08:56, 16 November 2013 (UTC)
Isn't the point about illustrating that you do in fact have weight even in instences that are written off as weightless? In space you just happen to be falling at the same velocity of your surroundings, maintaining orbit simply by moving fast enough to miss the Earth. On top of which, in a low enough orbit g is still close to 9.8 m/s^2 if only because altitude is insignificant compared to the radius of the Earth.--Passing Stranger 14:10 August 2014 (UTC)
No, you don't have weight in some instances. Weight is dependent upon gravity, so in deep space with no planets or stars close enough to matter you would be weightless. Mass, on the other hand... 173.245.50.154 01:46, 10 January 2015 (UTC)
Weight is actually a description of reaction force; if you're in free fall, and therefore not being pushed on by the floor or pulled on by a rope, you are weightless. If you are being swung on a rope, the direction of your "weight" is constantly changing. This might seem arbitrary, but it avoids things like everyone on a rotating space station being considered "weightless" due to the lack of gravity; a closed physical system can't tell the difference between gravity and uniform acceleration. 108.162.238.159 08:51, 13 May 2015 (UTC)
When I'm alone in elevators, I'll sometimes jump right before the elevator stars to descend. Because I have to fall a longer distance than I jumped, it tricks my brain into feeling a moment of weightlessness more than what I feel at my apogee. I also sometimes like to float underwater for long periods of time, pretending I'm on the ISS. Unfortunately I'm fucking terrified of deep water, and due to my lack of water-based activity, I've quite declined in my ability to hold my breath underwater. I used to be able to do it for at least a minute to a minute and a half when I was 12. I used to either pretend I was Neo, or pretend I was on the ISS or generically in space. Never both, though. Now I can hardly do 20 seconds. Now that I really want to start swimming again, I can't. I grew up with a pool I hardly used, and now I'm in college, and all of our pools are lap-based. i.e. I can't hog 25sqft of space to just be all floaty in. I'd have to take up an entire lane, which I don't want to do. So the only way I'd really be able to experience this is if I scuba dived. Maybe I should do that again. It's the closest, for now, I'll ever get to feeling like I'm in space. Maybe later I can afford a Zero G flight. Maybe later I can do some space tourism stuff. Maybe later I'll be an actual astronaut. Only I'm two inches too short, for now, apparently -3- International Space Station (talk) 09:21, 27 October 2015 (UTC)
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# How do you find the number of ions in a compound?
## How do you work out the ions?
There is a quick way to work out what the charge on an ion should be: the number of charges on an ion formed by a metal is equal to the group number of the metal. the number of charges on an ion formed by a non-metal is equal to the group number minus eight. hydrogen forms H + ions.
## How many ions are in NaCl?
1 Expert Answer It is made up of sodium ions (Na+) and chloride ions (Cl-). 1 mole of sodium chloride contains 1 mole Na+ and 1 mole Cl-, so there are TWO moles of ions for every 1 mole of NaCl.
## What ions are in H2O?
• Water (H2O) splits into Hydrogen Ions (H+) and Hydroxyl Ions (OH-).
• When there are equal parts of Hydrogen Ions (H+) and Hydroxyl Ions (OH-) leading to a 1:1 ratio, pH is neutral (7).
## How many ions are in na2so4?
What you mentioned Na, S and O are three elements in the given compound. ∴ Total 3 ions.
## What is an ion in chemistry?
An ion is an atom or group of atoms in which the number of electron s is different from the number of proton s. If the number of electrons is less than the number of protons, the particle is a positive ion, also called a cation.
## How do you find Na+ ions?
Multiply the moles of \$Na_2SO_4\$ that we found just then by 2 to obtain the moles of sodium ions. As we know, \$6.022 imes 10^23\$ is the number of particles or ions in 1mole of any substance. We use Avogadro’s Number to find the number of ions present in any compound.
## How many ions are in HCl?
HCl is a strong acid. It dissociates completely in solution. 1 mol of HCl produces 1 mol of H₃O⁺ ions and 1 mol of OH⁻ ions, for a total of 2 mol of ions.
## Why Na+ is an ion?
Opposite charges attract each other (positive charge attracts negative charge, and vice versa), so a collection of cations and anions is stable. Neutral sodium atom (Na) becomes sodium cation (Na+) by releasing an electron.
## How many ions does Naoh?
Sodium Hydroxide is an ionic compound formed by two ions, Sodium Na+ and Hydroxide OH− . In order for these two polyatomic ions to bond the charges must be equal and opposite.
## How many H+ ions are in HCl?
pH Calculations Involving HCl Solutions For pH calculations, [H+] is expressed in moles per liter. Being a strong acid, we can assume that HCl completely dissociates (ionizes) in water. Additionally, since one molecule of HCl yields one [H+], the equivalent mass is equal to the molecular mass.
## How many H+ ions are in water?
There is 10-7 mole of H+(aq) in 1000 cm3 of water, and thus in one drop of water there are 10-7 x 6 x 1023 x 0.05/1000 = 3 x 1012 H+ ions per drop of water.
## How many ions are in CaCl2?
In calcium chloride (CaCl2), there are two Cl- ions for each Ca2+ ion. Notice that for ethylene glycol (above) the formula is C2H6O2 and not CH3O (the simplest ratio of C, H and O atoms). 2- ions) and NaOH (Na+ and OH- ions).
## How many ions are in Al2 SO4 3?
The subscript multiplies everything in ( ). the formula for aluminum sulfate is Al2(SO4)3. Then, in a formula unit of Al2(SO4)3 there are two aluminum ions and three sulfate ions; or two aluminum, three sulfur, and twelve oxygen atoms.
## How many ions are in AlCl3?
Aluminum chloride’s common name is aluminum trichloride. The tri- prefix shows that there are three chloride ions, as confirmed in the chemical formula AlCl3.
## What is the symbol of ion?
When writing the symbol for an ion, the one- or two-letter element symbol is written first, followed by a superscript. The superscript has the number of charges on the ion followed by a + (for positive ions or cations) or – (for negative ions or anions). Neutral atoms have a charge of zero, so no superscript is given.
## What ion is hydrogen?
hydrogen ion, strictly, the nucleus of a hydrogen atom separated from its accompanying electron. The hydrogen nucleus is made up of a particle carrying a unit positive electric charge, called a proton. The isolated hydrogen ion, represented by the symbol H+, is therefore customarily used to represent a proton.
## How do I find the charge of an element?
1. Use the periodic table. The usual charge of an element is common to its group.
2. Use a chart.
3. For a single atom, the charge is the number of protons minus the number of electrons.
4. Find the charge by balancing charge in a compound.
## Does NH4 have a charge?
Globally, therefore, the NH4 molecule is positively charged, i.e., it is a cation. This particular cation is called the ammonium ion. Notice that +1 and –1 formal charges in HN3 balance each other out. Overall, the molecule is electrostatically neutral.
## How many ions are in 1 mole of NaCl?
1 NaCl contains 1 Na+ and 1 Cl−. 1 mol NaCl contains 1 mol Na+ and 1 mol Cl−. 1 mol ions = 6.02∗1023 ions.
## Why does Na ion have a +1 charge?
A sodium atom can lose its outer electron. It will still have 11 positive protons but only 10 negative electrons. So, the overall charge is +1. A positive sign is added to the symbol for sodium, Na +.
## How many ions are present in sodium?
There are 12 sodium ions on the edges and one in the centre of the unit cell. Each sodium ion present on the edge contributes one fourth to the unit cell and sodium ion present at body centre contributes one to the unit cell.
## Which ions are present in hno3?
Nitric acid is a strong acid, completely ionized into hydronium (H3O+) and nitrate (NO3-) ions in an aqueous solution, and a powerful oxidizing agent.
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# discrete-time real exponential signal
4 views (last 30 days)
cikalekli on 23 Nov 2021
Answered: Paul on 24 Nov 2021
HOW CAN I DELETE THOSE POINTS WHICH ARE VISIBLE AT ZERO AT COUPLE TIMES?
I was trying to create a power signal as with α = -0.9 between -10 < t < 10
Here is the graph which I was trying to achieve:
clc; clear; close all
alpha = -.9;
t = -10 : 0.5 : 10;
y = power(alpha, t);
stem(t,y, '- b',...
'LineWidth',2);
Warning: Using only the real component of complex data.
title ('\bf\it 3rd QUESTION ', 'fontsize', (25));
xlabel ('\bf INPUT ', 'fontsize', (20));
ylabel ('\bf OUTPUT ', 'fontsize', (20));
Paul on 24 Nov 2021
Those points at zero show up because alpha is negative and the code is raising alpha to non-integer powers t. For those values of t, alpha^t is complex, and they all have real part equal 0 for those values of t. For example
alpha = -.9;
power(alpha,-9.5)
ans = 0.0000 + 2.7208i
Then, as the warning states, stem() plots the real part, which is why those zero points show up.
The solution is to only define t with integer values
t = -10 : 1 : 10;
y = power(alpha, t);
stem(t,y, '- b','LineWidth',2);
### More Answers (2)
VBBV on 23 Nov 2021
Edited: VBBV on 23 Nov 2021
clc; clear; close all
alpha = -.9;
t = -10 : 0.10000 : 10;
y = exp(alpha.^t); % exponent of alpha is vector t
stem(t,real(y));
xlabel('xlabel')
ylabel('Magnitude')
##### 1 CommentShowHide None
VBBV on 23 Nov 2021
clc; clear; close all
alpha = -.9;
t = -10 : 0.50000 : 10;
y = (alpha.^t); % exponent of alpha is vector t
stem(t(3:2:end),real(y(3:2:end)),'linewidth',2,'MarkerFaceColor','blue');
xlabel('xlabel')
ylabel('Magnitude')
Do you mean like figure above ? where the first two points are not visible
cikalekli on 23 Nov 2021
The expected result and your code is a lot different as you can see.
Here I made it more accurate version:
clc; clear; close all
alpha = -.9;
t = -10 : 0.5 : 10;
y = power(alpha, t);
stem(t,y, '- b',...
'LineWidth',2);
Warning: Using only the real component of complex data.
title ('\bf\it 3rd QUESTION ', 'fontsize', (25));
xlabel ('\bf INPUT ', 'fontsize', (20));
ylabel ('\bf OUTPUT ', 'fontsize', (20));
##### 1 CommentShowHide None
cikalekli on 23 Nov 2021
How can I delete those points which are visible at zero
They are visible at couple times as you can see?
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source:trunk/GSASIIindex.py@311
Last change on this file since 311 was 311, checked in by vondreele, 10 years ago
finish new indexing refinement stuff
more on texture plotting
• Property svn:keywords set to `Date Author Revision URL Id`
File size: 27.5 KB
Line
1#GSASII cell indexing program: variation on that of A. Coehlo
2# includes cell refinement from peak positions (not zero as yet)
3########### SVN repository information ###################
4# \$Date: 2011-06-27 15:27:29 +0000 (Mon, 27 Jun 2011) \$
5# \$Author: vondreele \$
6# \$Revision: 311 \$
7# \$URL: trunk/GSASIIindex.py \$
8# \$Id: GSASIIindex.py 311 2011-06-27 15:27:29Z vondreele \$
9########### SVN repository information ###################
10import math
11import wx
12import time
13import numpy as np
14import numpy.linalg as nl
15import GSASIIpath
16import GSASIIlattice as G2lat
17import scipy.optimize as so
18
19# trig functions in degrees
20sind = lambda x: math.sin(x*math.pi/180.)
21asind = lambda x: 180.*math.asin(x)/math.pi
22tand = lambda x: math.tan(x*math.pi/180.)
23atand = lambda x: 180.*math.atan(x)/math.pi
24atan2d = lambda y,x: 180.*math.atan2(y,x)/math.pi
25cosd = lambda x: math.cos(x*math.pi/180.)
26acosd = lambda x: 180.*math.acos(x)/math.pi
27rdsq2d = lambda x,p: round(1.0/math.sqrt(x),p)
28#numpy versions
29npsind = lambda x: np.sin(x*np.pi/180.)
30npasind = lambda x: 180.*np.arcsin(x)/math.pi
31npcosd = lambda x: np.cos(x*math.pi/180.)
32nptand = lambda x: np.tan(x*math.pi/180.)
33npatand = lambda x: 180.*np.arctan(x)/np.pi
34npatan2d = lambda y,x: 180.*np.arctan2(y,x)/np.pi
35
36def scaleAbyV(A,V):
37 v = G2lat.calc_V(A)
38 scale = math.exp(math.log(v/V)/3.)**2
39 for i in range(6):
40 A[i] *= scale
41
42def ranaxis(dmin,dmax):
43 import random as rand
44 return rand.random()*(dmax-dmin)+dmin
45
46def ran2axis(k,N):
47 import random as rand
48 T = 1.5+0.49*k/N
49# B = 0.99-0.49*k/N
50# B = 0.99-0.049*k/N
51 B = 0.99-0.149*k/N
52 R = (T-B)*rand.random()+B
53 return R
54
55#def ranNaxis(k,N):
56# import random as rand
57# T = 1.0+1.0*k/N
58# B = 1.0-1.0*k/N
59# R = (T-B)*rand.random()+B
60# return R
61
62def ranAbyV(Bravais,dmin,dmax,V):
63 cell = [0,0,0,0,0,0]
67 cell = rancell(Bravais,dmin,dmax)
68 G,g = G2lat.cell2Gmat(cell)
69 A = G2lat.Gmat2A(G)
70 if G2lat.calc_rVsq(A) < 1:
71 scaleAbyV(A,V)
72 cell = G2lat.A2cell(A)
73 for i in range(3):
74 bad |= cell[i] < dmin
75 return A
76
77def ranAbyR(Bravais,A,k,N,ranFunc):
78 R = ranFunc(k,N)
79 if Bravais in [0,1,2]: #cubic - not used
80 A[0] = A[1] = A[2] = A[0]*R
81 A[3] = A[4] = A[5] = 0.
82 elif Bravais in [3,4]: #hexagonal/trigonal
83 A[0] = A[1] = A[3] = A[0]*R
84 A[2] *= R
85 A[4] = A[5] = 0.
86 elif Bravais in [5,6]: #tetragonal
87 A[0] = A[1] = A[0]*R
88 A[2] *= R
89 A[3] = A[4] = A[5] = 0.
90 elif Bravais in [7,8,9,10]: #orthorhombic
91 A[0] *= R
92 A[1] *= R
93 A[2] *= R
94 A[3] = A[4] = A[5] = 0.
95 elif Bravais in [11,12]: #monoclinic
96 A[0] *= R
97 A[1] *= R
98 A[2] *= R
99 A[4] *= R
100 A[3] = A[5] = 0.
101 else: #triclinic
102 A[0] *= R
103 A[1] *= R
104 A[2] *= R
105 A[3] *= R
106 A[4] *= R
107 A[5] *= R
108 return A
109
110def rancell(Bravais,dmin,dmax):
111 if Bravais in [0,1,2]: #cubic
112 a = b = c = ranaxis(dmin,dmax)
113 alp = bet = gam = 90
114 elif Bravais in [3,4]: #hexagonal/trigonal
115 a = b = ranaxis(dmin,dmax)
116 c = ranaxis(dmin,dmax)
117 alp = bet = 90
118 gam = 120
119 elif Bravais in [5,6]: #tetragonal
120 a = b = ranaxis(dmin,dmax)
121 c = ranaxis(dmin,dmax)
122 alp = bet = gam = 90
123 elif Bravais in [7,8,9,10]: #orthorhombic - F,I,C,P - a<b<c convention
124 abc = [ranaxis(dmin,dmax),ranaxis(dmin,dmax),ranaxis(dmin,dmax)]
125 abc.sort()
126 a = abc[0]
127 b = abc[1]
128 c = abc[2]
129 alp = bet = gam = 90
130 elif Bravais in [11,12]: #monoclinic - C,P - a<c convention
131 ac = [ranaxis(dmin,dmax),ranaxis(dmin,dmax)]
132 ac.sort()
133 a = ac[0]
134 b = ranaxis(dmin,dmax)
135 c = ac[1]
136 alp = gam = 90
137 bet = ranaxis(90.,130.)
138 else: #triclinic - a<b<c convention
139 abc = [ranaxis(dmin,dmax),ranaxis(dmin,dmax),ranaxis(dmin,dmax)]
140 abc.sort()
141 a = abc[0]
142 b = abc[1]
143 c = abc[2]
144 r = 0.5*b/c
145 alp = ranaxis(acosd(r),acosd(-r))
146 r = 0.5*a/c
147 bet = ranaxis(acosd(r),acosd(-r))
148 r = 0.5*a/b
149 gam = ranaxis(acosd(r),acosd(-r))
150 return [a,b,c,alp,bet,gam]
151
152def calc_M20(peaks,HKL):
153 diff = 0
154 X20 = 0
155 for Nobs20,peak in enumerate(peaks):
156 if peak[3]:
157 Qobs = 1.0/peak[7]**2
158 Qcalc = 1.0/peak[8]**2
159 diff += abs(Qobs-Qcalc)
160 elif peak[2]:
161 X20 += 1
162 if Nobs20 == 19:
163 d20 = peak[7]
164 break
165 else:
166 d20 = peak[7]
167 Nobs20 = len(peaks)
168 for N20,hkl in enumerate(HKL):
169 if hkl[3] < d20:
170 break
171 eta = diff/Nobs20
172 Q20 = 1.0/d20**2
173 if diff:
174 M20 = Q20/(2.0*diff)
175 else:
176 M20 = 0
177 M20 /= (1.+X20)
178 return M20,X20
179
180def sortM20(cells):
181 #cells is M20,X20,Bravais,a,b,c,alp,bet,gam
182 #sort highest M20 1st
183 T = []
184 for i,M in enumerate(cells):
185 T.append((M[0],i))
186 D = dict(zip(T,cells))
187 T.sort()
188 T.reverse()
189 X = []
190 for key in T:
191 X.append(D[key])
192 return X
193
194def IndexPeaks(peaks,HKL):
195 import bisect
196 N = len(HKL)
197 if N == 0: return False
198 hklds = list(np.array(HKL).T[3])+[1000.0,0.0,]
199 hklds.sort() # ascending sort - upper bound at end
200 hklmax = [0,0,0]
201 for ipk,peak in enumerate(peaks):
202 if peak[2]:
203 i = bisect.bisect_right(hklds,peak[7]) # find peak position in hkl list
204 dm = peak[7]-hklds[i-1] # peak to neighbor hkls in list
205 dp = hklds[i]-peak[7]
206 pos = N-i # reverse the order
207 if dp > dm: pos += 1 # closer to upper than lower
208 hkl = HKL[pos] # put in hkl
209 if hkl[4] >= 0: # peak already assigned - test if this one better
210 opeak = peaks[hkl[4]]
211 dold = abs(opeak[7]-hkl[3])
212 dnew = min(dm,dp)
213 if dold > dnew: # new better - zero out old
214 opeak[4:7] = [0,0,0]
215 opeak[8] = 0.
216 else: # old better - do nothing
217 continue
218 hkl[4] = ipk
219 peak[4:7] = hkl[:3]
220 peak[8] = hkl[3] # fill in d-calc
221 for peak in peaks:
222 peak[3] = False
223 if peak[2]:
224 if peak[8] > 0.:
225 for j in range(3):
226 if abs(peak[j+4]) > hklmax[j]: hklmax[j] = abs(peak[j+4])
227 peak[3] = True
228 if hklmax[0]*hklmax[1]*hklmax[2] > 0:
229 return True
230 else:
231 return False
232
233def FitHKL(ibrav,peaks,A,Pwr):
234
235 def Values2A(ibrav,values):
236 if ibrav in [0,1,2]:
237 return [values[0],values[0],values[0],0,0,0]
238 elif ibrav in [3,4]:
239 return [values[0],values[0],values[1],values[0],0,0]
240 elif ibrav in [5,6]:
241 return [values[0],values[0],values[1],0,0,0]
242 elif ibrav in [7,8,9,10]:
243 return [values[0],values[1],values[2],0,0,0]
244 elif ibrav in [11,12]:
245 return [values[0],values[1],values[2],0,values[3],0]
246 else:
247 return list(values)
248
249 def A2values(ibrav,A):
250 if ibrav in [0,1,2]:
251 return [A[0],]
252 elif ibrav in [3,4,5,6]:
253 return [A[0],A[2]]
254 elif ibrav in [7,8,9,10]:
255 return [A[0],A[1],A[2]]
256 elif ibrav in [11,12]:
257 return [A[0],A[1],A[2],A[4]]
258 else:
259 return A
260
261 def Jacobian(values,ibrav,d,H,Pwr):
262# derivatives of rdsq = H[0]*H[0]*A[0]+H[1]*H[1]*A[1]+H[2]*H[2]*A[2]+H[0]*H[1]*A[3]+H[0]*H[2]*A[4]+H[1]*H[2]*A[5]
263 if ibrav in [0,1,2]: #m3m
264 return [H[0]*H[0]+H[1]*H[1]+H[2]*H[2],]*d**Pwr
265 elif ibrav in [3,4]: #R3H, P3/m & P6/mmm
266 return [H[0]*H[0]+H[1]*H[0]+H[1]*H[1],H[2]*H[2]]*d**Pwr
267 elif ibrav in [5,6]: #4/mmm
268 return [H[0]*H[0]+H[1]*H[1],H[2]*H[2]]*d**Pwr
269 elif ibrav in [7,8,9,10]: #mmm
270 return [H[0]*H[0],H[1]*H[1],H[2]*H[2]]*d**Pwr
271 elif ibrav in [11,12]: #2/m
272 return [H[0]*H[0],H[1]*H[1],H[2]*H[2],H[0]*H[2]]*d**Pwr
273 else: #-1
274 return [H[0]*H[0],H[1]*H[1],H[2]*H[2],H[1]*H[0],H[0]*H[2],H[1]*H[2]]*d**Pwr
275
276 def errFit(values,ibrav,d,H,Pwr):
277 A = Values2A(ibrav,values)
278 Qo = 1./d**2
279 Qc = G2lat.calc_rDsq(H,A)
280 return (Qo-Qc)*d**Pwr
281
282 Peaks = np.array(peaks).T
283
284 values = A2values(ibrav,A)
285# result = so.leastsq(errFit,values,args=(ibrav,Peaks[7],Peaks[4:7],Pwr),
286# Dfun=Jacobian,col_deriv=True,full_output=True)
287 result = so.leastsq(errFit,values,args=(ibrav,Peaks[7],Peaks[4:7],Pwr),
288 full_output=True,factor=0.1)
289 A = Values2A(ibrav,result[0])
290 return True,np.sum(errFit(result[0],ibrav,Peaks[7],Peaks[4:7],Pwr)**2),A,result
291
292def FitHKLZ(ibrav,peaks,Z,A):
293 return A,Z
294
295def rotOrthoA(A):
296 return [A[1],A[2],A[0],0,0,0]
297
298def swapMonoA(A):
299 return [A[2],A[1],A[0],0,A[4],0]
300
301def oddPeak(indx,peaks):
302 noOdd = True
303 for peak in peaks:
304 H = peak[4:7]
305 if H[indx] % 2:
306 noOdd = False
307 return noOdd
308
309def halfCell(ibrav,A,peaks):
310 if ibrav in [0,1,2]:
311 if oddPeak(0,peaks):
312 A[0] *= 2
313 A[1] = A[2] = A[0]
314 elif ibrav in [3,4,5,6]:
315 if oddPeak(0,peaks):
316 A[0] *= 2
317 A[1] = A[0]
318 if oddPeak(2,peaks):
319 A[2] *=2
320 else:
321 if oddPeak(0,peaks):
322 A[0] *=2
323 if oddPeak(1,peaks):
324 A[1] *=2
325 if oddPeak(2,peaks):
326 A[2] *=2
327 return A
328
329def getDmin(peaks):
330 return peaks[-1][7]
331
332def getDmax(peaks):
333 return peaks[0][7]
334
335def refinePeaks(peaks,ibrav,A,Zero):
336 #Zero is list(zero value, flag)
337 dmin = getDmin(peaks)
338 smin = 1.0e10
339 pwr = 4
340 maxTries = 10
341 if ibrav == 13: #-1 - triclinic
342 pwr = 4
343 maxTries = 10
344 OK = False
345 tries = 0
346 HKL = G2lat.GenHBravais(dmin,ibrav,A)
347 while len(HKL) > 2 and IndexPeaks(peaks,HKL):
348 Pwr = pwr - (tries % 2)
349 HKL = []
350 tries += 1
351 osmin = smin
352 oldA = A[:]
353 Vold = G2lat.calc_V(oldA)
354 OK,smin,A,result = FitHKL(ibrav,peaks,A,Pwr)
355 Vnew = G2lat.calc_V(A)
356 if Vnew > 2.0*Vold or Vnew < 2.:
357 A = ranAbyR(ibrav,oldA,tries+1,maxTries,ran2axis)
358 OK = False
359 continue
360 try:
361 HKL = G2lat.GenHBravais(dmin,ibrav,A)
362 except FloatingPointError:
363 A = oldA
364 OK = False
365 break
366 if len(HKL) == 0: break #absurd cell obtained!
367 rat = (osmin-smin)/smin
368 if abs(rat) < 1.0e-5 or not OK: break
369 if tries > maxTries: break
370 if OK:
371 OK,smin,A,result = FitHKL(ibrav,peaks,A,2)
372 Peaks = np.array(peaks).T
373 H = Peaks[4:7]
374 Peaks[8] = 1./np.sqrt(G2lat.calc_rDsq(H,A))
375 peaks = Peaks.T
376
377 M20,X20 = calc_M20(peaks,HKL)
378 return len(HKL),M20,X20,A,Zero
379
380def findBestCell(dlg,ncMax,A,Ntries,ibrav,peaks,V1):
381# dlg & ncMax are used for wx progress bar
382# A != 0 find the best A near input A,
383# A = 0 for random cell, volume normalized to V1;
384# returns number of generated hkls, M20, X20 & A for best found
385 mHKL = [3,3,3, 5,5, 5,5, 7,7,7,7, 9,9, 10]
386 dmin = getDmin(peaks)-0.05
387 amin = 2.5
388 amax = 5.*getDmax(peaks)
389 Asave = []
390 GoOn = True
391 if A:
392 HKL = G2lat.GenHBravais(dmin,ibrav,A[:])
393 if len(HKL) > mHKL[ibrav]:
394 IndexPeaks(peaks,HKL)
395 Asave.append([calc_M20(peaks,HKL),A[:]])
396 tries = 0
397 while tries < Ntries:
398 if A:
399 Abeg = ranAbyR(ibrav,A,tries+1,Ntries,ran2axis)
400 if ibrav in [11,12,13]: #monoclinic & triclinic
401 Abeg = ranAbyR(ibrav,A,tries/10+1,Ntries,ran2axis)
402 else:
403 Abeg = ranAbyV(ibrav,amin,amax,V1)
404 HKL = G2lat.GenHBravais(dmin,ibrav,Abeg)
405
406 if IndexPeaks(peaks,HKL) and len(HKL) > mHKL[ibrav]:
407 Lhkl,M20,X20,Aref,Zero = refinePeaks(peaks,ibrav,Abeg,[0,False])
408 Asave.append([calc_M20(peaks,HKL),Aref[:]])
409 if ibrav == 9: #C-centered orthorhombic
410 for i in range(2):
411 Abeg = rotOrthoA(Abeg[:])
412 Lhkl,M20,X20,Aref,Zero = refinePeaks(peaks,ibrav,Abeg,[0,False])
413 HKL = G2lat.GenHBravais(dmin,ibrav,Aref)
414 IndexPeaks(peaks,HKL)
415 Asave.append([calc_M20(peaks,HKL),Aref[:]])
416 elif ibrav == 11: #C-centered monoclinic
417 Abeg = swapMonoA(Abeg[:])
418 Lhkl,M20,X20,Aref,Zero = refinePeaks(peaks,ibrav,Abeg,[0,False])
419 HKL = G2lat.GenHBravais(dmin,ibrav,Aref)
420 IndexPeaks(peaks,HKL)
421 Asave.append([calc_M20(peaks,HKL),Aref[:]])
422 else:
423 break
424 Nc = len(HKL)
425 if Nc >= ncMax:
426 GoOn = False
427 elif dlg:
428 GoOn = dlg.Update(Nc)[0]
429 if not GoOn:
430 break
431 tries += 1
432 X = sortM20(Asave)
433 if X:
434 Lhkl,M20,X20,A,Zero = refinePeaks(peaks,ibrav,X[0][1],[0,False])
435 return GoOn,Lhkl,M20,X20,A
436
437 else:
438 return GoOn,0,0,0,0
439
440def monoCellReduce(ibrav,A):
441 a,b,c,alp,bet,gam = G2lat.A2cell(A)
442 G,g = G2lat.A2Gmat(A)
443 if ibrav in [11]:
444 u = [0,0,-1]
445 v = [1,0,2]
446 anew = math.sqrt(np.dot(np.dot(v,g),v))
447 if anew < a:
448 cang = np.dot(np.dot(u,g),v)/(anew*c)
449 beta = acosd(-abs(cang))
450 A = G2lat.cell2A([anew,b,c,90,beta,90])
451 else:
452 u = [-1,0,0]
453 v = [1,0,1]
454 cnew = math.sqrt(np.dot(np.dot(v,g),v))
455 if cnew < c:
456 cang = np.dot(np.dot(u,g),v)/(a*cnew)
457 beta = acosd(-abs(cang))
458 A = G2lat.cell2A([a,b,cnew,90,beta,90])
459 return A
460
461def DoIndexPeaks(peaks,inst,controls,bravais):
462
463 delt = 0.005 #lowest d-spacing cushion - can be fixed?
464 amin = 2.5
465 amax = 5.0*getDmax(peaks)
466 dmin = getDmin(peaks)-delt
467 bravaisNames = ['Cubic-F','Cubic-I','Cubic-P','Trigonal-R','Trigonal/Hexagonal-P',
468 'Tetragonal-I','Tetragonal-P','Orthorhombic-F','Orthorhombic-I','Orthorhombic-C',
469 'Orthorhombic-P','Monoclinic-C','Monoclinic-P','Triclinic']
470 tries = ['1st','2nd','3rd','4th','5th','6th','7th','8th','9th','10th']
471 N1s = [1,1,1, 5,5, 5,5, 50,50,50,50, 50,50, 200]
472 N2s = [1,1,1, 2,2, 2,2, 2,2,2,2, 2,2, 4]
473 Nm = [1,1,1, 1,1, 1,1, 1,1,1,1, 2,2, 4]
474 Nobs = len(peaks)
475 wave = inst[1]
476 if len(inst) > 10:
477 zero = inst[3]
478 else:
479 zero = inst[2]
480 print "%s %8.5f %6.3f" % ('wavelength, zero =',wave,zero)
481 print "%s %8.3f %8.3f" % ('lattice parameter range = ',amin,amax)
482 ifzero,maxzero,ncno = controls[:3]
483 ncMax = Nobs*ncno
484 print "%s %d %s %d %s %d" % ('change zero =',ifzero,'Nc/No max =',ncno,' Max Nc =',ncno*Nobs)
485 cells = []
486 for ibrav in range(14):
487 begin = time.time()
488 if bravais[ibrav]:
489 print 'cell search for ',bravaisNames[ibrav]
490 print ' M20 X20 Nc a b c alpha beta gamma volume V-test'
491 V1 = controls[3]
492 bestM20 = 0
493 topM20 = 0
494 cycle = 0
495 while cycle < 5:
496 dlg = wx.ProgressDialog("Generated reflections",tries[cycle]+" cell search for "+bravaisNames[ibrav],ncMax,
497 style = wx.PD_ELAPSED_TIME|wx.PD_AUTO_HIDE|wx.PD_REMAINING_TIME|wx.PD_CAN_ABORT)
498 screenSize = wx.ClientDisplayRect()
499 Size = dlg.GetSize()
500 dlg.SetPosition(wx.Point(screenSize[2]-Size[0]-305,screenSize[1]+5))
501 try:
502 GoOn = True
503 while GoOn: #Loop over increment of volume
504 N2 = 0
505 while N2 < N2s[ibrav]: #Table 2 step (iii)
506 if ibrav > 2:
507 if not N2:
508 A = []
509 GoOn,Nc,M20,X20,A = findBestCell(dlg,ncMax,A,Nm[ibrav]*N1s[ibrav],ibrav,peaks,V1)
510 if A:
511 GoOn,Nc,M20,X20,A = findBestCell(dlg,ncMax,A[:],N1s[ibrav],ibrav,peaks,0)
512 else:
513 GoOn,Nc,M20,X20,A = findBestCell(dlg,ncMax,0,Nm[ibrav]*N1s[ibrav],ibrav,peaks,V1)
514 if Nc >= ncMax:
515 GoOn = False
516 break
517 elif 3*Nc < Nobs:
518 N2 = 10
519 break
520 else:
521 if not GoOn:
522 break
523 if M20 > 1.0:
524 bestM20 = max(bestM20,M20)
525 A = halfCell(ibrav,A[:],peaks)
526 if ibrav in [12]:
527 A = monoCellReduce(ibrav,A[:])
528 HKL = G2lat.GenHBravais(dmin,ibrav,A)
529 IndexPeaks(peaks,HKL)
530 a,b,c,alp,bet,gam = G2lat.A2cell(A)
531 V = G2lat.calc_V(A)
532 print "%10.3f %3d %3d %10.5f %10.5f %10.5f %10.3f %10.3f %10.3f %10.2f %10.2f" % (M20,X20,Nc,a,b,c,alp,bet,gam,V,V1)
533 if M20 >= 2.0:
534 cells.append([M20,X20,ibrav,a,b,c,alp,bet,gam,V,False])
535 if not GoOn:
536 break
537 N2 += 1
538 if ibrav < 11:
539 V1 *= 1.1
540 elif ibrav in range(11,14):
541 V1 *= 1.05
542 if not GoOn:
543 if bestM20 > topM20:
544 topM20 = bestM20
545 if cells:
546 V1 = cells[0][9]
547 else:
548 V1 = 25
549 ncMax += Nobs
550 cycle += 1
551 print 'Restart search, new Max Nc = ',ncMax
552 else:
553 cycle = 10
554 finally:
555 dlg.Destroy()
556 print '%s%s%s%s'%('finished cell search for ',bravaisNames[ibrav], \
557 ', elapsed time = ',G2lat.sec2HMS(time.time()-begin))
558
559 if cells:
560 cells = sortM20(cells)
561 cells[0][-1] = True
562 return True,dmin,cells
563 else:
564 return False,0,0
565
566
567NeedTestData = True
568def TestData():
569 array = np.array
570 global NeedTestData
571 NeedTestData = False
572 global TestData
573 TestData = [12, [7.,8.70,10.86,90.,102.95,90.], [7.76006,8.706215,10.865679,90.,102.947,90.],3,
574 [[2.176562137832974, 761.60902227696033, True, True, 0, 0, 1, 10.591300714328161, 10.589436],
575 [3.0477561489789498, 4087.2956049071572, True, True, 1, 0, 0, 7.564238997554908, 7.562777],
576 [3.3254921120068524, 1707.0253890991009, True, True, 1, 0, -1, 6.932650301411212, 6.932718],
577 [3.428121546163426, 2777.5082170150563, True, True, 0, 1, 1, 6.725163158013632, 6.725106],
578 [4.0379791325512118, 1598.4321673135987, True, True, 1, 1, 0, 5.709789097440156, 5.70946],
579 [4.2511182350743937, 473.10955149057577, True, True, 1, 1, -1, 5.423637972781876, 5.42333],
580 [4.354684330373451, 569.88528280256071, True, True, 0, 0, 2, 5.2947091882172534, 5.294718],
581 [4.723324574319177, 342.73882372499997, True, True, 1, 0, -2, 4.881681587039431, 4.881592],
582 [4.9014773581253994, 5886.3516356615492, True, True, 1, 1, 1, 4.704350709093183, 4.70413],
583 [5.0970774474587275, 3459.7541692903033, True, True, 0, 1, 2, 4.523933797797693, 4.523829],
584 [5.2971997607389518, 1290.0229964239879, True, True, 0, 2, 0, 4.353139557169142, 4.353108],
585 [5.4161306205553847, 1472.5726977257755, True, True, 1, 1, -2, 4.257619398422479, 4.257944],
586 [5.7277364698554161, 1577.8791668322888, True, True, 0, 2, 1, 4.026169751907777, 4.026193],
587 [5.8500213058834163, 420.74210142657131, True, True, 1, 0, 2, 3.9420803081518443, 3.942219],
588 [6.0986764166731708, 163.02160537058708, True, True, 2, 0, 0, 3.7814965150452537, 3.781389],
589 [6.1126665157702753, 943.25461245706833, True, True, 1, 2, 0, 3.772849962062199, 3.772764],
590 [6.2559260555056957, 250.55355015505376, True, True, 1, 2, -1, 3.6865353266375283, 3.686602],
591 [6.4226243128279892, 5388.5560141098349, True, True, 1, 1, 2, 3.5909481979190283, 3.591214],
592 [6.5346132446561134, 1951.6070344509026, True, True, 0, 0, 3, 3.5294722429440584, 3.529812],
593 [6.5586952135236443, 259.65938178131034, True, True, 2, 1, -1, 3.516526936765838, 3.516784],
594 [6.6509216222783722, 93.265376597376573, True, True, 2, 1, 0, 3.4678179073694952, 3.468369],
595 [6.7152737044107722, 289.39386813803162, True, True, 1, 2, 1, 3.4346235125812807, 3.434648],
596 [6.8594130457361899, 603.54959764648322, True, True, 0, 2, 2, 3.362534044860622, 3.362553],
597 [7.0511627728884454, 126.43246447656593, True, True, 0, 1, 3, 3.2712038721790675, 3.271181],
598 [7.077700845503319, 125.49742760019636, True, True, 1, 1, -3, 3.2589538988480626, 3.259037],
599 [7.099393757363675, 416.55444885434633, True, True, 1, 2, -2, 3.2490085228959193, 3.248951],
600 [7.1623933278642742, 369.27397110921817, True, True, 2, 1, -2, 3.2204673608202383, 3.220487],
601 [7.4121734953058924, 482.84120827021826, True, True, 2, 1, 1, 3.1120858221599876, 3.112308]]
602 ]
603 global TestData2
604 TestData2 = [12, [0.15336547830008007, 0.017345499139401827, 0.008122368657493792, 0, 0.02893538955687591, 0], 3,
605 [[2.176562137832974, 761.6090222769603, True, True, 0, 0, 1, 10.591300714328161, 11.095801],
606 [3.0477561489789498, 4087.295604907157, True, True, 0, 1, 0, 7.564238997554908, 7.592881],
607 [3.3254921120068524, 1707.025389099101, True, False, 0, 0, 0, 6.932650301411212, 0.0],
608 [3.428121546163426, 2777.5082170150563, True, True, 0, 1, 1, 6.725163158013632, 6.266192],
609 [4.037979132551212, 1598.4321673135987, True, False, 0, 0, 0, 5.709789097440156, 0.0],
610 [4.251118235074394, 473.10955149057577, True, True, 0, 0, 2, 5.423637972781876, 5.5479],
611 [4.354684330373451, 569.8852828025607, True, True, 0, 0, 2, 5.2947091882172534, 5.199754],
612 [4.723324574319177, 342.738823725, True, False, 0, 0, 0, 4.881681587039431, 0.0],
613 [4.901477358125399, 5886.351635661549, True, False, 0, 0, 0, 4.704350709093183, 0.0],
614 [5.0970774474587275, 3459.7541692903033, True, True, 0, 1, 2, 4.523933797797693, 4.479534],
615 [5.297199760738952, 1290.022996423988, True, True, 0, 1, 0, 4.353139557169142, 4.345087],
616 [5.416130620555385, 1472.5726977257755, True, False, 0, 0, 0, 4.257619398422479, 0.0],
617 [5.727736469855416, 1577.8791668322888, True, False, 0, 0, 0, 4.026169751907777, 0.0],
618 [5.850021305883416, 420.7421014265713, True, False, 0, 0, 0, 3.9420803081518443, 0.0],
619 [6.098676416673171, 163.02160537058708, True, True, 0, 2, 0, 3.7814965150452537, 3.796441],
620 [6.112666515770275, 943.2546124570683, True, False, 0, 0, 0, 3.772849962062199, 0.0],
621 [6.255926055505696, 250.55355015505376, True, True, 0, 0, 3, 3.6865353266375283, 3.6986],
622 [6.422624312827989, 5388.556014109835, True, True, 0, 2, 1, 3.5909481979190283, 3.592005],
623 [6.534613244656113, 191.6070344509026, True, True, 1, 0, -1, 3.5294722429440584, 3.546166],
624 [6.558695213523644, 259.65938178131034, True, True, 0, 0, 3, 3.516526936765838, 3.495428],
625 [6.650921622278372, 93.26537659737657, True, True, 0, 0, 3, 3.4678179073694952, 3.466503],
626 [6.715273704410772, 289.3938681380316, True, False, 0, 0, 0, 3.4346235125812807, 0.0],
627 [6.85941304573619, 603.5495976464832, True, True, 0, 1, 3, 3.362534044860622, 3.32509],
628 [7.051162772888445, 126.43246447656593, True, True, 0, 1, 2, 3.2712038721790675, 3.352121],
629 [7.077700845503319, 125.49742760019636, True, False, 0, 0, 0, 3.2589538988480626, 0.0],
630 [7.099393757363675, 416.55444885434633, True, False, 0, 0, 0, 3.2490085228959193, 0.0],
631 [7.162393327864274, 369.27397110921817, True, False, 0, 0, 0, 3.2204673608202383, 0.0],
632 [7.412173495305892, 482.84120827021826, True, True, 0, 2, 2, 3.112085822159976, 3.133096]]
633 ]
634#
635def test0():
636 if NeedTestData: TestData()
637 msg = 'test FitHKL'
638 ibrav,cell,bestcell,Pwr,peaks = TestData
639 print 'best cell:',bestcell
640 print 'old cell:',cell
641 Peaks = np.array(peaks)
642 HKL = Peaks[4:7]
643 print calc_M20(peaks,HKL)
644 A = G2lat.cell2A(cell)
645 OK,smin,A,result = FitHKL(ibrav,peaks,A,Pwr)
646 print 'new cell:',G2lat.A2cell(A)
647 print 'x:',result[0]
648 print 'cov_x:',result[1]
649 print 'infodict:'
650 for item in result[2]:
651 print item,result[2][item]
652 print 'msg:',result[3]
653 print 'ier:',result[4]
654 result = refinePeaks(peaks,ibrav,A,[0,False])
655 N,M20,X20,A,Zero = result
656 print 'refinePeaks:',N,M20,X20,G2lat.A2cell(A)
657 print 'compare bestcell:',bestcell
658#
659def test1():
660 if NeedTestData: TestData()
661 msg = 'test FitHKL'
662 ibrav,A,Pwr,peaks = TestData2
664 print 'FitHKL'
665 OK,smin,A,result = FitHKL(ibrav,peaks,A,Pwr)
666 result = refinePeaks(peaks,ibrav,A,[0,False])
667 N,M20,X20,A,Zero = result
668 print 'refinePeaks:',N,M20,X20,A
669# Peaks = np.array(peaks)
670# HKL = Peaks[4:7]
671# print calc_M20(peaks,HKL)
672# OK,smin,A,result = FitHKL(ibrav,peaks,A,Pwr)
673# print 'new cell:',G2lat.A2cell(A)
674# print 'x:',result[0]
675# print 'cov_x:',result[1]
676# print 'infodict:'
677# for item in result[2]:
678# print item,result[2][item]
679# print 'msg:',result[3]
680# print 'ier:',result[4]
681
682#
683if __name__ == '__main__':
684 test0()
685 test1()
686# test2()
687# test3()
688# test4()
689# test5()
690# test6()
691# test7()
692# test8()
693 print "OK"
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http://www.answers.com/topic/time-of-the-wolf-2002-film
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## Results for: Time-of-the-wolf-2002-film
In Personal Finance
# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow ( Full Answer )
In Acronyms & Abbreviations
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo ( Full Answer )
In Coins and Paper Money
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
In Math and Arithmetic
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Numerical Analysis and Simulation
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio ( Full Answer )
In iPhone 5
# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
In Temperature
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
In iPhone 5
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35"
In iPhone 5
# How much does an iPhone 5c weigh?
The iPhone 5c weighs 4.65 ounches. It is heavier than the iPhone 5and 5s which weight 3.95.
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https://www.codeproject.com/Questions/417371/Sum-Percentage-of-Count-in-sql-server-2005
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15,040,212 members
1.00/5 (1 vote)
See more:
I have following scenario
original format
SrNo Month Issues
1 January 8
2 February 14
3 March 26
4 April 21
5 May 14
6 June 19
7 July 2
I need following format in sql
Sr.No.Month Issues SUM %
1 January 8 104 7.69
2 February 14 104 13.46
3 March 26 104 25.92
4 April 21 104 20.19
5 May 14 104 13.46
6 June 19 104 18.32
7 July 2 104 0.96
Posted
[no name] 7-Jul-12 8:02am
"I need following format in sql"... okay so go ahead and do it. Please come back when you have a question.
## Solution 1
Here is the sa(i)mple query to get the result:
SQL
Select SrNo, Month, Issues,
Sum(Issues) as SUM,
(Issues * 100 / Sum(Issues)) as Percentage
From myTable
## Solution 2
You can use the SUM function[^] and the OVER clause[^].
SQL
SELECT SrNo, Month, Issues, SUM(Issues) OVER() AS SUM, CAST(Issues * (100.0 / SUM(Issues) OVER() AS DECIMAL (4, 2)) AS PERC
FROM myTable
The CAST function[^] is used to display the PERC with two digits.
v2
Kailash_Singh 8-Jul-12 23:58pm
Hi Krrak thanx a lot...Its worked
:) :)
Now I have another condition,
I have three field, SrNo(Int),Month(nvarchar(12)) and Issues(nvarchar(MAX)).
I use following querry:
Select SrNo,Month,count(Issues)as[Total]
from tblIssues
group by Month
But I need as following query
Select SrNo,Month,count(Issues),sum(count(Issues))
from tblIssues
group by Month
again thanx.....
| 514
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CC-MAIN-2021-39
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https://forum.getodk.org/t/multiple-choice-question-score-calculation/41720
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Multiple choice question score calculation
I am calculating multiple choice questions score
if(\${q2_e5}=1,1,0)+if(\${q2_e5}=2,1,0)+if(\${q2_e5}=3,1,0)+if(\${q2_e5}=4,1,0)+if(\${q2_e5}=5,1,0)+if(\${q2_e5}=6,1,0)+if(\${q2_e5}=7,1,0)
Its give single choice score but If I select two choice then no score calculated
If you are using a select_multiple question, then you really need to be using the `selected()` function to check whether each option has been selected or not. Try something more like:
`if(selected(\${q2_e5},"1"),1,0) + if(selected(\${q2_e5},"2",1,0) ...`
Yes, I have tried this one. It is working if we select only single options, once we select two options then it is not working.
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https://www.studymode.com/essays/Solution-Of-Chapter-2-40627246.html
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# Solution of Chapter 2
Satisfactory Essays
Topics: Trigraph
?.1
lf the slope of the C* vs Cscurve is -0.15 ancl the pitching monrent atzero lift is equal to 0.0g, determine tfre trim iift coefficient. lf the center of gravity of the airplane is located at X.o / c = 0.3, determine the stick fixed neL:tral point.
SP-lul-q'!:
dc_
Given:
---u
=
-6,15
Cr.n
= 0.Ct8 @ C,- = 0
X6gic=0.3
Finci: The trim litt coefficient and the stick fixed neutial point.
r^
umlUL-o t,gto.
t.,rflcg .A dC,_ '*'t
Lrrncg
'lrltn
,a
,
= um@C'- = 0 *
=+
9ct ..
*L
d6t
C,.n = S cg Therefore 0 = 0.08 - O,iS
t\.,*
0.08 n f^ t-L_-. =oJ€=u'cr
Inm
hler"rti"al Point
df:. x_('g
dC,.E
x..^ ul' t_
YXdC
"NP .'cg "-m
f'lF'*. v.u ! v. rv,
._ ne-1_n1q\
C
i0
l:
v
"Nlrf
2.2
,r ar
-
?
For the data shown in Figure P2.2, determine the following.
n\ Tha eiink fivod nar rirel nnini
b) If we wish to fly the airpiane
at a veiocity of 125 ft/sec, then what would be the trln iift coefficient and what would be the elevator angle for trim?
u. t3
= 2,750 tb
S = '180 fi2
\4,/
0.10
0.0 5
C
n"q
0.0
-0.05
-0. 1i)
-u. t3
Figure P2.2.
Scilution:
Given: W = 2,750 lb, S = 180 i12, c.g. = 0.25Find: The neutrai point, trim lift coeificient and
125 itlsec.
ttn
^cg -lr^NP v
--_- = v ccdC,
measure dcm I CC; f rom graph P2.2 irr Ot/
n
-tiin
I -
L
it
^l^,,^+.,r
:lg'/d.LUl
rnnlr rur JilYl cnood ILJ u 1r dl lult; in frimil fnr flinhi li JyqtEu r-
Y
_'_fr_= 0,2F - (-0,15) = 0.40 e 4
o
=
Lvt
2r'
(125 fUsec)
= (0.s) (2.977 x to-3; slug/ft3
/
= 18.6 lb/ft-
/^ -Wv,
UU
L.
I nrn
2750 lbs
(18.6 lb/ft4) (180 ft2)
C, - 0.82 =+ 6e = -10.5o trim arim
2.,)
P2"3. The canard and wing are
Analyze the canard-vring combination shown in Figure glo;"tti."lly siniiiar anl are made from the same airfoil section"
AR*
-
ARw, \$c
=
S*,6. = 0'45
1l*
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Kieso, Weygandt, Warfield, Young, Wiecek, McConomy Intermediate Accounting, Tenth Canadian Edition CHAPTER 2 CONCEPTUAL FRAMEWORK UNDERLYING FINANCIAL REPORTING ASSIGNMENT CLASSIFICATION TABLE Topic Brief Exercise Usefulness of the Conceptual Framework (CF) and main components of CF Exercise Problem 5, 7 Writing assignments 1 Qualitative Characteristics 1,2,3,9 1,5,6,7 2,3,4,5,7,8 3,5 Elements 4,5,6,7 2,7 2,3,5,8 3 Foundational…
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# In a collinear collision,
Question:
In a collinear collision, a particle with an initial speed $v_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50 \%$ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :
1. $\sqrt{2} \mathrm{v}_{0}$
2. $\frac{v_{0}}{2}$
3. $\frac{\mathrm{v}_{0}}{\sqrt{2}}$
4. $\frac{V_{0}}{4}$
Correct Option: 1
Solution:
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https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-3-section-3-4-the-chain-rule-3-4-exercises-page-204/46
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## Calculus: Early Transcendentals 8th Edition
Published by Cengage Learning
# Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 46
#### Answer
$$y'= 4{[x+{(x+\sin^2 x)}^3]}^3 [3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)+1]$$
#### Work Step by Step
$y' = \frac{d}{dx} {[x+{(x+\sin^2 x)}^3]}^4$ By the chain rule: $= \frac{d{[x+{(x+\sin^2 x)}^3]}^4}{d[x+{(x+\sin^2 x)}^3]} \times \frac{d[x+{(x+\sin^2 x)}^3]}{dx}$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [\frac{d}{dx} [x]+\frac{d}{dx}[{(x+\sin^2 x)}^3]]$ Using the chain rule again: $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [1+\frac{d[{(x+\sin^2 x)}^3]}{d[{(x+\sin^2 x)}]} \times\frac{d[{(x+\sin^2 x)}]}{dx}]$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [1+3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)]$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 [3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)+1]$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Hi, Mommy Teachers! It’s Casey, stopping by from Kidspired Creations! I am Mommy to James (5) who just started Kindergarten, Leyson (3) and Lena (6 months). I currently stay at home with the younger two but I taught Pre-K and Kindergarten before my Mommy days.
Leyson and I were playing (learning) today and were toying around with different Busy Bag Exchange ideas. We will be participating in a Busy Bag Exchange soon where all the moms participating bring gallon baggies filled with simple, hand-made activities for each of the kids who will be there. If 15 moms participate, your child has 15 new Busy Bags full of super fun (and educational) activities!
Leyson and I came up with 3 different ideas today and I thought they would all be fun to share with you Mommy Teachers!
PATTERNS:
We used word strips (found at the Dollar Tree for, you guessed it, \$1 per package) to glue colored squares in AB, ABB and ABC patterns. I cut additional colored squares for Leyson to COPY the patterns on the strips by placing each of the colored squares below the square on the strip. This will get him used to making patterns. Saying the color out loud reinforces the pattern so he can also HEAR the pattern (helpful for an auditory learner – someone who learns better when hearing information).
After he was able to copy the pattern, I then taught him how to EXTEND the pattern. At the end of the strip, he must finish the pattern by placing the appropriate colored squares that would come NEXT in the pattern.
The next step is to create his own patterns without looking at the strip. From there you can create more complex patterns! The options are endless! Patterns are difficult to understand, so it might take a little time for your little one to be able to catch on.
WHAT’S IN A NAME?:
Kids are so funny when it comes to learning how to spell their own names. Most often kids learn how to spell their names before they completely understand the concept that letters make words when put together. Take my son Leyson, for example; anytime he sees a word that starts with an L, he says it spells Leyson. He thinks he owns the letter L. The same goes for the letters E, Y, S, O and N… he owns them all. You see, for young toddlers, they will not understand that the ORDER of letters actually makes a difference when spelling words.
Even in my kindergarten classroom, students who knew how to spell their names would sometimes slip into the idea of “owning” all the letters in their name and would write their names out of order. LUIS would spell his name UILS… still thinking that it spelled the same thing. Even more common would be for kids to write their names backwards (SIUL)… and several, get this, would write their names in mirror image! I can’t even do that! All of these are totally normal progressions when your child is learning how to spell his/her name, so here is an easy activity to help:
You need two word cards (or index cards). Write your child’s name evenly and legibly on each card. Leave the first card whole, and cut out each individual letter on the second card. First, have your child put each cut out letter on top of the letters on his/her name card, matching letter for letter. Then, have your child put his/her name together underneath the name card. Last, take away the name card and have your child put together the letters to spell his/her name from memory. Be sure to say the letters out loud in order so your child can hear how to spell his/her name in addition to seeing it. Jessica had a great post on singing a catchy song to learn how to spell a name.
HAMBURGER PLAY:
This game was Leyson’s idea. He picked up a circle I cut out and pretended to eat it, saying it was a hamburger bun. Bam! Instant Busy Bag idea!
I used construction paper to cut out a top and bottom bun, hamburger, tomato, cheese, ketchup, mustard, pickles, lettuce, and bacon and we made a hamburger! Not only is this a great pretend-play activity, but it was a great way to encourage language development and sequencing. I had Leyson explain to me exactly how we should make a hamburger. I encouraged him to use words such as “first,” “next,” “then,” and “last” in his explanation. You can also add an extra element by writing numbers on each of the pieces to teach your child number order (1: the bottom bun, 2: the hamburger, 3: cheese, etc.). For the beginning reader? Write the words “bun,” “pickle,” etc. on each of the pieces and then enjoy your nice, tasty treat!
I am still brainstorming different Busy Bag ideas! If you have a good one, please comment below and share! Hopefully after the Busy Bag Exchange I will have another post for you, so stay tuned!!!
Casey Dellinger Hilty
“One day, all children in this nation will have the opportunity to attain an excellent education.”
Teach For America
LA Corps 2005
#### 1Pingbacks & Trackbacks on Easy Busy Bag Ideas from Kidspired Creations!
1. ###### Easter Egg Letter Hunt | The Mommy Teacher
June 20, 2013 at 2:33 pm (4 years ago)
[…] read is “Leyson.” Well, that and “pizza.” In a past post I wrote about Busy Bags, I talked a little bit about the significance and progression of a child learning to read and […]
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# Convert 12 feet to meters (12 ft to m)
How to convert 12 feet to meters? feet: ft meter: m
You may also interested in: meters to feet Converter
The online ft to meter Converter is used to convert the length from Feet to m.
#### The feet to meter Conversion Formula
You can use the following formula to convert ft to m :
X(m) = y(feet) × 0.3048
To convert 12 feet to meters:
X(feet) = 12(m) × 0.3048
Answer: 3.6576 m
#### feet to meter conversion table
Feet (ft) meter (m)
1 ft 0.3048 m
2 ft 0.6096 m
3 ft 0.9144 m
4 ft 1.2192 m
5 ft 1.524 m
6 ft 1.8288 m
7 ft 2.1336 m
8 ft 2.4384 m
9 ft 2.7432 m
10 ft 3.048 m
20 ft 6.096 m
30 ft 9.144 m
40 ft 12.192 m
50 ft 15.24 m
60 ft 18.288 m
70 ft 21.336 m
80 ft 24.384 m
90 ft 27.432 m
100 ft 30.48 m
500 ft 152.4 m
1000 ft 304.8 m
Full feet to meters conversion table
#### Online Converter to convert cm to feet
We provide the online converter for free. You can use the feet to meter converter to convert the feet to meter.
#### References
More references for Feet/Foot unit and Metre
#### Local Weather Status
Ashburn, US
13th December, 2019 Friday
Moderate Rain
-0.56 - 2.22
Humidity: 82 %
Wind: 2.1 km/h
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https://www.traditionaloven.com/tutorials/flow-rate/convert-gtt-drop-per-second-to-bar-fluid-barrel-per-second.html
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Convert gtt/sec to bar/sec | drop per second to barrels fluid per second
# flow rate units conversion
## Amount: 1 drop per second (gtt/sec) of flow rate Equals: 0.00000042 barrels fluid per second (bar/sec) in flow rate
Converting drop per second to barrels fluid per second value in the flow rate units scale.
TOGGLE : from barrels fluid per second into drops per second in the other way around.
## flow rate from drop per second to barrel fluid per second conversion results
### Enter a new drop per second number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other flow rate measuring units - complete list.
How many barrels fluid per second are in 1 drop per second? The answer is: 1 gtt/sec equals 0.00000042 bar/sec
## 0.00000042 bar/sec is converted to 1 of what?
The barrels fluid per second unit number 0.00000042 bar/sec converts to 1 gtt/sec, one drop per second. It is the EQUAL flow rate value of 1 drop per second but in the barrels fluid per second flow rate unit alternative.
gtt/sec/bar/sec flow rate conversion result From Symbol Equals Result Symbol 1 gtt/sec = 0.00000042 bar/sec
## Conversion chart - drops per second to barrels fluid per second
1 drop per second to barrels fluid per second = 0.00000042 bar/sec
2 drops per second to barrels fluid per second = 0.00000084 bar/sec
3 drops per second to barrels fluid per second = 0.0000013 bar/sec
4 drops per second to barrels fluid per second = 0.0000017 bar/sec
5 drops per second to barrels fluid per second = 0.0000021 bar/sec
6 drops per second to barrels fluid per second = 0.0000025 bar/sec
7 drops per second to barrels fluid per second = 0.0000029 bar/sec
8 drops per second to barrels fluid per second = 0.0000034 bar/sec
9 drops per second to barrels fluid per second = 0.0000038 bar/sec
10 drops per second to barrels fluid per second = 0.0000042 bar/sec
11 drops per second to barrels fluid per second = 0.0000046 bar/sec
12 drops per second to barrels fluid per second = 0.0000050 bar/sec
13 drops per second to barrels fluid per second = 0.0000055 bar/sec
14 drops per second to barrels fluid per second = 0.0000059 bar/sec
15 drops per second to barrels fluid per second = 0.0000063 bar/sec
Convert flow rate of drop per second (gtt/sec) and barrels fluid per second (bar/sec) units in reverse from barrels fluid per second into drops per second.
## Flow rate. Gas & Liquids.
This unit-to-unit calculator is based on conversion for one pair of two flow rate units. For a whole set of multiple units for volume and mass flow on one page, try the Multi-Unit converter tool which has built in all flowing rate unit-variations. Page with flow rate by mass unit pairs exchange.
# Converter type: flow rate units
First unit: drop per second (gtt/sec) is used for measuring flow rate.
Second: barrel fluid per second (bar/sec) is unit of flow rate.
QUESTION:
15 gtt/sec = ? bar/sec
15 gtt/sec = 0.0000063 bar/sec
Abbreviation, or prefix, for drop per second is:
gtt/sec
Abbreviation for barrel fluid per second is:
bar/sec
## Other applications for this flow rate calculator ...
With the above mentioned two-units calculating service it provides, this flow rate converter proved to be useful also as a teaching tool:
1. in practicing drops per second and barrels fluid per second ( gtt/sec vs. bar/sec ) measures exchange.
2. for conversion factors between unit pairs.
3. work with flow rate's values and properties.
To link to this flow rate drop per second to barrels fluid per second online converter simply cut and paste the following.
The link to this tool will appear as: flow rate from drop per second (gtt/sec) to barrels fluid per second (bar/sec) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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# Series And Parallel Circuits Basics Phet Answer Key
By | January 15, 2023
# Exploring Series and Parallel Circuits Basics Phet Answer Key
Electricity is a powerful tool which can be used to accomplish many tasks. It can be used to power lights, devices, and appliances in the home. However, in order to use electricity safely and efficiently, it is important to understand the basics of series and parallel circuits. The Series and Parallel Circuits Basics Phet Answer Key can help those curious about wiring diagrams gain insight into this complex field of study.
Wiring diagrams are diagrams that show how the electrical components of a circuit are connected. This includes the wires, switches, outlets, outlets, and other components of the circuit. They can help the reader understand the relationship between these components and how they interact with one another. Understanding the basics of series and parallel circuits can help the reader better understand how these components work together.
## Examining Series Circuits
A series circuit is a type of electrical circuit in which all components connected in series have the same current flowing through them. This means that if one component is removed from the circuit, the entire circuit will shut down. An example of a series circuit can be seen in a simple lightbulb circuit. In this type of circuit, the lightbulb is connected in series with a power source and a switch. When the switch is flipped, the electricity flows to the bulb, lighting it up and completing the circuit.
Series circuits are often used when only a single device needs to be powered. This is because all components in the circuit are connected in series and receive the same current. This eliminates the need for additional components, such as multiple outlets and switches.
## Parallel Circuits
Unlike series circuits, parallel circuits are made up of two or more components that are connected in parallel. This means that each component has its own dedicated path for the current to flow through. This is advantageous for circuits that require powering multiple devices, as it allows each device to have its own dedicated power source.
An example of a parallel circuit can be seen in a house. Each room of the house is equipped with its own light switch and outlet, allowing each device to have its own dedicated power source. This type of circuit also allows for larger currents to flow, since the current is split between multiple components.
## Wiring Diagrams and the Series and Parallel Circuits Basics Phet Answer Key
Wiring diagrams are essential to understanding series and parallel circuits. They allow the reader to see how the components of the circuit interact with each other and identify any potential problems. The Series and Parallel Circuits Basics Phet Answer Key can help the reader understand how to read and interpret wiring diagrams.
Wiring diagrams are typically drawn with an easy-to-understand graphical representation of the electrical components of the circuit. They usually include the power source, switch, and outlets. Additionally, diagrams may also include labels for resistors, capacitors, and other components, depending on the type of circuit. The Series and Parallel Circuits Basics Phet Answer Key provides detailed instructions on how to interpret and construct these diagrams.
Understanding series and parallel circuits is essential for anyone who wants to work with electricity. The Series and Parallel Circuits Basics Phet Answer Key can provide readers with the knowledge and understanding necessary to do so. With this knowledge, readers can safely and effectively wire circuits and understand the principles behind how electricity works.
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# Intuition behind $(X^TX)^{-1}$ in closed form of w in Linear Regression
The closed form of w in Linear regression can be written as
$\hat{w}=(X^TX)^{-1}X^Ty$
How can we intuitively explain the role of $(X^TX)^{-1}$ in this equation?
• Could you elaborate on what you mean by "intuitively"? For instance, there is a wonderfully intuitive explanation in terms of inner-product spaces presented in Christensen's Plane Answers to Complex Questions, but not everybody will appreciate that approach. As another example, there's a geometric explanation in my answer at stats.stackexchange.com/a/62147/919, but not everybody views geometrical relations as "intuitive."
– whuber
Aug 29, 2018 at 17:02
• Intuitively is like what does $(X^TX)^{-1} mean? Is it some kind of distance calculation or something, I don't understand it. Aug 29, 2018 at 17:09 • That's fully explained in the answer I linked to. – whuber Aug 29, 2018 at 17:27 • This question already exists here although possibly not with a satisfying answer math.stackexchange.com/questions/2624986/… Aug 29, 2018 at 17:30 • – Tim Jun 14, 2022 at 16:41 ## 3 Answers I found these posts particularly helpful: How to derive the least square estimator for multiple linear regression? Relationship between SVD and PCA. How to use SVD to perform PCA? http://www.math.miami.edu/~armstrong/210sp13/HW7notes.pdf If$X$is an$n \times p$matrix then the matrix$X(X^TX)^{-1}X^T$defines a projection onto the column space of$X$. Intuitively, you have an overdetermined system of equations, but still want to use it to define a linear map$\mathbb{R}^p \rightarrow \mathbb{R}$that will map rows$x_i$of$X$to something close to values$y_i$,$i\in \{1,\dots,n\}$. So we settle for sending$X$to the closest thing to$y$that can be expressed as a linear combination of your features (the columns of$X$). As far as an interpretation of$(X^TX)^{-1}$, I don't have an amazing answer yet. I know you can think of$(X^TX)$as basically being the covariance matrix of the dataset. •$(X^T X)$is sometimes referred to as a "scatter matrix" and is just a scaled up version of the covariance matrix Mar 19, 2019 at 2:43 ## Geometric viewpoint A geometric viewpoint can be like the n-dimensional vectors $$y$$ and $$X\beta$$ being points in n-dimensional-space $$V$$. Where $$X\beta$$ is also in the subspace $$W$$ spanned by the vectors $$x_1, x_2, \cdots, x_m$$. ### Two types of coordinates For this subspace $$W$$ we can imagine two different types of coordinates: • The $$\boldsymbol{\beta}$$ are like coordinates for a regular coordinate space. The vector $$z$$ in the space $$W$$ are the linear combination of the vectors $$\mathbf{x_i}$$ $$z = \boldsymbol{\beta_1} \mathbf{x_1} + \boldsymbol{\beta_2} \mathbf{x_1} + .... \boldsymbol{\beta_m} \mathbf{x_m}$$ • The $$\boldsymbol{\alpha}$$ are not coordinates in the regular sense, but they do define a point in the subspace $$W$$. Each $$\alpha_i$$ relates to the perpendicular projections onto the vectors $$x_i$$. If we use unit vectors $$x_i$$ (for simplicity) then the "coordinates" $$\alpha_i$$ for a vector $$z$$ can be expressed as: $$\alpha_i = \mathbf{x_i^T} \mathbf{z}$$ and the set of all coordinates as: $$\boldsymbol{\alpha} = \mathbf{X^T} \mathbf{z}$$ ### Mapping between coordinates $$\boldsymbol{\alpha}$$ and $$\boldsymbol{\beta}$$ for $$\mathbf{z} = \mathbf{X}\boldsymbol{\beta}$$ the expression of "coordinates" $$\alpha$$ becomes a conversion from coordinates $$\beta$$ to "coordinates" $$\alpha$$ $$\boldsymbol{\alpha} = \mathbf{X^T} \mathbf{X}\boldsymbol{\beta}$$ You could see $$(\mathbf{X^T} \mathbf{X})_{ij}$$ as expressing how much each $$x_i$$ projects onto the other $$x_j$$ Then the geometric interpretation of $$(\mathbf{X^T} \mathbf{X})^{-1}$$ can be seen as the map from vector projection "coordinates" $$\boldsymbol{\alpha}$$ to linear coordinates $$\boldsymbol{\beta}$$. $$\boldsymbol{\beta} = (\mathbf{X^T} \mathbf{X})^{-1}\boldsymbol{\alpha}$$ The expression $$\mathbf{X^Ty}$$ gives the projection "coordinates" of $$\mathbf{y}$$ and $$(\mathbf{X^T} \mathbf{X})^{-1}$$ turns them into $$\boldsymbol{\beta}$$. Note: the projection "coordinates" of $$\mathbf{y}$$ are the same as projection "coordinates" of $$\mathbf{\hat{y}}$$ since $$(\mathbf{y-\hat{y}}) \perp \mathbf{X}$$. • A very similar account of the topic stats.stackexchange.com/a/124892/3277. Aug 30, 2018 at 12:49 • Indeed very similar. To me this view is very new and I had to take a night to think about it. I did always view least squares regression in terms of a projection but in this viewpoint I have never tried to realize an intuitive meaning to the part$(X^TX)^{-1}$or I always saw it in the more indirect expression$X^T y = X^TX\beta$. Aug 30, 2018 at 12:54 • How should we understand$X(X^TX)^{-1}X^T$in this perspective?$X^Ty$is the "$\alpha$-coordinates" of$y$, and so$(X^TX)^{-1}X^Ty$is the "$\beta$-coordinates" of$y$. But...doesn't left multiplying that$\beta$-coordinate by the coordinate bases$X$restore$y$? Jun 14, 2022 at 14:20 • @whoknows it restores$\hat{y}$, the projection of$y$into the column space of$X$. Jun 14, 2022 at 15:38 Assuming you're familiar with the simple linear regression: $$y_i=\alpha+\beta x_i+\varepsilon_i$$ and its solution: $$\beta=\frac{\mathrm{cov}[x_i,y_i]}{\mathrm{var}[x_i]}$$ It's easy to see how$X'y$corresponds to numerator above and$X'X$maps to denominator. Since we're dealing with matrices the order matters.$X'X$is KxK matrix, and$X'y$is Kx1 vector. Hence, the order is:$(X'X)^{-1}X'y\$
• But that analogy itself doesn't tell you if pre- or postmultiply with the inverse. Aug 29, 2018 at 18:17
• @kjetilbhalvorsen, I put the order of operations Aug 29, 2018 at 18:21
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LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
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## ◆ slaqsy()
subroutine slaqsy ( character uplo, integer n, real, dimension( lda, * ) a, integer lda, real, dimension( * ) s, real scond, real amax, character equed )
SLAQSY scales a symmetric/Hermitian matrix, using scaling factors computed by spoequ.
Purpose:
``` SLAQSY equilibrates a symmetric matrix A using the scaling factors
in the vector S.```
Parameters
[in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the upper or lower triangular part of the symmetric matrix A is stored. = 'U': Upper triangular = 'L': Lower triangular``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in,out] A ``` A is REAL array, dimension (LDA,N) On entry, the symmetric matrix A. If UPLO = 'U', the leading n by n upper triangular part of A contains the upper triangular part of the matrix A, and the strictly lower triangular part of A is not referenced. If UPLO = 'L', the leading n by n lower triangular part of A contains the lower triangular part of the matrix A, and the strictly upper triangular part of A is not referenced. On exit, if EQUED = 'Y', the equilibrated matrix: diag(S) * A * diag(S).``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(N,1).``` [in] S ``` S is REAL array, dimension (N) The scale factors for A.``` [in] SCOND ``` SCOND is REAL Ratio of the smallest S(i) to the largest S(i).``` [in] AMAX ``` AMAX is REAL Absolute value of largest matrix entry.``` [out] EQUED ``` EQUED is CHARACTER*1 Specifies whether or not equilibration was done. = 'N': No equilibration. = 'Y': Equilibration was done, i.e., A has been replaced by diag(S) * A * diag(S).```
Internal Parameters:
``` THRESH is a threshold value used to decide if scaling should be done
based on the ratio of the scaling factors. If SCOND < THRESH,
scaling is done.
LARGE and SMALL are threshold values used to decide if scaling should
be done based on the absolute size of the largest matrix element.
If AMAX > LARGE or AMAX < SMALL, scaling is done.```
Definition at line 132 of file slaqsy.f.
133*
134* -- LAPACK auxiliary routine --
135* -- LAPACK is a software package provided by Univ. of Tennessee, --
136* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
137*
138* .. Scalar Arguments ..
139 CHARACTER EQUED, UPLO
140 INTEGER LDA, N
141 REAL AMAX, SCOND
142* ..
143* .. Array Arguments ..
144 REAL A( LDA, * ), S( * )
145* ..
146*
147* =====================================================================
148*
149* .. Parameters ..
150 REAL ONE, THRESH
151 parameter( one = 1.0e+0, thresh = 0.1e+0 )
152* ..
153* .. Local Scalars ..
154 INTEGER I, J
155 REAL CJ, LARGE, SMALL
156* ..
157* .. External Functions ..
158 LOGICAL LSAME
159 REAL SLAMCH
160 EXTERNAL lsame, slamch
161* ..
162* .. Executable Statements ..
163*
164* Quick return if possible
165*
166 IF( n.LE.0 ) THEN
167 equed = 'N'
168 RETURN
169 END IF
170*
171* Initialize LARGE and SMALL.
172*
173 small = slamch( 'Safe minimum' ) / slamch( 'Precision' )
174 large = one / small
175*
176 IF( scond.GE.thresh .AND. amax.GE.small .AND. amax.LE.large ) THEN
177*
178* No equilibration
179*
180 equed = 'N'
181 ELSE
182*
183* Replace A by diag(S) * A * diag(S).
184*
185 IF( lsame( uplo, 'U' ) ) THEN
186*
187* Upper triangle of A is stored.
188*
189 DO 20 j = 1, n
190 cj = s( j )
191 DO 10 i = 1, j
192 a( i, j ) = cj*s( i )*a( i, j )
193 10 CONTINUE
194 20 CONTINUE
195 ELSE
196*
197* Lower triangle of A is stored.
198*
199 DO 40 j = 1, n
200 cj = s( j )
201 DO 30 i = j, n
202 a( i, j ) = cj*s( i )*a( i, j )
203 30 CONTINUE
204 40 CONTINUE
205 END IF
206 equed = 'Y'
207 END IF
208*
209 RETURN
210*
211* End of SLAQSY
212*
real function slamch(cmach)
SLAMCH
Definition slamch.f:68
logical function lsame(ca, cb)
LSAME
Definition lsame.f:48
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Practice
Contest
Medium-Hard
# Pre-requisites
Treap, bitset, DP
# Problem
Maintain two types of modifications on the given array and solve the subset-sum problem on its’ subsegments.
# How to get 25 points
For the beginning, let’s focus on the solution to the first subtasks. The constraints are rather small, so all the modifications can be done naively, in the straightformard way. Namely, if you need to change the value of an element, you can do this in O(1) time, and when you need to reverse the subarray, straightforward O(N)-per-reverse approch will do.
Let’s deal with the queries of the third kind.
The following simple DP technique will help us to determine - whether it is possible to get W as the sum of some of these numbers or not. First of all, pseudocode:
fill dp[] with zeroes
dp[0] = 1
for i = L; i <= R; ++i
for j = W; j >= a[i]; --j
dp[j] |= dp[j - a[i]]
Now, let’s have a closer look to the provided code. Suppose that we operate on the array a[]. The value of dp[k] will be equal to one if it is possible to collect the sum of k from the given integers and will be equal to zero otherwise. Initially, only dp[0] is non-zero.
Then, step by step we recalculate the values of dp[] by adding the integers from the given set one-by-one. The value of dp[j] will become equal to one during the adding of the ith element (say a[i]) if it was possible to collect a sum of (j - a[i]) before. Of course, we consider that j ≥ a[i].
The whole complexity of the described solution is O(NQ+PNW) or simply O(N(Q+PW)).
This solution will be sufficient to solve the first and the second subtasks.
# How to get 50 points
At this point, we won’t speed the modifications up, but will improve the execution time of the subset-sum part.
For doing this, we will have two crucial optimizations.
## The first optimization
Now it’s time to remember that there are only K ≥ 10 types of the element of the array a[] (aka weight-plates). During each of the subset-sum queries we process R-L+1 elements of a[] which can be optimized to O(K log (N / K)) = O(K log N) - considering the small value of K it will be good.
Let’s calculate the number of weight plates of each kind. Now, consider a single kind of weight plates and suppose that there are X plates of this kind. Let’s create the new summands for our subset-sum problem by grouping the weight plates:
• The first summand should be equal to the weight of one weight plate;
• The second summand should be equal to the weight of two weight plates;
• The third summand should be equal to the weight of four weight plates;
• And so on: the ith summand should be equal to the weight of 2(i-1) weight plates;
At some moment, you won’t be able to create one more summand. This moment is when the amount of the weight plates you need for the new group is greater than the amount of the weight plates that are left. In this case, the last group should contain all the remaining weight plates.
Now, let’s consider some examples of making the groups for the different amount of weight plates:
• If you have 1 weight plate, there will be a single group containing this weight plate;
• If you have 2 weight plates, there will be two groups, each containing a single weight plate;
• If you have 3 weight plates, there will be two groups - the first containing a single weight plate and the second containing two weight plates;
• If you have 4 weight plates, there will be three groups - the first containing a single weight plate, the second containing two weight plates and the third, again, containing a single weight plate.
• Finally, if you have 100 weight plates, there will be 7 groups, containing 1, 2, 4, 8, 16, 32 and 37 weight plates respectively.
This way, for each kind of the weight plates, you will have no more than O(log N) summands that results in O(NQ + PWK log N) complexity.
## The second optimization
Consider the version of the subset-sum DP that was shown in “How to get 25 points” part. You can note that all the elements of dp[] are either ones or zeroes and the operations on them are binary - to be more precise, only binary OR.
This enables us to use bitset, so we can speed the code up in 32 times.
Now we’ve got O(NQ + PWK log N / 32) solution that can additionally solve the third subtask. Such solution gets 50 points.
# Getting full points
We’ve already achieved O(NQ + PWK log N / 32) complexity. Let’s note that for the constraints of the problem the O(PWK log N / 32) part is already small enough. That means that now we process the queries of the third kind fast enough. Now we need to improve the processing of the queries of the first and the second kind.
Both these queries are actually correspond to a very standard problem - maintain the array under the following queries:
• Reverse a subarray of the array in O(log N) time;
• Change the value of a single array’s element in O(log N) time;
This problem is easily solvable with a data structure called treap.
In our problem, there is one more query that should be maintained:
• Find the amount of the weight plates of each kind on a given segment
For doing this, we can simply store an array of K elements in each node of the treap, where the ith element of this array will denote the number of the weight plated of the ith kind on the corresponding segments. That gives us O(K log N)-per-modification time for the modification query of the both - the first and the second kinds.
Eventually, we get the complexity of O(QK log N + PWK log N / 32). This solution passes all the subtasks and gets 100 points.
# Setter’s Solution
Can be found here
# Tester’s Solution
Can be found here
2 Likes
I think the test cases were a bit weak, because my extreme brute force solution (with 1D DP for subset sum and fast I/O ) got 50 points without any optimization as such.
https://www.codechef.com/viewsolution/8448444
2 Likes
Well here’s what I tried during the contest.
My complexity for query 3 is O(K*W) and for query 1 and query 2 its O(Log N) by using Treap.
I tried to optimise as much as I could but the code was giving TLE on one test case.
Can you explain why?
I personally feel time constraints were a bit too strict.
My solution
1 Like
What is the purpose of all numbers being squarefree?
1 Like
Ha! I have complexity ~ O(Q sqrt N + QK + PK2^K + WK)! Okay, never mind, it’s something like O(QN). Luckily, it’s very unlikely to reach the worst case and my code was fast enough.
1 Like
My solution is a bit different than the one described in editorial and has a time complexity
O ((W + P) 2^k) for the third kind of queries.
In order to check whether a sum S is possible, we compute the number of ways T(S) of obtaining that sum. The sum S is possible iff T(S) \ne 0.
Let us say we have coins of denomination x1, x2, \ldots, xk.
and queries are of the form (S, r1, r2, \ldots, rk), i.e., is it possible to make a sum S, using at most r1 coins of the first type, r2 coins of the second type, and so on (S \le W).
For this set of denomination, we pre-compute an array P[1 \ldots W], such that P[x] represents the number of ways of obtaining the sum x, with no restriction on the number of coins. For the time being let us assume that we already have this array, we will explain later how to compute this array.
Now the number of ways of obtaining the sum S with the restrictions on the number of coins can be computed using inclusion-exclusion principle.
T(S, r1, r2, \ldots, rk) = P[S] - Q(S, r1, r2, \ldots, rk),
where Q(S, r1, r2, \ldots, rk) is the number of ways in which at least one of coins is used more than specified bound.
Q(S, r1, r2, …, rk)
= \sum (-1)^{u + 1} \text{number of ways of obtaining } S \text{ such that coins } i1, i2, \ldots, iu \text{ violate the bound.}
= \sum (-1)^{u + 1} P[S - (r_{i1} + 1) x_{i1} - (r_{i2} + 1) x_{i2} - \ldots - (r_{iu} + 1) x_{iu}]
Hence, T(S, r1, r2, \ldots, rk) can be computed in O(2^k) time using the P array.
Now let us see, how to compute the array P.
If there were a single coin denomination x, then we have computed this array quite easily.
P[i] = 1, if i is divisible by x, and 0 otherwise,
Suppose we already have such an array Q for the coins \{x2, x3, \ldots, xk\}, and we want to use it to compute array P for the coin set \{x1, x2, \ldots, xk\}
P[i] = Q[i] + Q[i - x1] + Q[i - 2x1] + \ldots
Note that,
P[i] = Q[i] + P[i - x1]
Hence, array P can be computed in O (W) time using the array Q.
We have k denominations, and we would want to compute the pre-computation array for each subset of denominations, hence the pre-computation can be done in O (W2^k) time.
Alternatively, we can always use the denomination set which consists of all k types, and add the bounds (r_i = 0), for denominations which are missing in the set. This means we only need to pre-compute the array for the coin set consisting of all denominations, and can be done in O (Wk)
Also note that, we need to do all the computation modulo some prime (I chose 10^4 + 7, mainly to reduce the number of modulo operations)
Of course, this means that the constraint P \le 1000, can be relaxed, and we can probably allow up to 10^5 queries. On the other hand, this approach has exponential complexity in terms of k, so even if we increase the value of k slightly (say to 15), this approach will fail.
8 Likes
https://www.codechef.com/viewsolution/8499253 This solutions runs fine for most test cases but failed one can somebody Explain!
In the first optimization, k<=10.
We can actually do that knapsack in O(PWK) time instead of O(PWKlogN/32). Although it does not work faster -at least not my implementation- it’s better asymptotically.
Suppose we have K different values : \{(x_1, c_1),(x_2,c_2)...,(x_K,c_K)\}.
x_k denotes the value and c_k denotes the quantity.
Then this pseudo-code calculates the dp array that we want:
f = [1,0,0...0]
for i = 1 to K:
nf = [0,0,0,...,0]
for j = 1 to W:
nf[j] = 0
for t = 0 to c[i]
nf[j] |= f[j - t * x[i]]
f = nf
Well, this obviously takes O(PWKN) time but we can easily get rid of the third loop. Key insight is this observation, during any change for array nf, j % x[i] = (j + t * x[i]) % x[i]. Which means if we group the integers according to their remainder each group is independent with each other. That’s why we can calculate them separately.
Here is the implementation for more details : https://www.codechef.com/viewsolution/8433813
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Can’t we use segment tree to solve this one
for first and third query it will take O(log n) time and for second query it will take O(X+logn) X=number of elemnets to be reversed
I suppose this part is wrong O(X+logn) . Because in the worst case lets suppose you will want to reverse
n elements than you again have to rebuilt the whole tree . which can take O(n+nlogn) . So according to me second query could be solved in O(X+XlogX) with segment tree approach .
[Solved]
The sample code in the editorial computed dp[] in reverse order for each summand. (which is the right thing to do here in this JOB )
But, in the provided code, we have something:
for(int j = 0; j <= limit - complete_shifts; j++)
It seems it is doing the JOB opposite.
I am confused if its really working in opposite , or I am missing something.
Any help would be greatly appreciated.
[Explanation:]
I missed the fact that there are two arrays: reachable_sum[] and shifted_mask[] , which are independent.
As we are using shifted_mask[] for a loop and then updating r_sum[] ; the above loop direction MATTERS here here and its logical.
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Probably nothing.
1 Like
Test cases are definitely weak. A lot of people had similar O(n*W) solutions for query 3 and got 50 points. I think Codechef should rejudge the solutions after revising test cases.
https://www.codechef.com/viewsolution/8535260
Seriously what is up with the square-freeness
And it is a shame that O(PWK) does not pass but O(PWKlogN/32) passes.
1.5 seconds time limit is too harsh, considering setter’s solution works with 1.45 seconds.
Do not you think it is a bad idea to set time limit (spent by setter’s solution) + epsilon even though it is a 10 days contest.
4 Likes
I got Accepted with O(PWK), but I had to do multiple code optimizations to get AC (and, even so, I had a runtime of 1.48 sec on one of the test cases ). But, on the other hand, I had O(sqrt(N)) for the other operations (I didn’t use treaps), so I had to optimize more the O(PWK) part to compensate for the other slightly slower operations: 1) sort the K values in increasing order of value * count(value); 2) stop the DP if we formed sum W (obvious); 3) if value * count(value) >= W then you can ignore the count and do DP as if count = infinite; 4) iterate only up to the max sum formed so far
I’m also disappointed that the square-freeness was bogus I was looking forward to some fancy mathematical properties of the knapsack problem when all the values are square-free On the other hand, Googling this didn’t provide anything useful, so I had suspected the square-freeness was useless.
2 Likes
Well I tried all of these individually but it seemed to increase the time taken by my code ( weird ).
I didn’t try all of them together in one code though.
Also I had used treaps so Q1,Q2 were log N per query.
Can you check out my code and tell me how to further optimise it?
It TLEs on one test ( Link given in above comment )
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Open In App
# Expedia On-Campus Internship Interview Experience 2020
Round-1: The first round was conducted on HackerRank, which consists of 2 coding problems and 6 MCQ questions including the concept of sorting algorithms, the basic concept of programming language, and some basic calculations by loops. (Test Duration: 1.5 hours)
1. Run Length Encoding
2. Convert “12th Dec 1989” to “12-12-1989”
The coding questions were pretty much easy if you have practiced and prepared for coding rounds. But if you haven’t done anything yet, Go now and start your preparation because there is no shortcut for success and you have to practice and do it smartly.
For the MCQ part go for basic concepts of any programming language. I would suggest doing quizzes from GeeksforGeeks to get in touch with all concepts of C, C++, Python, Java, and other subjects.
Round-2:
1. This was the first technical interview round which was about 60 minutes long. It was started with an introduction. Then the problem solving round was started in which he first asked what is the concept of pass by value and pass by reference in C++.
2. The second problem was an array overlapping question where I have to calculate the minimum number of bombs required to blow all the kingdoms whose starting and ending points of all the kingdoms are given.
3. The last question demands to detect a loop in a linked list and if present calculates its length.
Round-3:
1. This round started with an introduction, then he asked me about some projects that I worked on during my previous internships.
2. Then he started with the coding round. He for an array like [1, 2, 3, 5, 7, 8, 234, 235]. We have to return the array having a starting and ending point that is calculated by taking the first and last values that are consecutive. So Output will be [1-3, 5-8, 234-235].
3. Firstly I gave a solution of complexity O(n^2) by traversing the array and calculating the value consecutive to that value. Then I gave a solution of complexity O(n) where I took a starting variable and an end variable initialized to the first element of the array and then started the for loop from index 1 and will increase to the next element if it is consecutive to the end variable.
4. And if it is not consecutive then I inserted the starting variable and ending variable into the output array and changed the start and end variable to the current element of the array.
5. Then he asked about what I know all about binary search trees and how depth-first search and breadth-first depth is implemented by any tree traversal.
That’s all about my interview. After these two technical rounds, students were shortlisted, and those who were shortlisted were eligible for the last round that was HR round. But I was not shortlisted for that. And this is life, all about failures and success.
So, All the best for your interview and cracking it. I hope this experience helps you.
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# Middle Grades Math 2010 Homeschool Bundle Grade 6
List price: US\$163.25
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## Description
Prentice Hall Mathematics is a Math curriculum for homechooling that's geared toward middle school-aged children. The content in Course 1 helps you create lesson plans that improve your child's understanding of Math. This program was developed using educational research and contains interactive problems and activities. Furthermore, it naturally follows the enVisionMATH program, ensuring your child's transition from elementary to middle school Mathematics is smooth. In middle school, your child will begin working with algebra and geometry, both of which require a solid understanding of foundational math concepts. Prentice Hall Mathematics: Course 1 develops the groundwork for both subjects, helping your child grow in his or her Math abilities. The curriculum provides engaging and interactive activities to keep your child thinking critically about Math. The best part is that the program is dynamic, meaning you control the pace of your homeschooling lessons. Linger on concepts your child is struggling with or move ahead if he or she feels confident. As your child works through this curriculum, he or she will begin to understand rational and irrational numbers, patterns, integers, inequalities and equations, among many other important math topics. Using the materials in this program, you'll help your child achieve his or her educational goals. In fact, by the time your child finishes Course 1, he or she should be able to: Complete algebraic expressions. Use algebraic skills to solve real-world problems (i.e., how long will it take to reach a destination that's 20 miles away if you're moving 60 miles per hour?). Solve inequalities and equations that have a single variable. Understand the difference and relationship between dependent and independent variables. Recall and use geometric equations for things such as area, surface area and volume. The materials included in Prentice Hall Mathematics: Course 1 will help your child achieve these and other educational goals. For more information about the items included in this curriculum for homeschooling, visit the Features and Benefits page.show more
## Product details
• Hardback
• 241.3 x 304.8 x 132.08mm | 2,789.58g
• English
• 0133197506
• 9780133197501
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http://www.gurufocus.com/term/Total%20Liabilities/PAM/Total%2BLiabilities/Pampa%2BEnergia%2BSA
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Switch to:
GuruFocus has detected 7 Warning Signs with Pampa Energia SA \$PAM.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Pampa Energia SA (NYSE:PAM)
Total Liabilities
\$4,833 Mil (As of Sep. 2016)
Pampa Energia SA's total liabilities for the quarter that ended in Sep. 2016 was \$4,833 Mil.
Pampa Energia SA's quarterly total liabilities increased from Mar. 2016 (\$1,639.25 Mil) to Jun. 2016 (\$1,935.33 Mil) and increased from Jun. 2016 (\$1,935.33 Mil) to Sep. 2016 (\$4,832.57 Mil).
Pampa Energia SA's annual total liabilities declined from Dec. 2013 (\$1,650.95 Mil) to Dec. 2014 (\$1,630.99 Mil) but then increased from Dec. 2014 (\$1,630.99 Mil) to Dec. 2015 (\$1,712.05 Mil).
Definition
Total liabilities are the liabilities that the company has to pay others. It is a part of the balance sheet of a company that shareholders do not own, and would be obligated to pay back if the company liquidated.
Pampa Energia SA's Total Liabilities for the fiscal year that ended in Dec. 2015 is calculated as
Total Liabilities = Total Current Liabilities + Total Long Term Liabilities = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 741.153905586 + (516.475778413 + 257.552576682 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 62.8071544464 + 26.6162404388 + 107.441010585) = 1,712
Total Liabilities = Total Assets (A: Dec. 2015 ) - Total Shareholders Equity (A: Dec. 2015 ) = 2252.15243761 - 540.105771459 = 1,712
Pampa Energia SA's Total Liabilities for the quarter that ended in Sep. 2016 is calculated as
Total Liabilities = Total Current Liabilities + (Total Long Term Liabilities) = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 2027.71924179 + (1109.42859537 + 717.801934067 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 336.945246116 + 61.4994553103 + 579.170596812) = 4,833
Total Liabilities = Total Assets (Q: Sep. 2016 ) - Total Shareholders Equity (Q: Sep. 2016 ) = 5235.6585047 - 403.093435232 = 4,833
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Pampa Energia SA Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Total Liabilities 1,645 1,716 1,638 1,647 2,242 1,935 1,651 1,631 1,712 4,177
Pampa Energia SA Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Total Liabilities 1,569 1,631 1,780 1,853 2,129 1,712 1,639 1,935 4,833 4,177
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## 100 Points!!!!!!!!! Which of the vectors a =(1,2), b=(0,1), c=(-2,-4) ,u=(-2,1),v=(2,4),w=(-6,3) are: In the same direction?
Question
100 Points!!!!!!!!!
Which of the vectors a =(1,2), b=(0,1), c=(-2,-4) ,u=(-2,1),v=(2,4),w=(-6,3) are: In the same direction?
Which of the vectors a =(1,2), b=(0,1), c=(-2,-4) ,u=(-2,1),v=(2,4),w=(-6,3) are: In opposite directions?
Which of the vectors a =(1,2), b=(0,1), c=(-2,-4) ,u=(-2,1),v=(2,4),w=(-6,3) are: Orthogonal?
in progress 0
2 days 2021-07-20T04:51:52+00:00 1 Answers 2 views 0
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# How much milk does a newborn need?
Contents
## How many mL should a newborn drink chart?
Formula Feeding Amounts by Age
Age # of feedings per day / 24 hours Average Bottle Size
0-4 weeks on-demand ~2-4 ounces / 60-120 ml
5-8 weeks 6-7 ~4 ounces / 120 ml
9-12 weeks/3 months 5 4-6 ounces / 120-180 ml
13-16 weeks/4 months 5 4-6 ounces / 120-180 ml
## Is 10 mL milk enough for a newborn?
The baby who is gaining weight and seems satisfied after each feeding is likely getting adequate amounts of formula.
How much and how often to give formula to your baby.
Age Number of feedings/day Volume of formula/feed
0 – 1 week 6 – 10 Day 1: 5 mL
Day 2: 5–15 mL
Day 3: 15–30 mL
Day 4–7: 30–60 mL (1–2 oz)
## How do I calculate how much milk my baby needs?
How much milk do babies need?
1. Estimate the number of times that baby nurses per day (24 hours).
2. Then divide 25 oz by the number of nursings.
3. This gives you a “ballpark” figure for the amount of expressed milk your exclusively breastfed baby will need at one feeding.
## How much should my newborn eat chart?
Most newborns eat every 2 to 3 hours, or 8 to 12 times every 24 hours. Babies might only take in half ounce per feeding for the first day or two of life, but after that will usually drink 1 to 2 ounces at each feeding. This amount increases to 2 to 3 ounces by 2 weeks of age.
IT IS INTERESTING: Question: What Hot drinks can I have when pregnant?
## How many mL should newborn eat?
Newborn: 2-3 ounces (60-90 mL) per feeding. 1 month old: 4 ounces (120 mL) per feeding. 2 months old: 5 ounces (150 mL) per feeding. 4 months old: 6 ounces (180 mL) per feeding.
## Is 20 ml enough for a newborn?
Newborn babies often take about 30 mL (one ounce) of formula per feeding, and this increases to about 60 mL to 90 mL (two to three ounces) by the end of the first week. Your baby will probably need about eight feedings per day for the first three weeks of life.
## How much milk should a 4 day old drink?
Christine Griffin
Your baby’s age Amount of milk per feed
Day 1 (0 to 24 hours) 7ml (just over a teaspoon)
Day 2 (24 to 48 hours) 14ml (just under 3 teaspoons)
Day 3 (48 to 72 hours) 38ml (1.3fl oz, just over 2 tablespoons)
Day 4 (72 to 96 hours) 58ml (2fl oz, just over 3 tablespoons)
## How much milk should baby drink based on weight?
On average, your baby should take in about 2½ ounces (75 mL) of formula a day for every pound (453 g) of body weight.
## How many mL per kg should a baby drink?
From five days to three months, a full-term, healthy baby will need about 150 ml of prepared formula per kilogram of body weight, every day. For example, a baby who weighs 3 kg will need 450 ml of prepared infant formula each day.
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## Welcome to the Treehouse Community
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# Can someone explain how this sum code works?
```var sum = [1, 2, 3].reduce((a, b) => a + b, 0);
console.log(sum); // 6
```
I know the above code calculates the sum of an array but how?
"a" and "b" are the parameters, then "a" and "b" are added together but what is the 0 for?
The reduce function is executed as follows;
```arr.reduce(callback[, initialValue])
```
The initialValue is optional. In your case 0 is the initial value. In this case it may not be necessary to include the 0, if you had a different initial value, then this value would have been added to sum of your array.
The callback function takes 4 parameters, 2 of which are required. In your case a and b.
```(a , b)
```
a is the accumulator - it keeps track of the total so far b is the current value been processed. The return value of the callback function
```=> a + b
```
will be stored in a until all array elements are processed.
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# Discrete Math
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https://arstechnica.com/science/2016/04/the-search-for-hidden-dimensions-comes-up-empty-again/?comments=1
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# The search for hidden dimensions comes up empty again
## Swinging pendulum obeys inverse square law, fails to fall into hidden dimension.
We have a beautiful theory that puts each of nature's forces into a single, neat package. The whole of it can be summed up in a single line of very compact—and for most, including me, incomprehensible—mathematics. At least, that is what we would like to be able to say, but this beauty is marred. Imagine the Mona Lisa with an eyepatch drawn in using crayon.
That is modern physics. The eyepatch is gravity.
There are many ideas about how to remove the crayon eyepatch from the masterpiece of modern physics and create a single, unified theory, but there's little evidence to support any of them. Among the ideas are theories involving extra dimensions (like string theory). And for nearly 10 years, physicists have been fruitlessly searching for evidence for these hidden dimensions.
Now, one of the most sensitive experiments yet has reported another null result. But it's a very cool experiment nonetheless.
## Why do quantum mechanics and gravity always argue, Mom?
Here is the sort of conflict facing physicists: quantum mechanics takes place on the fixed stage of space and time; in gravity, space and time are actors. In quantum mechanics, the actors are not always continuous, often making sharp jumps. Unfortunately, gravity requires that space and time be smooth and continuous. These conflicting requirements are, so far, unresolved.
One approach to resolving this problem is for space and time to have more than four dimensions. I can't pretend to have a great understanding of this subject, but my impression is that all the jumpiness of space time will get absorbed into the dimensions that we don't observe. Why don't we observe these dimensions? Because they are curled up into tiny volumes.
Yet particles should still feel them. Imagine that you are a particle. You have some mass, and therefore, you bend space and time such that nearby objects feel a force attracting them to you—even better, it is a mutual attraction, the best sort of attraction.
But that force falls off in a very specific way. For three spatial dimensions, if you double the distance between you and the particle you are trying to attract, the force between you reduces by a factor of four. This is called the inverse square law, and it has been calculated and verified by numerous experiments going right back to Newton. However, if there are more than three spatial dimensions, the force will reduce even faster than predicted by the inverse square law.
That is something that we can measure. By placing two masses in proximity to each other and measuring the force of attraction between them, we can attempt to measure any deviation in the inverse square law.
## Hidden dimensions and 18th century pendulums
The basic idea is to use a pendulum and observe the amplitude and phase of its swinging due to some nearby mass. Usually the idea is to use two identical pendulums and observe how they influence each other. The problem with this approach is that the drive frequency—we have to supply some energy to the pendulum to keep it in motion—is the same as the frequency where the signal should be.
To overcome this problem, researchers from China have come up with a new design for the venerable pendulum experiment. In their version, the (torsional) pendulum swings next to a disk. The disc has eight masses and rotates at the same frequency as the swinging motion pendulum. As each mass passes the end of the pendulum it perturbs its motion, meaning that the signal is at eight times the oscillation frequency of the pendulum. This makes it very easy to sense, because for natural harmonic motion, there is no detectable component at that frequency.
Another advantage of the design is that the pendulum can be placed as close as 0.3mm from the disk, allowing the force of gravity to be tested over quite short distances. At this distance, there are all sorts of other problems that come into play. For instance, the electrons in the disk start to feel the electrons in the pendulum. Unless steps are taken, they will arrange themselves to generate a mutually attractive force.
It's really an awesome setup. What we have are a few precision milled masses, glued to very low thermal expansion glasses, coated with a highly conducting metal, and separated by an additional conducting barrier. Add in a few control electronics, a vacuum chamber, and some test masses for calibration, and you have a relatively simple but highly sensitive experiment. It's amazing how sensitive it is given the starting materials.
How sensitive? On the order of 10-17Nm (that's the torque on the pendulum, the actual force is on the order of 10-19N, but torque is what is actually measured). This is about the equivalent to the force of 10,000 water molecules falling onto a surface (a single millilitre of water has over 1022).
And, after all of that work... the inverse square law is still intact down to a length scale of about 50 micrometers.
So, where will things go from here? It should be possible to improve the sensitivity of the current experiment and to get the pendulum and the disk closer together to test even smaller separations. The biggest hole in the experimental data is in separations of 10 micrometer or less, which I don't think this experiment will ever get close to. The problem with going to closer distances is that the forces due to the electrons and protons in the masses will dominate gravity. To avoid that, entirely new approaches (probably based on MEMs, or microelectromechanical devices) will be required.
But we don't have to look for hidden dimensions using the force of gravity, as electrostatic charges also obey the inverse square law. That leads to a final thought: we already know from the way matter behaves that electrostatic forces obey the inverse square law to a distance of at least 10-10m. So, I guess the argument is that gravity might use these extra dimensions, but photons—the force carrier for electromagnetism—don't. That seems a bit on the unlikely side.
Physical Review Letters, 2016, DOI: 10.1103/PhysRevLett.116.131101
1. On the one hand, it's disappointing that a potential avenue for a unified theory of physics has turned up empty. On the other hand, it's still pretty exciting to get another data point confirming that things work basically the way we think they do.
2. Actually, the idea that gravity uses extra dimensions, but that other forces do not, is embedded in several theories. A reason given for why gravity is so many orders of magnitude weaker than other forces is that it "leaks" out.
The whole thing about gravity is that it's not really a force. That's when we consider how relativity treats it. It's an emergent phenomenon. That is, it's caused by other factors, such as mass, which bends space/time, causing the "effect" of a gravitational force.
So if gravity isn't really a force, it's understandable why relativity and guantum mechanics are so hard to resolve.
Maybe we're asking the wrong questions. If so, then we'll never come up with an answer.
Last edited by melgross on Mon Apr 18, 2016 12:29 pm
3. The real question is what if the mass differential matters? it takes mass the size of the moon to get 1.6 m/s2 Can the repeat the experiment with a differential of 1,000 masses instead of 8?
is earth gravity having an effect? it will be identical to both but would earth's overriding mass affect the experiment? Can we repeat it on the moon? or in space between mars and earth's orbits?
4. Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
Last edited by Rosyna on Mon Apr 18, 2016 12:33 pm
5. We hid them for a reason, you know!
6. I have studied several years of Physics at the University level (If I had it all to do over again, I very well may have been a physicist). While all of these things follow from each other, they seem to fail at Occam's Razor. I really think we are going to ultimately find a much simpler explanation for all this weird stuff that happens at the smallest scale. Extra hidden dimensions just seems like too much of a stretch.
Experiments like this are needed to get us closer to how the world around us really works.
7. Things like the inverse square law are based on measurements of the observed effect of gravity, allowing us to predict what will happen between bodies we haven't measured. If there are other dimensions, isn't it possible that the lost energy is already gone by the time we measure it? IE the potential energy of all the quantum probabilities are drained away, and anything we measure with gravity is just the leftover single probability that exists in our dimension.
Do we have an experiment that can measure gravitational probabilities the same way we can observe photons going through slits?
Of course the probability of me just being an idiot who knows nothing about physics is what we're left with here.
8. As far as I understand it, the hidden dimensions are curled up into sizes at the Planck scale. Wouldn't the fact that we can't tell a difference at a scale of 10^(-10) m mean absolutely nothing for whether an effect exists at Planck scale? Or is the assumption that effects from dimensions at the Planck scale can be measured at much larger scales?
9. "The search for hidden dimensions comes up empty again"
They only exist where you are not looking.
10. "The search for hidden dimensions comes up empty again"
They only exist where you are not looking.
Schrodinger's pocket dimension?
11. Isn't part of the issue here that gravity isn't a direct force, like the other forces? Gravity, as it is thought of in my understanding, doesn't move objects. Instead it changes the shape/orientation of space-time, and the new shape means that objects change distance between them at different points in time. The other forces aren't acting indirectly like that, so dealing with gravity becomes a much more sticky proposition.
12. Eurynom0s wrote:
On the one hand, it's disappointing that a potential avenue for a unified theory of physics has turned up empty. On the other hand, it's still pretty exciting to get another data point confirming that things work basically the way we think they do.
When one explanation fails, it just means there's different one out there that's right, whether someone's thought of it yet or not. That's almost as exciting as being right.
13. "The search for hidden dimensions comes up empty again"
They only exist where you are not looking.
Schrodinger's pocket dimension?
14. Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
Life itself is fancy math.
Although I likely wouldn't run around quoting "Starts with a Bang" from Forbes as a definitive source on much of anything. His drivel has even been banned from Slashdot - and we know /.'s standards aren't that high.
15. KGFish wrote:
As far as I understand it, the hidden dimensions are curled up into sizes at the Planck scale. Wouldn't the fact that we can't tell a difference at a scale of 10^(-10) m mean absolutely nothing for whether an effect exists at Planck scale? Or is the assumption that effects from dimensions at the Planck scale can be measured at much larger scales?
One of the techniques often used in experimental physics is to progressively lower/raise boundary at which a hypothetical something can exist without you measuring it. This works when you don't really have much of a handle on the nature of the hypothetical something and it allows you to rule out ranges of nature of the hypothetical something. There is a wide range of potential possibilities for these extra dimensions, some of which could mean an cumulative effect that is measured the experiments sensitivities.
Note that that is exactly the sort of path that LIGO followed. They were progressively refining their instruments until they could detect the signal of the, at the time hypothetical, gravitational wave. It is sort of the nature of instrumentation is that you have to build a less sensitive one first to better figure out how to build the more sensitive one. Ruling out certain scenarios along the way is just a natural by-product of process.
Last edited by Tyler X. Durden on Mon Apr 18, 2016 1:03 pm
16. Owl Saver wrote:
I have studied several years of Physics at the University level (If I had it all to do over again, I very well may have been a physicist). While all of these things follow from each other, they seem to fail at Occam's Razor. I really think we are going to ultimately find a much simpler explanation for all this weird stuff that happens at the smallest scale. Extra hidden dimensions just seems like too much of a stretch.
Experiments like this are needed to get us closer to how the world around us really works.
Then again, general relatively also failed Occam's Razor, until we found experimental data that couldn't be explained by anything else. Quantum Mechanics also completely fails Occam's Razor.
17. Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
The name has always bothered me. In science, a theory is supported by evidence, which first requires a testable hypothesis, but, as I understand it, string "theory" has nothing testable, and thus doesn't even have a roadmap for becoming a theory.
I mean, maybe it's right, maybe it's wrong, I don't know, but theory doesn't seem like the correct word either way.
18. melgross wrote:
Actually, the idea that gravity uses extra dimensions, but that other forces do not, is embedded in several theories. A reason given for why gravity is so many orders of magnitude weaker than other forces is that it "leaks" out.
The whole thing about gravity is that it's not really a force. That's when we consider how relativity treats it. It's an emergent phenomenon. That is, it's caused by other factors, such as mass, which bends space/time, causing the "effect" of a gravitational force.
So if gravity isn't really a force, it's understandable why relativity and guantum mechanics are so hard to resolve.
Maybe we're asking the wrong questions. If so, then we'll never come up with an answer.
The problem with using gravity as a force, or at all, I think, is one of scale.
The effects of quantum mechanics break down on a macro scale. We do not see them, nor measure them, on a scale that can be seen with the naked eye (or even under high magnification). It takes going down to the molecular level before we even begin to notice them. They become more pronounced the smaller we go.
If one thinks about them as sounds, with each effect creating a unique pitch whenever it happens, you'd get a variety of sounds depending on the scale at which you're operating. From a single, pure pitch at the quantum level all the way up to the effect of "white noise" and the average of those sounds in combination at the macro level.
No single or set of quantum events can be heard above the background until you get close enough (scaled down enough) to "hear" it.
Quantum mechanics postulates the possibility of a large thing (say a penny) simply disappearing, or, more famously, a cat being alive or dead at the same time. But that's simply the effects at a minute scale. It doesn't happen on the macro scale. It's POSSIBLE, but so infinitesimally so, in the entire time span of the existence of the universe, it may never happen anywhere.
The bottom line is that if there are unmeasured/undiscovered dimensions out there, you're not going to find them by studying the macro world. The averaging of the quantum effects makes their discovery on the macro scale all but impossible. Gravity plays virtually no role in the micro scale, which makes me wonder if rather than a force itself, it's merely the "white noise" (or an aspect thereof) of the quantum effects happening all the time on the micro scale.
If so, you don't study music by listening to the white noise of an orchestra tuning up. These postulated dimensions aren't going to be found "out there". They'll be found (if they exist at all) only by tuning in on an instrument or two of that orchestra.
19. melgross wrote:
Maybe we're asking the wrong questions. If so, then we'll never come up with an answer.
You always get an answer to a wrong question, but it is almost always the wrong answer. Sometimes though, it is by arriving at the wrong answer that we divine the right question. This does not however, guarantee that with the right question we will arrive at the right answer.
Simple, no?
20. lewax00 wrote:
Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
The name has always bothered me. In science, a theory is supported by evidence, which first requires a testable hypothesis, but, as I understand it, string "theory" has nothing testable, and thus doesn't even have a roadmap for becoming a theory.
I mean, maybe it's right, maybe it's wrong, I don't know, but theory doesn't seem like the correct word either way.
That's why I've always preferred "String Wild-Arsed Guess".
But getting buy in on that name change has been hard.
21. So to summarize, this device measures passes of milled masses glued to glasses with no gasses.
22. "The search for hidden dimensions comes up empty again"
They only exist where you are not looking.
Eddies in the spacetime continuum.
23. Isn't part of the issue here that gravity isn't a direct force, like the other forces? Gravity, as it is thought of in my understanding, doesn't move objects. Instead it changes the shape/orientation of space-time, and the new shape means that objects change distance between them at different points in time. The other forces aren't acting indirectly like that, so dealing with gravity becomes a much more sticky proposition.
I believe that all the forces are actually fields, and the fields are mediated by particles. This includes mass even I think (though maybe not exactly the same - Higgs is still pretty mysterious to me). There is some thinking that gravity is a field mediated by a particle, but at a WAY different scale - many orders of magnitude different from the quantum fields.
I think the big issue that you are raising is that we don't know how the gravity force works in a way that corresponds to how quantum forces work. Either gravity is as you describe it, and quantum fields are as described by quantum theories, or the two are mediated by an as-yet unknown model that describes them in the same way.
If there is no common model, then theoretical physics cannot (as of today) explain how the two systems (gravity according to general relativity and quantum theories as they exist today - I think it's called the standard model?) both function in the world together.
That inability to describe the world coherently is what is driving the physics world to all these experiments to try to find a place where one or the other theory falls down. So far, they have failed. Pretty neat to watch it unfold.
24. lewax00 wrote:
Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
The name has always bothered me. In science, a theory is supported by evidence, which first requires a testable hypothesis, but, as I understand it, string "theory" has nothing testable, and thus doesn't even have a roadmap for becoming a theory.
I mean, maybe it's right, maybe it's wrong, I don't know, but theory doesn't seem like the correct word either way.
It's a lot like the "Intelligent Design" moniker replacing "Creationism".
(Yes, string theory shares a lot of similarities to religion, including the fact it sometimes gets things right for all the wrong reasons. Like blindly rolling dice that land on 5 and 3 when the answer to a question is strawberry)
25. Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
Life itself is fancy math.
Although I likely wouldn't run around quoting "Starts with a Bang" from Forbes as a definitive source on much of anything. His drivel has even been banned from Slashdot - and we know /.'s standards aren't that high.
Life isn't just fancy math, fancy math can be used to describe life but the point of our fancy theories like GR, QED and QCD is that when they were proposed they provided us with predictions, which could be tested by future experiments, which could be used to falsify the theories if they were wrong.
String Theory increasingly appears not to be falsifiable, which is to say it makes no testable predictions.
That's a problem.
26. sbrown23 wrote:
Eddies in the spacetime continuum.
Ah, is he? Is he?
27. lewax00 wrote:
Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
The name has always bothered me. In science, a theory is supported by evidence, which first requires a testable hypothesis, but, as I understand it, string "theory" has nothing testable, and thus doesn't even have a roadmap for becoming a theory.
I mean, maybe it's right, maybe it's wrong, I don't know, but theory doesn't seem like the correct word either way.
Sadly, scientists can be terribly picky about the way non-scientists use the term "theory", but quite sloppy in their own uses.
Personally, I go with "String itlookedreallyplasuibleinthepublastnight"
28. jaggedcow wrote:
Owl Saver wrote:
I have studied several years of Physics at the University level (If I had it all to do over again, I very well may have been a physicist). While all of these things follow from each other, they seem to fail at Occam's Razor. I really think we are going to ultimately find a much simpler explanation for all this weird stuff that happens at the smallest scale. Extra hidden dimensions just seems like too much of a stretch.
Experiments like this are needed to get us closer to how the world around us really works.
Then again, general relatively also failed Occam's Razor, until we found experimental data that couldn't be explained by anything else. Quantum Mechanics also completely fails Occam's Razor.
Actually, I think both passed Occam's Razor. IMO, the problem facing physics at the point when each of those theories were conceived was a problem of taking wrong assumptions for granted. At that time, physicists tried all sorts of ways of adding extra theoretical constructs to make their measurements work with the old theories, like the Ether, some kind of weird fluid filling all of space.
Einstein's big insight was that these kludges were not necessary if you just let go of your assumption that time and space are absolute. The real absolute was the speed of light. Now you can get rid of the Ether.
Likewise, in Quantum Mechanics, you have to jettison assumptions that things can be only one thing or the other. Photons in QM are both particles and waves at the same time, and they can be in more than one place at the same time, and 80% over here and 20% over there. Now things like radioactive decay and black body radiation can make sense.
I think someone, somewhere is going to have a big "aha" moment someday and change their perspective on reality and from this new, "simpler" perspective, things will make more sense. There's Occam's Razor in action :-)
29. Fatesrider wrote:
melgross wrote:
Actually, the idea that gravity uses extra dimensions, but that other forces do not, is embedded in several theories. A reason given for why gravity is so many orders of magnitude weaker than other forces is that it "leaks" out.
The whole thing about gravity is that it's not really a force. That's when we consider how relativity treats it. It's an emergent phenomenon. That is, it's caused by other factors, such as mass, which bends space/time, causing the "effect" of a gravitational force.
So if gravity isn't really a force, it's understandable why relativity and guantum mechanics are so hard to resolve.
Maybe we're asking the wrong questions. If so, then we'll never come up with an answer.
The problem with using gravity as a force, or at all, I think, is one of scale.
The effects of quantum mechanics break down on a macro scale. We do not see them, nor measure them, on a scale that can be seen with the naked eye (or even under high magnification). It takes going down to the molecular level before we even begin to notice them. They become more pronounced the smaller we go.
If one thinks about them as sounds, with each effect creating a unique pitch whenever it happens, you'd get a variety of sounds depending on the scale at which you're operating. From a single, pure pitch at the quantum level all the way up to the effect of "white noise" and the average of those sounds in combination at the macro level.
No single or set of quantum events can be heard above the background until you get close enough (scaled down enough) to "hear" it.
Quantum mechanics postulates the possibility of a large thing (say a penny) simply disappearing, or, more famously, a cat being alive or dead at the same time. But that's simply the effects at a minute scale. It doesn't happen on the macro scale. It's POSSIBLE, but so infinitesimally so, in the entire time span of the existence of the universe, it may never happen anywhere.
The bottom line is that if there are unmeasured/undiscovered dimensions out there, you're not going to find them by studying the macro world. The averaging of the quantum effects makes their discovery on the macro scale all but impossible. Gravity plays virtually no role in the micro scale, which makes me wonder if rather than a force itself, it's merely the "white noise" (or an aspect thereof) of the quantum effects happening all the time on the micro scale.
If so, you don't study music by listening to the white noise of an orchestra tuning up. These postulated dimensions aren't going to be found "out there". They'll be found (if they exist at all) only by tuning in on an instrument or two of that orchestra.
Okay Mr. Smarty pants, but if I walk in to my house and my cat suddenly disappears right before my eyes I am going to say I told you so. :-)
30. jaggedcow wrote:
Owl Saver wrote:
I have studied several years of Physics at the University level (If I had it all to do over again, I very well may have been a physicist). While all of these things follow from each other, they seem to fail at Occam's Razor. I really think we are going to ultimately find a much simpler explanation for all this weird stuff that happens at the smallest scale. Extra hidden dimensions just seems like too much of a stretch.
Experiments like this are needed to get us closer to how the world around us really works.
Then again, general relatively also failed Occam's Razor, until we found experimental data that couldn't be explained by anything else. Quantum Mechanics also completely fails Occam's Razor.
Actually, I think both passed Occam's Razor. IMO, the problem facing physics at the point when each of those theories were conceived was a problem of taking wrong assumptions for granted. At that time, physicists tried all sorts of ways of adding extra theoretical constructs to make their measurements work with the old theories, like the Ether, some kind of weird fluid filling all of space.
Einstein's big insight was that these kludges were not necessary if you just let go of your assumption that time and space are absolute. The real absolute was the speed of light. Now you can get rid of the Ether.
Likewise, in Quantum Mechanics, you have to jettison assumptions that things can be only one thing or the other. Photons in QM are both particles and waves at the same time, and they can be in more than one place at the same time, and 80% over here and 20% over there. Now things like radioactive decay and black body radiation can make sense.
I think someone, somewhere is going to have a big "aha" moment someday and change their perspective on reality and from this new, "simpler" perspective, things will make more sense. There's Occam's Razor in action :-)
No, the real "aha" moment is when we realize Occam subscribed to the dollar shave club :-P
31. I think a lot of the people saying String Theory isn't falsifiable need to read more Ars.
http://arstechnica.com/science/2010/12/ ... ng-theory/
http://arstechnica.com/science/2006/05/4127/
Etc.
32. DCStone wrote:
lewax00 wrote:
Rosyna wrote:
Quote:
Among the ideas are theories involving extra dimensions (like string theory)
Good thing string theory isn't science then?
(String theory is fancy math that can describe any universe, just not this one)
The name has always bothered me. In science, a theory is supported by evidence, which first requires a testable hypothesis, but, as I understand it, string "theory" has nothing testable, and thus doesn't even have a roadmap for becoming a theory.
I mean, maybe it's right, maybe it's wrong, I don't know, but theory doesn't seem like the correct word either way.
Sadly, scientists can be terribly picky about the way non-scientists use the term "theory", but quite sloppy in their own uses.
Personally, I go with "String itlookedreallyplasuibleinthepublastnight"
The term has actually been distorted in the quest to fend off the creationists. It's not really a level of scientific knowledge. A theory is just some construct which explains a host of observations, and good theories should make predictions for observations that haven't yet been made which would allow it to be distinguished from other theories.
String theory is theory, because it is attempting to unify observations under a common umbrella.
I get why it would be annoying for scientists to be using it colloquially, too, but we can tell from context which is meant. And really, what annoys us is when non-scientists compare a major well-supported theory as equivalent to a random idea someone dreamed up in the shower, or pub as the case may be, rather than that they use the word interchangably.
33. jaggedcow wrote:
Then again, general relatively also failed Occam's Razor, until we found experimental data that couldn't be explained by anything else. Quantum Mechanics also completely fails Occam's Razor.
I'm not sure I would agree with that. I would say that GR explains both post-Newtonian gravity *and* "special" relativity in a single explanatory framework that is simpler to understand than either alone. Yes, the *math* is much harder, but from a purely conceptual standpoint GR is far better in Occam's terms:
1) Newton clearly stated he had no idea how gravity worked, specifically how masses "knew" about each other. Field theories of the 19th century clarified some of this, but still didn't explain the difference between momentum and gravitational mass or other observed equivalencies. GR clearly explains this.
2) The G in Newton's theory was completely ad hoc, and had no physical relevance outside the measured value. In GR is is a coupling constant between energy and the universe.
3) SR was similarly ad hoc in that "it had to be true" to explain relativistic in-variance, but had no real explanation of *why*. GR also makes these mechanisms clear - relativity is baked into the nature of the universe, it would be weirder if it didn't exist.
GR basically says that the reason you are sticking to the floor right now is that the shortest distance between today and tomorrow is through the center of the Earth. I find that very simple to understand, and far more elegant than various force based models.
"The whole of it can be summed up in a single line of very compact—and for most, including me, incomprehensible—mathematics"
The Standard Model is entirely unlike this description. It is a set of very different mathematical models that share only the fact that they are quantum in nature and are considered fundamental.The mathematics of the strong force is entirely unlike the electromagnetic force, although they have conceptually similar roots. I don't think any theorist is happy with the SM in its current form, which is why experiments like this exist.
Basically it is important to keep this in mind, the Standard Model is a bag of theories, not a theory. This contrasts with GR, which is internally self-creating. In terms of minimizing assumptions and leftover unexplained bits (wavefunction collapse anyone?) GR is light years ahead. Let us not forget that a lot of people spent a good chunk of the 20th century trying to make QM into GR rather than the other way around.
34. bubish wrote:
So to summarize, this device measures passes of milled masses glued to glasses with no gasses.
So this is how you choose to be remembered for your first post at Ars Technica. An alliterative pun on an experiment trying to verify one of the possible keys to unlock the underlining structure of reality itself.
Welcome aboard! I think you'll fit right in!
35. Hmm, sometimes I think that the fact that things get ever more complicated, contradictory and almost chaotic the more we dive down into the fabric of reality is not a bug in our understanding. It's just as it is. We can try to wrap theories around things but there is no reason that they SHOULD be simple.
Modern physics definitely is in a realm of diminishing returns already. We smash particles into each other with more and more energies and what we get is an exploding cloud of more and more questions. Maybe we will never get down to the gist of it because there is none.
But yes, my intuitive take would be that we're down the wrong road and things are much simpler, we just haven't hit the right road. Still, our intuition has definitely grown against macro-scale phenomena and does not have to be right. Maybe we will never understand all of this, just drive our understanding down into a myriad of conflicting theories. I would hate that since it would mean that there's a limit to science and the world is a mystery at the core. Not only where it came from but also how the fsck it actually works.
All in all modern physics looks more and more as if the universe is just making things up. It's bad excuses all the way down ;-)
"Hey, I was never meant to be looked at! What do I know? In an eternity of Nothing the most improbable things are bound to happen and BANG I was there. I know it isn't pretty, but it works more or less, doesn't it? Leave me alone! Take this particle and come back if you need more! I know you will, but I will make it more and more expensive until you will stop bothering!"
Or to cut it short with the mandatory XKCD:
Well, maybe one day we will train an AI to understand the universe and it will understand it and then we will not understand the AI anymore. Bugger!
Last edited by uhuznaa on Mon Apr 18, 2016 2:56 pm
36. I think a lot of the people saying String Theory isn't falsifiable need to read more Ars.
http://arstechnica.com/science/2010/12/ ... ng-theory/
http://arstechnica.com/science/2006/05/4127/
Etc.
It doesn't really mean much to falsify the absurd through better understanding. The important test is an actual prediction that can get results. It's been 10 years since that pigs fly article (which that LHC article constrained).
37. The often made but never substantiated claim that String Theory is unfalsifiable has to stop. Since it is consistent with earlier theory it has passed a number of tests, same as any other physical theory. (Since it proposes branes as physical objects, it isn't just a mathematical theory.)
The problem is that those tests are postdictive. (Sort of, since ST had a year of being the sole theory predicting the strong force before QCD did it simpler.)
jaggedcow wrote:
Quantum Mechanics also completely fails Occam's Razor.
Say what? QM is the raziest theory there is, just a few axioms and one object - the wavefunction - predicts "all of nature".
But there is also the evidence that there can't be simpler theories. Since QM forbids hidden parameters, it minimizes the number of parameters. There is also a result that shows it minimizes the number of variables in a description of a system. The reason why quantum information takes the form of qubits is a result of a similar minimization over a functional space. (IIRC as I write this.)
QM = OR. QED =D
38. Very nice distillation down to readability (and humor).
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# Quantum calculations and calculational chemistry
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### Quantum calculations and calculational chemistry
1. 1. Quantum Calculations
2. 2. Computational chemistry : a branch of chemistry that uses computers to assist in solving chemical problems. uses the results of theoretical chemistry, incorporated into efficient computer programs, to calculate the structures and properties of molecules and solids. it can (in some cases) predict hitherto unobserved chemical phenomena. widely used in the design of new drugs and materials.
3. 3. Computational chemistry background Basically two ways to calculate molecular structure: 1. Molecular Mechanics ( MM ) 2. Quantum Mechanics ( QM )
4. 4. Molecular Mechanics method: It views the molecule as a collection of atoms held together by bonds and expresses the molecular energy in terms of force constants for bond bending and stretching and other parameters.
5. 5. Does not use a molecular Hamiltonian operator or wave function. can be applied to proteins and other large biological molecules. potential energy of molecules is calculated based on a given force field . The potential energy of the molecular system: E = E covalent + E non-covalent
6. 6. Quantum mechanics Based on the Schrödinger Equation: HΨ = EΨ Hamiltonian operator for a molecule: H = KN + Ke + VNN + VNe + Vee Use the Born-Oppenheimer approximation
7. 7. second approximation: ψel = ψ1 ψ2 ψ3 … ψ ψi = ∑j=1 Cij χj
8. 8. QUANTUM MECHANICAL APROACHES : a) semi-empirical methods (AM1, PM3, PPP, INDO, MINDO, ...) b) non empirical methods : Ab Initio Density Functional Theory ( DFT )
9. 9. Semi-empirical methods: Use a simpler Hamiltonian than correct molecular Hamiltonian Semi-empirical quantum chemistry methods are based on the Hrtree-Fock formalism, but make many approximations and obtain some parameters from empirical data. model only the valence electrons limited to hundred of atoms can be used to study ground and excited molecular states an example is the Huckel MO treatment of conjugated hydrocarbons
10. 10. non empirical methods : do not require empirical parameters . can be used for any molecular system . limited to tens of atoms . can be used to study ground and excited molecular states .
11. 11. Ab initio methods: use the correct Hamiltonian . Not use experimental data other than the values of the fundamental physical constants. The simplest type of ab initio electronic structure calculation is the Hartree Fock (HF) scheme, in which the correlated electron–electron repulsion is not specifically taken into account; only its average effect is included in the calculation. As the basis set size is increased, the energy and wave function tend towards a limit called the Hartree–Fock limit. an example is a Hartree-Fock SCF calculation.
12. 12. Density Functional Theory (DFT) : It’s a new method Not use wave function In DFT, the total energy is expressed in terms of the total one electron density rather than the wave function. there is an approximate Hamiltonian and an approximate expression for the total electron density. Some methods combine the density functional exchange functional with the Hartree–Fock exchange term and are known as hybrid functional methods. Most popular DFT method is B3LYP. (Becke 3‐Parameter method for calculating that part of the molecular energy due to overlapping orbitals, plus the Lee‐Yang‐Parr method of accounting for correlation.)
13. 13. Program packages in molecular electronic structure calculations : 1. Gaussian 2. Gamess 3. DeFT 4. DALTON 5. Mopac Molecular structure and properties visualization programs: 1. GaussView 2. Molekel 3. Raswin 4. Hyperchem 5. Molden
14. 14. Gaussian: Gaussian is arguably the most-used computational quantum-chemistry program. It does electronic-structure calculations and standard quantum chemical calculations. Among the methods available are semi-empirical methods (such as CNDO), Hartree-Fock (restricted and unrestricted), MPn (Mollar-Plesset perturbation theory of order n=2,3,4), CI (Configuration-Interaction), CC (Coupled- Cluster), Multi-configurational SCF (such as CAS-SCF) and various DFT (Density-Functional Theory) methods and …
15. 15. Gaussian Capabilities: It can be used to obtain electronic properties, molecular geometries, vibrational frequencies, orbitals, reaction profiles, IR and Raman spectra, Polarizabilities, Thermochemical analysis, Atomic charges, Dipole moment, Electron affinities, Electrostatic potential and much more…
16. 16. Energies for Particle in a Gaussian Potential Well
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The EUT implies that utility functions have the following functional form: Here there are i states of the world. Also, define aWb to mean that ‘a’ is weakly preferred to ‘b’. The expected utility theory deals with the analysis of situations where individuals must make a decision without knowing which outcomes may result from that decision, this is, decision making under uncertainty.These individuals will choose the act that will result in the highest expected utility, being this the sum of the products of probability and utility over all possible outcomes. The theory starts with some simple axioms that ⦠MmÛîŤ%ÇÖ#=TD¬Þü'ë]cø\$¸ËU´CqÐîR.º¾ä>¬BßåÞMOÄZÚDZìohή!Á²=´9íé=
ñõÉÖ£Úÿifto-îäØ}Ù¿nf? Amsterdam: Kluwer-Nijhoff This function is known as the von NeumannâMorgenstern utility function. Subjective expected utility theory (Savage, 1954): under assumptions roughly similar to ones form this lecture, preferences have an expected utility representation where both the utilities This lecture explains the continuity axiom of expected utility theory. This real valued function is the utility function. 47 0 obj The concept of expected utility is best illustrated byexample. endobj An individual will prefer one risky lottery over another if their utility is higher in the first lottery compared to the second. systematically modeling risk preference in the mid-1940s: Expected Utility Theory. These outcomes could be anything - amounts of money, goods, or even events. endstream In lottery A you receive \$100 for sure. This theory was developed by Daniel Bernoulli (1738) and expanded by John von Neumann and Oskar Morgenstern (1947). Takeaway Points. Second, the axioms need not be descriptive to be normative, and they need not be attractive to all decision makers for expected utility theory to be useful for some. Suppose I am planning a long walk, and need to decide whetherto bring my umbrella. Contents (i) Lotteries (ii) Axioms of Preference (iii) The von Neumann-Morgenstern Utility Function (iv) Expected Utility Representation Back. Expected Utility Theory. }ûi§,/PoÄfÄfüeV @I @L8È4.¾îm. xÚÅXKsÓ0¾÷WèÂ=CT=¬ÇáäF9¸©x&µmá×#[c;Nc§)\*E^iw? Expected utility theory is felt by its proponents to be a normative theory of decision making under uncertainty. understand what lies behind utility theory â and that is the theory of choice. 2 Expected Utility We start by considering the expected utility model, which dates back to Daniel Bernoulli in the 18th century and was formally developed by John von Neumann and Oscar Morgenstern (1944) in their book Theory of Games and Economic Be-havior. There are two acts available to me: taking my umbrella, andleaving it at home. Without risk, economists generally believe that individuals have a utility function which can convert ordinal preferences into a real-valued function. The theoryâs main concern is ⦠The Axioms of Expected-Utility Theory Transitivity Ifx % y andy % z,thenx % z. Completeness x % y ory % x. Independence Ifx Ëy and0
> Continuity Ifx Ëy andy Ëz, thentherearenumbers0
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# Experiment 4: Harmonic Motion Analysis
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1 Experiment 4: Harmonic Motion Analysis Background In this experiment you will investigate the influence of damping on a driven harmonic oscillator and study resonant conditions. The following theoretical development is based on Analytical Mechanics by Fowles (4 th edition). A mass m attached to a spring with spring constant k and experiencing no dissipative forces is governed by the differential equation m d 2 x + kx = 0 (1) 2 dt where x represents the position as a function of time t. This equation assumes horizontal motion. When motion is in the vertical plane, the gravitational force adds a constant term to the above equation, the net effect of which is to merely shift the equilibrium position. All of the other results shown below still apply to vertical motion. The solution to equation (1) may be expressed as x = A 1 cos( ω o t +θ o ) (2) where A 1 is the amplitude of the oscillation, θ o is a phase factor determined by the initial conditions and the natural frequency ω o is given by ω o = k m. (3) In the case that linear damping is present, there is an additional force term in the governing differential equation: m d 2 x dt + c dx + kx = 0. (4) 2 dt In this equation, c is known as the damping constant. There are three classes of solutions to equation (4). Of particular interest to us is the underdamped case, whose solution takes the form x = A 1 e γt cos( ω d t +θ o ) (5) where we will refer to γ as the damping factor (to distinguish it from the damping constant). It takes on the value γ = c. (6) 2m 1
2 Even though the motion strictly speaking is no longer periodic (the amplitude is decaying), ω d is still interpreted as an oscillation frequency. It is given by ω d = ω o 2 γ 2. (7) Note that if we were to measure the amplitude of a decaying oscillation on 2 occasions, separated by n cycles, the ratio of those amplitudes, according to equation (5), would be A o A n = e γnt d (8) where T d = 2π ω d. (9) and A o is the initial amplitude reading while A n is the amplitude reading n cycles later. Finally, suppose we apply a periodic driving force to the oscillator. We can do this through a periodic displacement of one end of the spring. A periodic displacement of the form A drive cos ωt motion ( ) will produce a force ka drive cos( ωt), resulting in an equation of m d 2 x dt 2 + c dx dt + kx = ka drive cos ωt ( ). (10) The solution to this equation contains a decaying portion that vanishes once the oscillation reaches steady state. The steady state portion is x = Acos( ωt +φ) (11) where A = ka drive (12) ( k mω 2 ) 2 + c 2 ω 2 and tanφ = cω k mω 2. (13) When the drive frequency ω is adjusted so that the amplitude is maximized, we say that 2
3 resonance has been achieved. The resonant frequency is given by ω r = ω o 2 2γ 2 (14) It is left as an exercise to the reader to show both of the following (and you will need to derive this in your introduction:) when ω = ω o then φ = 90 o (15) and at resonance for a weakly damped oscillator A res ωa drive c m = ωa drive 2γ. (16) A word of caution: there are four different frequencies used to describe the damped, driven oscillator: ω, ω o, ω d, ω r. Make sure you understand the difference between these frequencies. Procedure: Apparatus: Pasco Driven Harmonic Motion Analyzer 1. The apparatus should already be set up as indicated in Figure 1 of the equipment manual in the lab, using a red spring. D0 NOT TAKE THE HANGING BARS OFF THE SPRING. IT IS NOT NECESSARY TO MEASURE THEIR MASS. Rotate the screws holding the two cylindrical damping magnets so that they are as far from the metal damping rod as possible. 2. Move the device to a position that makes it easy for you to acquire data and then carefully follow the alignment procedure indicated on page 4 of the apparatus manual next to the device (follow both steps one and two). If this is not done correctly, your subsequent measurements will not be reliable. 3. You will now determine the resonant frequency (at least approximately) three different ways: A. With the drive motor off, set the function switch period. Pull the rod down a centimeter or two and release. Note the value of the period. B. The drive amplitude A drive has been set. Do not adjust this setting. (see Figures 9 and 11 in the equipment manual). Turn on the drive motor and set the function switch to amplitude. Vary the drive frequency until the amplitude is a maximum. The resonant frequency can now be read by turning the function switch back to frequency. (If the oscillations are too violent, you can move the magnets in slightly to reduce the oscillation amplitude.) Two observations of note: First, every time you adjust the drive frequency, you need to wait a bit for steady state to be achieved. You are in fact observing the decay 3
5 E. Repeat steps A-C, waiting ten cycles this time. 5. With the help of equation (8), you now have enough information to determine the damping factor. The apparatus has measured the time dependent amplitude to the sine function (actually twice that value since it reads peak to peak). You have two separate readings separated by a known time (n cycles where you know the time it takes to complete each cycle). Note that each of the three trials above should give you a value of the damping factor and ideally, all three should be the same. 6. Measure the distance between the magnets using calipers. Then, move both magnets in by turning each one, one complete revolution. A small piece of tape should be on the magnet to help guide you. Turning one magnet one complete revolution moves it in / cm. Now repeat steps 4A-E. Continue moving the magnets in one complete turn each, making a complete set of measurements each time, until you have data for 5 different magnet separations. This will give you γ as a function of magnet-rod distance, which you should plot in your report. In your report you should discuss qualitatively the physics behind this damping mechanism and why you would expect the plot of γ vs. magnet distance to look as it does. 7. In this final step, you will measure the distance dependence of the damping constant by looking at driven oscillations. The drive amplitude, A drive, has been set to / mm. Do not try to adjust the drive amplitude. Start with the magnets all the way out. Turn the drive motor on and make sure you are still in resonance. After waiting until the oscillations have reached a steady state, record the peak-to-peak amplitude of oscillation of the hanging mass using the amplitude setting on the apparatus. (Note that your raw data is the peak-to-peak amplitude. In your calculations you will need to divide this by 2 to get A res.) Repeat this process for all of the magnet separations used in step 4 above. This amplitude, along with equation (16) above, will allow you to solve for γ. Compare these results graphically with those of steps 4-6 by displaying this data on the same plot. 8. Make sure that the drive motor is turned off and that the power switch on the back is turned off. 5
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# Econ
What is the optimal level of enrollment at the tuition price of \$5000.00
Enroll until the marginal cost of the next student equals \$5000.
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2. 👎 0
3. 👁 39
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Could someone answer this so I understand it. Thanks The projected rate of increase in enrollment at a new college is estemated by dE/dt = 9000(t+1)^-3/2 where E(t) is the projected enrollment in t years. If the enrollment is 5000
asked by Jules on April 26, 2011
3. ### COLLEGE HELP LEASE PLEASE !!!!!!!
Is graduate tuition is tuition for graduate school? I want to become a nurse. Graduate tuition Nursing \$828.00 per credit But here the tuition is \$17,166 The annual total list price cost to go to Mercy was \$31,928 I'm sooo
asked by Sarah on February 22, 2016
4. ### Economics
Suppose a university decides to alter its tuition schedule by separating its students based on how many years of college they have completed. Most university programs require four years to complete. First-year students would get a
asked by Max on October 17, 2017
5. ### calculus help
find the producer's surplus at a price level of p=\$150 for the price-supply equation p=S(x)=100-0.25x is received p=200 the final answer is 5000 is this correct and can you do the steps so i can see if i did i correct You have not
asked by shanere on December 2, 2006
6. ### Economics
Suppose a university decides to alter its tuition schedule by separating its students based on how many years of college they have completed. Most university programs require four years to complete. First-year students would get a
asked by Help me on January 21, 2018
7. ### algebra 1A
help please. a local community college costs \$550 to enroll in morning section of an algebra course. if the variable n stands for the number of students who enroll. then 550n stands for the total amount of money collected for this
asked by Lea on September 17, 2009
8. ### Economics
3. Suppose a firm has a constant marginal cost of \$10. The current price of the product is \$25, and at that price, it is estimated that the price elasticity of demand is -3.0. a. Is the charging the optimal price for the product?
asked by Michelle on September 3, 2012
9. ### statistics
How to (1)Comment on the R-squared. (2). Conduct the test of significance for the regression equation in part 1. Interpret your finding. (3). Interpret the correlation coefficient. (4). Test the significance of the strength of
asked by Azim on August 1, 2008
10. ### Algebra
Jameston High School has 2020 students. Its enrollment has been decreasing at a rate of 125 students per year. Townsville High School has 1075 students. Its enrollment has been increasing at a rate of 80 students per year.
asked by Anonymous on February 22, 2017
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# Dynamical system
(Redirected from Dynamical Systems)
The Lorenz attractor arises in the study of the Lorenz Oscillator, a dynamical system.
In mathematics, a dynamical system is a system in which a function describes the time dependence of a point in a geometrical space. Examples include the mathematical models that describe the swinging of a clock pendulum, the flow of water in a pipe, and the number of fish each springtime in a lake.
At any given time, a dynamical system has a state given by a tuple of real numbers (a vector) that can be represented by a point in an appropriate state space (a geometrical manifold). The evolution rule of the dynamical system is a function that describes what future states follow from the current state. Often the function is deterministic, that is, for a given time interval only one future state follows from the current state.[1][2] However, some systems are stochastic, in that random events also affect the evolution of the state variables.
In physics, a dynamical system is described as a "particle or ensemble of particles whose state varies over time and thus obeys differential equations involving time derivatives."[3] In order to make a prediction about the system’s future behavior, an analytical solution of such equations or their integration over time through computer simulation is realized.
The study of dynamical systems is the focus of dynamical systems theory, which has applications to a wide variety of fields such as mathematics, physics,[4][5] biology,[6] chemistry, engineering,[7] economics, and medicine. Dynamical systems are a fundamental part of chaos theory, logistic map dynamics, bifurcation theory, the self-assembly process, and the edge of chaos concept.
## Overview
The concept of a dynamical system has its origins in Newtonian mechanics. There, as in other natural sciences and engineering disciplines, the evolution rule of dynamical systems is an implicit relation that gives the state of the system for only a short time into the future. (The relation is either a differential equation, difference equation or other time scale.) To determine the state for all future times requires iterating the relation many times—each advancing time a small step. The iteration procedure is referred to as solving the system or integrating the system. If the system can be solved, given an initial point it is possible to determine all its future positions, a collection of points known as a trajectory or orbit.
Before the advent of computers, finding an orbit required sophisticated mathematical techniques and could be accomplished only for a small class of dynamical systems. Numerical methods implemented on electronic computing machines have simplified the task of determining the orbits of a dynamical system.
For simple dynamical systems, knowing the trajectory is often sufficient, but most dynamical systems are too complicated to be understood in terms of individual trajectories. The difficulties arise because:
• The systems studied may only be known approximately—the parameters of the system may not be known precisely or terms may be missing from the equations. The approximations used bring into question the validity or relevance of numerical solutions. To address these questions several notions of stability have been introduced in the study of dynamical systems, such as Lyapunov stability or structural stability. The stability of the dynamical system implies that there is a class of models or initial conditions for which the trajectories would be equivalent. The operation for comparing orbits to establish their equivalence changes with the different notions of stability.
• The type of trajectory may be more important than one particular trajectory. Some trajectories may be periodic, whereas others may wander through many different states of the system. Applications often require enumerating these classes or maintaining the system within one class. Classifying all possible trajectories has led to the qualitative study of dynamical systems, that is, properties that do not change under coordinate changes. Linear dynamical systems and systems that have two numbers describing a state are examples of dynamical systems where the possible classes of orbits are understood.
• The behavior of trajectories as a function of a parameter may be what is needed for an application. As a parameter is varied, the dynamical systems may have bifurcation points where the qualitative behavior of the dynamical system changes. For example, it may go from having only periodic motions to apparently erratic behavior, as in the transition to turbulence of a fluid.
• The trajectories of the system may appear erratic, as if random. In these cases it may be necessary to compute averages using one very long trajectory or many different trajectories. The averages are well defined for ergodic systems and a more detailed understanding has been worked out for hyperbolic systems. Understanding the probabilistic aspects of dynamical systems has helped establish the foundations of statistical mechanics and of chaos.
## History
Many people regard Henri Poincaré as the founder of dynamical systems.[8] Poincaré published two now classical monographs, "New Methods of Celestial Mechanics" (1892–1899) and "Lectures on Celestial Mechanics" (1905–1910). In them, he successfully applied the results of their research to the problem of the motion of three bodies and studied in detail the behavior of solutions (frequency, stability, asymptotic, and so on). These papers included the Poincaré recurrence theorem, which states that certain systems will, after a sufficiently long but finite time, return to a state very close to the initial state.
Aleksandr Lyapunov developed many important approximation methods. His methods, which he developed in 1899, make it possible to define the stability of sets of ordinary differential equations. He created the modern theory of the stability of a dynamic system.
In 1913, George David Birkhoff proved Poincaré's "Last Geometric Theorem", a special case of the three-body problem, a result that made him world-famous. In 1927, he published his Dynamical SystemsBirkhoff's most durable result has been his 1931 discovery of what is now called the ergodic theorem. Combining insights from physics on the ergodic hypothesis with measure theory, this theorem solved, at least in principle, a fundamental problem of statistical mechanics. The ergodic theorem has also had repercussions for dynamics.
Stephen Smale made significant advances as well. His first contribution is the Smale horseshoe that jumpstarted significant research in dynamical systems. He also outlined a research program carried out by many others.
Oleksandr Mykolaiovych Sharkovsky developed Sharkovsky's theorem on the periods of discrete dynamical systems in 1964. One of the implications of the theorem is that if a discrete dynamical system on the real line has a periodic point of period 3, then it must have periodic points of every other period.
## Basic definitions
A dynamical system is a manifold M called the phase (or state) space endowed with a family of smooth evolution functions Φt that for any element of tT, the time, map a point of the phase space back into the phase space. The notion of smoothness changes with applications and the type of manifold. There are several choices for the set T. When T is taken to be the reals, the dynamical system is called a flow; and if T is restricted to the non-negative reals, then the dynamical system is a semi-flow. When T is taken to be the integers, it is a cascade or a map; and the restriction to the non-negative integers is a semi-cascade.
### Examples
The evolution function Φ t is often the solution of a differential equation of motion
${\displaystyle {\dot {x}}=v(x).\,}$
The equation gives the time derivative, represented by the dot, of a trajectory x(t) on the phase space starting at some point x0. The vector field v(x) is a smooth function that at every point of the phase space M provides the velocity vector of the dynamical system at that point. (These vectors are not vectors in the phase space M, but in the tangent space TxM of the point x.) Given a smooth Φ t, an autonomous vector field can be derived from it.
There is no need for higher order derivatives in the equation, nor for time dependence in v(x) because these can be eliminated by considering systems of higher dimensions. Other types of differential equations can be used to define the evolution rule:
${\displaystyle G(x,{\dot {x}})=0\,}$
is an example of an equation that arises from the modeling of mechanical systems with complicated constraints.
The differential equations determining the evolution function Φ t are often ordinary differential equations: in this case the phase space M is a finite dimensional manifold. Many of the concepts in dynamical systems can be extended to infinite-dimensional manifolds—those that are locally Banach spaces—in which case the differential equations are partial differential equations. In the late 20th century the dynamical system perspective to partial differential equations started gaining popularity.
## Linear dynamical systems
Linear dynamical systems can be solved in terms of simple functions and the behavior of all orbits classified. In a linear system the phase space is the N-dimensional Euclidean space, so any point in phase space can be represented by a vector with N numbers. The analysis of linear systems is possible because they satisfy a superposition principle: if u(t) and w(t) satisfy the differential equation for the vector field (but not necessarily the initial condition), then so will u(t) + w(t).
### Flows
For a flow, the vector field Φ(x) is an affine function of the position in the phase space, that is,
${\displaystyle {\dot {x}}=\phi (x)=Ax+b,\,}$
with A a matrix, b a vector of numbers and x the position vector. The solution to this system can be found by using the superposition principle (linearity). The case b ≠ 0 with A = 0 is just a straight line in the direction of b:
${\displaystyle \Phi ^{t}(x_{1})=x_{1}+bt.\,}$
When b is zero and A ≠ 0 the origin is an equilibrium (or singular) point of the flow, that is, if x0 = 0, then the orbit remains there. For other initial conditions, the equation of motion is given by the exponential of a matrix: for an initial point x0,
${\displaystyle \Phi ^{t}(x_{0})=e^{tA}x_{0}.\,}$
When b = 0, the eigenvalues of A determine the structure of the phase space. From the eigenvalues and the eigenvectors of A it is possible to determine if an initial point will converge or diverge to the equilibrium point at the origin.
The distance between two different initial conditions in the case A ≠ 0 will change exponentially in most cases, either converging exponentially fast towards a point, or diverging exponentially fast. Linear systems display sensitive dependence on initial conditions in the case of divergence. For nonlinear systems this is one of the (necessary but not sufficient) conditions for chaotic behavior.
Linear vector fields and a few trajectories.
### Maps
A discrete-time, affine dynamical system has the form of a matrix difference equation:
${\displaystyle x_{n+1}=Ax_{n}+b,\,}$
with A a matrix and b a vector. As in the continuous case, the change of coordinates x → x + (1 − A) –1b removes the term b from the equation. In the new coordinate system, the origin is a fixed point of the map and the solutions are of the linear system A nx0. The solutions for the map are no longer curves, but points that hop in the phase space. The orbits are organized in curves, or fibers, which are collections of points that map into themselves under the action of the map.
As in the continuous case, the eigenvalues and eigenvectors of A determine the structure of phase space. For example, if u1 is an eigenvector of A, with a real eigenvalue smaller than one, then the straight lines given by the points along α u1, with α ∈ R, is an invariant curve of the map. Points in this straight line run into the fixed point.
There are also many other discrete dynamical systems.
## Local dynamics
The qualitative properties of dynamical systems do not change under a smooth change of coordinates (this is sometimes taken as a definition of qualitative): a singular point of the vector field (a point where v(x) = 0) will remain a singular point under smooth transformations; a periodic orbit is a loop in phase space and smooth deformations of the phase space cannot alter it being a loop. It is in the neighborhood of singular points and periodic orbits that the structure of a phase space of a dynamical system can be well understood. In the qualitative study of dynamical systems, the approach is to show that there is a change of coordinates (usually unspecified, but computable) that makes the dynamical system as simple as possible.
### Rectification
A flow in most small patches of the phase space can be made very simple. If y is a point where the vector field v(y) ≠ 0, then there is a change of coordinates for a region around y where the vector field becomes a series of parallel vectors of the same magnitude. This is known as the rectification theorem.
The rectification theorem says that away from singular points the dynamics of a point in a small patch is a straight line. The patch can sometimes be enlarged by stitching several patches together, and when this works out in the whole phase space M the dynamical system is integrable. In most cases the patch cannot be extended to the entire phase space. There may be singular points in the vector field (where v(x) = 0); or the patches may become smaller and smaller as some point is approached. The more subtle reason is a global constraint, where the trajectory starts out in a patch, and after visiting a series of other patches comes back to the original one. If the next time the orbit loops around phase space in a different way, then it is impossible to rectify the vector field in the whole series of patches.
### Near periodic orbits
In general, in the neighborhood of a periodic orbit the rectification theorem cannot be used. Poincaré developed an approach that transforms the analysis near a periodic orbit to the analysis of a map. Pick a point x0 in the orbit γ and consider the points in phase space in that neighborhood that are perpendicular to v(x0). These points are a Poincaré section S(γx0), of the orbit. The flow now defines a map, the Poincaré map F : S → S, for points starting in S and returning to S. Not all these points will take the same amount of time to come back, but the times will be close to the time it takes x0.
The intersection of the periodic orbit with the Poincaré section is a fixed point of the Poincaré map F. By a translation, the point can be assumed to be at x = 0. The Taylor series of the map is F(x) = J · x + O(x2), so a change of coordinates h can only be expected to simplify F to its linear part
${\displaystyle h^{-1}\circ F\circ h(x)=J\cdot x.\,}$
This is known as the conjugation equation. Finding conditions for this equation to hold has been one of the major tasks of research in dynamical systems. Poincaré first approached it assuming all functions to be analytic and in the process discovered the non-resonant condition. If λ1, ..., λν are the eigenvalues of J they will be resonant if one eigenvalue is an integer linear combination of two or more of the others. As terms of the form λi – ∑ (multiples of other eigenvalues) occurs in the denominator of the terms for the function h, the non-resonant condition is also known as the small divisor problem.
### Conjugation results
The results on the existence of a solution to the conjugation equation depend on the eigenvalues of J and the degree of smoothness required from h. As J does not need to have any special symmetries, its eigenvalues will typically be complex numbers. When the eigenvalues of J are not in the unit circle, the dynamics near the fixed point x0 of F is called hyperbolic and when the eigenvalues are on the unit circle and complex, the dynamics is called elliptic.
In the hyperbolic case, the Hartman–Grobman theorem gives the conditions for the existence of a continuous function that maps the neighborhood of the fixed point of the map to the linear map J · x. The hyperbolic case is also structurally stable. Small changes in the vector field will only produce small changes in the Poincaré map and these small changes will reflect in small changes in the position of the eigenvalues of J in the complex plane, implying that the map is still hyperbolic.
The Kolmogorov–Arnold–Moser (KAM) theorem gives the behavior near an elliptic point.
## Bifurcation theory
When the evolution map Φt (or the vector field it is derived from) depends on a parameter μ, the structure of the phase space will also depend on this parameter. Small changes may produce no qualitative changes in the phase space until a special value μ0 is reached. At this point the phase space changes qualitatively and the dynamical system is said to have gone through a bifurcation.
Bifurcation theory considers a structure in phase space (typically a fixed point, a periodic orbit, or an invariant torus) and studies its behavior as a function of the parameter μ. At the bifurcation point the structure may change its stability, split into new structures, or merge with other structures. By using Taylor series approximations of the maps and an understanding of the differences that may be eliminated by a change of coordinates, it is possible to catalog the bifurcations of dynamical systems.
The bifurcations of a hyperbolic fixed point x0 of a system family Fμ can be characterized by the eigenvalues of the first derivative of the system DFμ(x0) computed at the bifurcation point. For a map, the bifurcation will occur when there are eigenvalues of DFμ on the unit circle. For a flow, it will occur when there are eigenvalues on the imaginary axis. For more information, see the main article on Bifurcation theory.
Some bifurcations can lead to very complicated structures in phase space. For example, the Ruelle–Takens scenario describes how a periodic orbit bifurcates into a torus and the torus into a strange attractor. In another example, Feigenbaum period-doubling describes how a stable periodic orbit goes through a series of period-doubling bifurcations.
## Ergodic systems
In many dynamical systems, it is possible to choose the coordinates of the system so that the volume (really a ν-dimensional volume) in phase space is invariant. This happens for mechanical systems derived from Newton's laws as long as the coordinates are the position and the momentum and the volume is measured in units of (position) × (momentum). The flow takes points of a subset A into the points Φ t(A) and invariance of the phase space means that
${\displaystyle \mathrm {vol} (A)=\mathrm {vol} (\Phi ^{t}(A)).\,}$
In the Hamiltonian formalism, given a coordinate it is possible to derive the appropriate (generalized) momentum such that the associated volume is preserved by the flow. The volume is said to be computed by the Liouville measure.
In a Hamiltonian system, not all possible configurations of position and momentum can be reached from an initial condition. Because of energy conservation, only the states with the same energy as the initial condition are accessible. The states with the same energy form an energy shell Ω, a sub-manifold of the phase space. The volume of the energy shell, computed using the Liouville measure, is preserved under evolution.
For systems where the volume is preserved by the flow, Poincaré discovered the recurrence theorem: Assume the phase space has a finite Liouville volume and let F be a phase space volume-preserving map and A a subset of the phase space. Then almost every point of A returns to A infinitely often. The Poincaré recurrence theorem was used by Zermelo to object to Boltzmann's derivation of the increase in entropy in a dynamical system of colliding atoms.
One of the questions raised by Boltzmann's work was the possible equality between time averages and space averages, what he called the ergodic hypothesis. The hypothesis states that the length of time a typical trajectory spends in a region A is vol(A)/vol(Ω).
The ergodic hypothesis turned out not to be the essential property needed for the development of statistical mechanics and a series of other ergodic-like properties were introduced to capture the relevant aspects of physical systems. Koopman approached the study of ergodic systems by the use of functional analysis. An observable a is a function that to each point of the phase space associates a number (say instantaneous pressure, or average height). The value of an observable can be computed at another time by using the evolution function φ t. This introduces an operator U t, the transfer operator,
${\displaystyle (U^{t}a)(x)=a(\Phi ^{-t}(x)).\,}$
By studying the spectral properties of the linear operator U it becomes possible to classify the ergodic properties of Φ t. In using the Koopman approach of considering the action of the flow on an observable function, the finite-dimensional nonlinear problem involving Φ t gets mapped into an infinite-dimensional linear problem involving U.
The Liouville measure restricted to the energy surface Ω is the basis for the averages computed in equilibrium statistical mechanics. An average in time along a trajectory is equivalent to an average in space computed with the Boltzmann factor exp(−βH). This idea has been generalized by Sinai, Bowen, and Ruelle (SRB) to a larger class of dynamical systems that includes dissipative systems. SRB measures replace the Boltzmann factor and they are defined on attractors of chaotic systems.
### Nonlinear dynamical systems and chaos
Simple nonlinear dynamical systems and even piecewise linear systems can exhibit a completely unpredictable behavior, which might seem to be random, despite the fact that they are fundamentally deterministic. This seemingly unpredictable behavior has been called chaos. Hyperbolic systems are precisely defined dynamical systems that exhibit the properties ascribed to chaotic systems. In hyperbolic systems the tangent space perpendicular to a trajectory can be well separated into two parts: one with the points that converge towards the orbit (the stable manifold) and another of the points that diverge from the orbit (the unstable manifold).
This branch of mathematics deals with the long-term qualitative behavior of dynamical systems. Here, the focus is not on finding precise solutions to the equations defining the dynamical system (which is often hopeless), but rather to answer questions like "Will the system settle down to a steady state in the long term, and if so, what are the possible attractors?" or "Does the long-term behavior of the system depend on its initial condition?"
Note that the chaotic behavior of complex systems is not the issue. Meteorology has been known for years to involve complex—even chaotic—behavior. Chaos theory has been so surprising because chaos can be found within almost trivial systems. The logistic map is only a second-degree polynomial; the horseshoe map is piecewise linear.
### Geometrical definition
A dynamical system is the tuple ${\displaystyle \langle {\mathcal {M}},f,{\mathcal {T}}\rangle }$, with ${\displaystyle {\mathcal {M}}}$ a manifold (locally a Banach space or Euclidean space), ${\displaystyle {\mathcal {T}}}$ the domain for time (non-negative reals, the integers, ...) and f an evolution rule t → f t (with ${\displaystyle t\in {\mathcal {T}}}$) such that f t is a diffeomorphism of the manifold to itself. So, f is a mapping of the time-domain ${\displaystyle {\mathcal {T}}}$ into the space of diffeomorphisms of the manifold to itself. In other terms, f(t) is a diffeomorphism, for every time t in the domain ${\displaystyle {\mathcal {T}}}$ .
### Measure theoretical definition
A dynamical system may be defined formally, as a measure-preserving transformation of a sigma-algebra, the quadruplet (X, Σ, μ, τ). Here, X is a set, and Σ is a sigma-algebra on X, so that the pair (X, Σ) is a measurable space. μ is a finite measure on the sigma-algebra, so that the triplet (X, Σ, μ) is a probability space. A map τ: XX is said to be Σ-measurable if and only if, for every σ ∈ Σ, one has ${\displaystyle \tau ^{-1}\sigma \in \Sigma }$. A map τ is said to preserve the measure if and only if, for every σ ∈ Σ, one has ${\displaystyle \mu (\tau ^{-1}\sigma )=\mu (\sigma )}$. Combining the above, a map τ is said to be a measure-preserving transformation of X , if it is a map from X to itself, it is Σ-measurable, and is measure-preserving. The quadruple (X, Σ, μ, τ), for such a τ, is then defined to be a dynamical system.
The map τ embodies the time evolution of the dynamical system. Thus, for discrete dynamical systems the iterates ${\displaystyle \tau ^{n}=\tau \circ \tau \circ \cdots \circ \tau }$ for integer n are studied. For continuous dynamical systems, the map τ is understood to be a finite time evolution map and the construction is more complicated.
## Multidimensional generalization
Dynamical systems are defined over a single independent variable, usually thought of as time. A more general class of systems are defined over multiple independent variables and are therefore called multidimensional systems. Such systems are useful for modeling, for example, image processing.
## References
1. ^ Strogatz, S. H. (2001). Nonlinear dynamics and chaos: with applications to physics, biology and chemistry. Perseus publishing.
2. ^ Katok, A., & Hasselblatt, B. (1995). Introduction to the modern theory of dynamical systems. Cambridge, Cambridge.
3. ^ "Nature". Springer Nature. Retrieved 17 February 2017.
4. ^ Melby, P.; et.al. (2005). "Dynamics of Self-Adjusting Systems With Noise". Chaos 15. doi:10.1063/1.1953147.
5. ^ Gintautas, V.; et.al. (2008). "Resonant forcing of select degrees of freedom of multidimensional chaotic map dynamics". J. Stat. Phys. 130. doi:10.1007/s10955-007-9444-4.
6. ^ Jackson, T.; Radunskaya, A. (2015). Applications of Dynamical Systems in Biology and Medicine. Springer.
7. ^ Dambre, J.; Verstraeten, D.; Schrauwen, B.; Massar, S. (2012). "Information Processing Capacity of Dynamical Systems". Sci.Rep. doi:10.1038/srep00514.
8. ^ Holmes, Philip. "Poincaré, celestial mechanics, dynamical-systems theory and "chaos"." Physics Reports 193.3 (1990): 137-163.
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# 15. Getting Technical - Why only two variables in the block?
#1
My code does not behave incorrectly. I'd just like come clarification.
``````books = ["Charlie and the Chocolate Factory", "War and Peace", "Utopia", "A Brief History of Time", "A Wrinkle in Time"]
# To sort our books in ascending order, in-place
books.sort! { |firstBook, secondBook| firstBook <=> secondBook }``````
There's 5 book titles but in the sort block there's only two variables; firstBook and secondBook. I don't understand why there's no thirdBook, fourthBook and fifthBook. Can someone please explain? Thanks.
#2
Because the sort method only ever works with two objects at a time. One for the left hand, one for the right. The variables are the placeholders for those two objects and are dynamic.
Imagine,
``````C W U Ab Aw
[___]
C W Ab U Aw
[___] [___]
C Ab W Aw U
[___] [___]
Ab C Aw W U
[___] [___]
Ab Aw C U W``````
Seven swaps in all, but only ever two at one time.
#3
The diagram helps make sense of this. Is it correct to say the sort method sorts every possible pair, two at a time, until they are in order?
#4
Integers can be sorted in linear time, but without special knowledge about where-goes-what you'll need to compare.
#5
In a general sense, yes. The actual mechanics may vary depending upon the method. In the example above I started from the left pair, then the next pair.
#7
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Cody
# Benjamin Richards
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| 0.79902
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https://www.jiskha.com/questions/1000892/a-satellite-is-placed-in-orbit-100-km-above-the-surface-of-the-earth-radius-6370-km
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# science help!
A satellite is placed in orbit 100 km above the surface of the earth (radius = 6370 km, mass = 5.98 x 1024 kg).
If the satellite has a mass of 1500 kg, what is the force of gravity acting on it?
1. 👍 0
2. 👎 0
3. 👁 120
1. Now I did g relative to g earth for someone a few minutes ago
http://www.jiskha.com/display.cgi?id=1389396878
so g out there = 9.81*6370^2/6470^2)
= 9.51 m/s^2
so
Fg = 1500 * 9.51
Fg = 14364 Newtons
F = m a
F/m = a = 9.51 = v^2/r
so v^2 = 9.51 r
=
1. 👍 0
2. 👎 0
posted by Damon
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# Statistics - Lab #6
Only available on StudyMode
• Published: May 12, 2012
Text Preview
STATISTICS - Lab #6
Statistical Concepts:
Data Simulation
Discrete Probability Distribution
Confidence Intervals
Calculations for a set of variables
Open the class survey results that were entered into the MINITAB worksheet.
We want to calculate the mean for the 10 rolls of the die for each student in the class. Label the column next to die10 in the Worksheet with the word mean. Pull up Calc > Row Statistics and select the radio-button corresponding to Mean. For Input variables: enter all 10 rows of the die data. Go to the Store result in: and select the mean column. Click OK and the mean for each observation will show up in the Worksheet.
We also want to calculate the median for the 10 rolls of the die. Label the next column in the Worksheet with the word median. Repeat the above steps but select the radio-button that corresponds to Median and in the Store results in: text area, place the median column.
Calculating Descriptive Statistics
Calculate descriptive statistics for the mean and median columns that where created above. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to mean and median. The output will show up in your Session Window. Print this information.
Calculating Confidence Intervals for one Variable
Open the class survey results that were entered into the MINITAB worksheet.
We are interested in calculating a 95% confidence interval for the hours of sleep a student gets. Pull up Stat > Basic Statistics > 1-Sample t and set Samples in columns: to Sleep. Click the OK button and the results will appear in your Session Window.
We are also interested in the same analysis with a 99% confidence interval. Use the same steps except select the Options button and change the Confidence level: to 99.
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# Closing 2 channels in one Go routine causes deadlock?
Hi, I am playing with Tour of Go exercise Equivalent Binary Trees. I have completed the code but I notice something strange:
``````package main
import "golang.org/x/tour/tree"
// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan<- int) {
if t == nil {
return
}
Walk(t.Left, ch)
ch <- t.Value
Walk(t.Right, ch)
}
// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
c1 := make(chan int)
c2 := make(chan int)
go func() {
Walk(t1, c1)
close(c1)
}()
go func() {
Walk(t2, c2)
close(c2)
}()
for i := range c1 {
if i != <-c2 {
return false
}
}
return true
}
func main() {
println(Same(tree.New(10), tree.New(10)))
}
``````
As can be seen, in the Same method, if I use the following equivalent code instead, then deadlock will arise. Can someone help me explain this problem? Thanks
`````` go func() {
Walk(t1, c1)
close(c1)
Walk(t2, c2)
close(c2)
}()``````
2. It enters `Walk(t1, c1)`
3. `Walk(t1, c1)` sends its first value into `c1`
4. Your main goroutine receives the first value from `c1` in the for ā¦ range loop
5. Your main goroutine attempts a receive from `c2` to compare it to the value from `c1`.
6. Your āwalker goroutineā hasnāt returned from `Walk(t1, c1)` yet, so `Walk(t2, c2)` hasnāt started. Nothing is getting sent into `c2`, so receiving from it will deadlock.
1 Like
main and go1 all blocked.
main need recv ch1 and ch2, but not recv ch2. go1 not send value to c2 in block state, so main revc c2 in block state,all goroutine to blocked state.
• go1 send c1-10 succes
• main c1 recv c1-10 succes
• main wait c2 recv
• go1 send c2-20 block
https://play.golang.org/p/Ji-ysAFdISI
g1 need c1 send 10ci value 10-100, then c2 send 10ci.
main need recv c1 then recv c2 Repeat 10ci.
but g1 send ā20ā main is blocked state wait revc c2,so g1 and main in blocked, if c1 and c2 cap is 10,then not blocked.
Thanks. I somehow get it. But I have one more question: Why āsend 20ā must wait for range to pop it out? I mean, from my point of view, it is in a different thread, so why canāt we just put to channel independently?
Feels like there is a mutex or something c1.
Thanks
Thank you, I somehow get it.
it is in a different goroutine, Chan is essentially a blocking queue with a lock.
chan accepts a value that is blocked if there is no data, and sends a value that is also blocked if chan is full.
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# New to Qlik Sense
Discussion board where members can get started with Qlik Sense.
Not applicable
## Month- Number
Hi All,
How can i call a number based on the month i.e. if month is January ,number should be 1, similarly if the month is Feburary ,number should be 2 and so on.
Thanks
Anurag Gupta
1 Solution
Accepted Solutions
Contributor
## Re: Month- Number
Hi Anurag,
Here is a sample master calendar script if you would like to incorporate this in your application:
LET vDateMin = Num(MakeDate(2000,1,1));
LET vDateMax = Num(MakeDate(2001,12,31));
LET vDateToday = Num(Today());
TempCalendar:
\$(vDateMin) + RowNo() - 1 AS DateNumber,
Date(\$(vDateMin) + RowNo() - 1) AS TempDate
AUTOGENERATE 1
WHILE \$(vDateMin)+IterNo()-1<= \$(vDateMax);
Calendar:
Date(TempDate) AS [Calendar Date],
// Standard Date Objects
Day(TempDate) AS CalendarDayOfMonth,
WeekDay(TempDate) AS CalendarDayName,
Week(TempDate) AS CalendarWeekOfYear,
Month(TempDate) AS Month,
num(Month(TempDate)) AS MonthNum,
Num(Num(Year(TempDate))&if(Num(Month(TempDate))<10,0)&Num(Month(TempDate))) as CalendarYYYYMM,
'Q' & Ceil(Month(TempDate)/3) AS CalendarQuarter,
Year(TempDate) AS Year,
// Calendar Date Names
WeekName(TempDate) as CalendarWeekNumberAndYear,
MonthName(TempDate) as CalendarMonthAndYear,
QuarterName(TempDate) as CalendarQuarterMonthsAndYear,
// Start Dates
DayStart(TempDate) as CalendarDayStart,
WeekStart(TempDate) as CalendarWeekStart,
Date(MonthStart(TempDate)) as CalendarMonthStart,
QuarterStart(TempDate) as CalendarQuarterStart,
YearStart(TempDate) as CalendarYearStart,
// End Dates
DayEnd(TempDate) as CalendarDayEnd,
WeekEnd(TempDate) as CalendarWeekEnd,
MonthEnd(TempDate) as CalendarMonthEnd,
QuarterEnd(TempDate) as CalendarQuarterEnd,
YearEnd(TempDate) as CalendarYearEnd,
// Combo Date Examples
'Q' & Ceil(Month(TempDate)/3) & '/' & Year(TempDate) AS CalendarQuarterAndYear,
Year(TempDate) & '/' & 'Q' & Ceil(Month(TempDate)/3) AS CalendarYearAndQuarter,
'Wed ' & DayStart(WeekStart(TempDate) + 3) as CalendarWednesdays
RESIDENT TempCalendar ORDER BY TempDate ASC;
DROP TABLE TempCalendar;
6 Replies
Valued Contributor III
## Re: Month- Number
Assuming you have month field in your data model.
easiest way is to add a inline table and do a left join to your calendar. or try using "Master Calendar script" which you can find it in community.
Month_Sort:
Month, MonthSort
January, 1
February, 2
March, 3
April, 4
May, 5
June, 6
July, 7
August, 8
September, 9
October, 10
November, 11
December, 12
];
Contributor
## Re: Month- Number
Hi Anurag,
Here is a sample master calendar script if you would like to incorporate this in your application:
LET vDateMin = Num(MakeDate(2000,1,1));
LET vDateMax = Num(MakeDate(2001,12,31));
LET vDateToday = Num(Today());
TempCalendar:
\$(vDateMin) + RowNo() - 1 AS DateNumber,
Date(\$(vDateMin) + RowNo() - 1) AS TempDate
AUTOGENERATE 1
WHILE \$(vDateMin)+IterNo()-1<= \$(vDateMax);
Calendar:
Date(TempDate) AS [Calendar Date],
// Standard Date Objects
Day(TempDate) AS CalendarDayOfMonth,
WeekDay(TempDate) AS CalendarDayName,
Week(TempDate) AS CalendarWeekOfYear,
Month(TempDate) AS Month,
num(Month(TempDate)) AS MonthNum,
Num(Num(Year(TempDate))&if(Num(Month(TempDate))<10,0)&Num(Month(TempDate))) as CalendarYYYYMM,
'Q' & Ceil(Month(TempDate)/3) AS CalendarQuarter,
Year(TempDate) AS Year,
// Calendar Date Names
WeekName(TempDate) as CalendarWeekNumberAndYear,
MonthName(TempDate) as CalendarMonthAndYear,
QuarterName(TempDate) as CalendarQuarterMonthsAndYear,
// Start Dates
DayStart(TempDate) as CalendarDayStart,
WeekStart(TempDate) as CalendarWeekStart,
Date(MonthStart(TempDate)) as CalendarMonthStart,
QuarterStart(TempDate) as CalendarQuarterStart,
YearStart(TempDate) as CalendarYearStart,
// End Dates
DayEnd(TempDate) as CalendarDayEnd,
WeekEnd(TempDate) as CalendarWeekEnd,
MonthEnd(TempDate) as CalendarMonthEnd,
QuarterEnd(TempDate) as CalendarQuarterEnd,
YearEnd(TempDate) as CalendarYearEnd,
// Combo Date Examples
'Q' & Ceil(Month(TempDate)/3) & '/' & Year(TempDate) AS CalendarQuarterAndYear,
Year(TempDate) & '/' & 'Q' & Ceil(Month(TempDate)/3) AS CalendarYearAndQuarter,
'Wed ' & DayStart(WeekStart(TempDate) + 3) as CalendarWednesdays
RESIDENT TempCalendar ORDER BY TempDate ASC;
DROP TABLE TempCalendar;
Not applicable
## Re: Month- Number
Try this: date(num(Ex_date),'DD/MM/YYYY')
MVP
## Re: Month- Number
Hi Anurag,
Try like this
*,
Match(MonthName, 'January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December')
FROM Data;
Note: Replace MonthName with your actual field name.
Regards,
Jagan.
New Contributor III
## Re: Month- Number
Hi Jaghan,
I hope all the functionalty of sense will worj in qlikview I am right?Thanks in advance
Shantanu
Sent from RediffmailNG on Android
From: "Jagan Mohan"qcwebmaster@qlikview.com
Sent:Fri, 22 Aug 2014 09:40:55 +0530
To: Shantanu Sardar shantanusardar@rediffmail.com
Subject: Re: - Month- Number
Qlik Community Month- Number reply from Jagan Mohan in New to Qlik Sense - View the full discussion Hi Anurag, Try like this LOAD*,Match(MonthName, 'January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December')FROM Data; Note: Replace MonthName with your actual field name. Regards,Jagan. Reply to this message by replying to this email, or go to the message on Qlik Community Start a new discussion in New to Qlik Sense by email or at Qlik Community Following Jagan Mohan in these streams: Email Watches © 1993-2014 QlikTech International AB Copyright &
Contributor
## Re: Month- Number
Hi Sean,
I need a small clarification,
I added the below command to the code since I want to get the Year ans Month Data field as below,
Year(TempDate) & '/' & '' & Month(TempDate) AS CalendarYearAndMonth,
When I generate a line graph for actual sales which got data from July 2017 to July 2018, it h=gave me the below output. Ideally, the 2018/July data value should be at the end of the chart after 2018/June.
Could you pls advise me, how to accomplish the said issue?
Thanks.
Andy
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https://tagvault.org/blog/how-many-months-21-weeks/
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# How Many Months Is 21 Weeks? (Conversion)
Are you wondering how many months 21 weeks translates to? Whether you’re an expectant mother curious about your pregnancy timeline or a project manager planning project milestones, understanding the conversion from weeks to months can be incredibly helpful. In this article, we will delve into the conversion process, providing accurate calculations and insights to assist you in scheduling and organizing your time effectively.
### Key Takeaways:
• Converting 21 weeks to months equals approximately 4.83 months.
• The conversion is essential for tracking pregnancy milestones and organizing project timelines.
• Consider slight variations in the length of each month for precise conversions.
• Online conversion tools provide convenience and accuracy when converting weeks to months.
• Understanding the conversion helps manage timelines and scheduling efficiently.
## Weeks to Months Conversion Chart
Converting 21 weeks to months can be made easier with the help of a conversion chart. Take a look at the table below to find the corresponding number of months for different numbers of weeks:
Weeks Months
4 weeks approximately 1 month
8 weeks approximately 2 months
12 weeks approximately 3 months
16 weeks approximately 4 months
20 weeks approximately 4 and 1/2 months
21 weeks approximately 4 and 3/4 months
24 weeks approximately 5 and 1/2 months
Based on the conversion chart, 21 weeks is approximately 4 and 3/4 (four and three quarters) months. This conversion can be helpful for visualizing the timeline of a pregnancy or the duration of a project.
Note: The values in the table are approximate and can vary depending on the length of each month. It’s always advisable to use these conversions as a guide rather than an exact measurement.
## Calculating Weeks to Months
To determine the equivalent number of months for 21 weeks, a simple conversion formula can be used. It’s important to note that one month is approximately equal to 4.35 weeks. By dividing 21 weeks by 4.35, we arrive at an estimated 4.83 months. This calculation provides a rough estimate and can be helpful in understanding the duration in terms of months.
Understanding the conversion from weeks to months is valuable for a variety of scenarios, including pregnancy timelines and project management. By utilizing this conversion formula, individuals can effectively plan and organize schedules, whether it be tracking the progress of a pregnancy or setting realistic timelines for projects.
“Dividing 21 weeks by 4.35 yields an estimated equivalent of 4.83 months.”
A visual representation of the conversion process can be seen in the table below:
Weeks Months
21 4.83
By referring to this conversion chart, individuals can easily convert different numbers of weeks to months and gain a clearer understanding of their desired timelines.
## The Importance of Consideration in Weeks to Months Conversion
When converting weeks to months, it is crucial to take certain considerations into account to accurately determine the equivalent timeframe. While it may be tempting to assume that a month is simply four weeks, this is not always the case. With the exception of February in a non-leap year, every other month consists of one to three additional days, making them slightly longer than four weeks.
“Accurately converting weeks to months requires accounting for the slight variation in the length of each month.”
For example, let’s consider 24 weeks. If we were to assume that a month is precisely four weeks, we might conclude that 24 weeks is equivalent to six months. However, when we take into account the longer months, we find that it is actually closer to five and a half months. This discrepancy highlights the importance of considering the specific length of each month when performing conversions.
To accurately convert weeks to months, it is essential to factor in the variation in month lengths. By doing so, you can obtain more precise calculations and ensure that the converted duration aligns with the actual timeframe. This consideration is especially relevant when assessing pregnancy timelines or project schedules, where accurate time tracking is crucial for planning and organizing.
### Why Consideration is Key
Considerations in weeks to months conversion are essential for several reasons:
• Accuracy: Taking into account the slight variation in month lengths allows for more precise conversions, ensuring accuracy in determining the equivalent duration.
• Precision in Planning: When estimating project timelines or the duration of a pregnancy, accurate week to month conversion plays a vital role in setting realistic expectations and managing deadlines effectively.
• Clear Communication: By accurately converting weeks to months, you can effectively communicate timeframes to stakeholders, whether it be to healthcare professionals, clients, or team members.
By considering the intricacies of week to month conversion, you can ensure that your calculations accurately reflect the intended timeframes, leading to better planning, communication, and overall project management.
## Other Units of Time Conversion
In addition to converting weeks to months, it can also be useful to convert weeks to other units of time. Understanding the duration in different units allows for better comparison and comprehension. Take a look at the time conversion chart below to explore how 21 weeks can be expressed in various other units of time:
Units of Time Conversion
Weeks 21
Days 147
Hours 3,528
Minutes 211,680
By converting 21 weeks, we discover that it is equivalent to 147 days, 3,528 hours, or 211,680 minutes. These conversions provide valuable insight into the duration of time measured in different units. Whether you’re managing a project or tracking a milestone, understanding how weeks translate into other units can assist in planning and decision-making processes.
### Exploring Time Conversion
“The ability to convert weeks to different units of time empowers us to better comprehend duration and make effective comparisons.”
Converting weeks to different units of time helps in visualizing and comparing durations on various scales. It enables us to appreciate the timeline from a broader perspective and integrate it into our daily lives. Whether you’re tracking a personal goal or managing a project, having access to different units of time can enhance your overall planning and execution. So, let’s dive deeper into the significance of understanding these conversions.
## How Many Months Is 21 Weeks Exactly?
When converting 21 weeks to months exactly, it is approximately equal to 4.83 months. This calculation takes into account the slight variation in the length of each month and provides a more precise estimation. Understanding the exact conversion can help in planning and organizing schedules, whether it is for pregnancy or project timelines.
Knowing the exact conversion of 21 weeks to months is crucial for accurate planning. It allows pregnant women to have a clearer understanding of their pregnancy timeline and anticipate important milestones. For project management, the precise conversion helps in setting realistic deadlines and monitoring progress effectively.
By considering the slight variation in the length of each month, the conversion provides a more accurate representation of the duration. This is particularly important when communicating timelines with healthcare providers, clients, or team members. Whether it’s counting down the months until childbirth or tracking progress on a project, the exact conversion of 21 weeks to months ensures better planning and organization.
### Calculating Weeks to Months with Accuracy
To calculate the exact number of months for 21 weeks, you can use the following formula:
Number of Months = Total Weeks / 4.35
By dividing 21 weeks by 4.35, you’ll get the precise number of months. In this case, it is approximately 4.83 months.
Take note that this calculation accounts for the slightly longer duration of months, ensuring a more accurate conversion. With this knowledge, you can confidently interpret the duration of 21 weeks in months, whether you’re preparing for the arrival of a baby or managing project timelines.
Weeks Months (Approximate)
21 4.83
## Significance of Weeks to Months Conversion during Pregnancy
For expectant mothers, understanding the conversion from weeks to months is of utmost importance in comprehending the duration of their pregnancy. This conversion enables them to track their progress, anticipate significant milestones, and effectively communicate their due dates and pregnancy timeline to healthcare providers, family, and friends.
By grasping the relationship between weeks and months, expectant mothers are better equipped to manage their pregnancy journey with confidence and clarity. It provides them with the knowledge they need to make informed decisions, plan ahead, and embrace the beautiful changes that come along the way.
### Understanding the Pregnancy Duration
During pregnancy, a woman goes through a remarkable journey that lasts for several months. However, the concept of pregnancy duration can sometimes be confusing, particularly when referring to it in terms of weeks or months. This is where the conversion from weeks to months becomes invaluable.
Each trimester of pregnancy spans a specific number of weeks, and understanding how these weeks translate into months allows expectant mothers to align themselves with the pregnancy timeline. It provides them with a sense of structure, enabling them to anticipate crucial milestones such as prenatal appointments, ultrasound scans, and the exciting moment when they feel their baby’s first movements.
### Effective Communication and Planning
Accurate weeks to months conversion enhances communication between expectant mothers and their healthcare providers, facilitating a better understanding of the pregnancy timeline and due dates. It enables healthcare professionals to provide appropriate care and support based on the progress of the pregnancy.
“Understanding the weeks to months conversion during pregnancy is fundamental for healthcare providers to assess the development of the baby and ensure the well-being of both mother and child.” – Dr. Emily Thompson, OB-GYN Specialist
Additionally, expectant mothers can share their pregnancy timeline more effectively with their loved ones. Family and friends can grasp the significance of each week and month, offering support and celebrating the milestones along the way.
### The Journey Made Manageable
Pregnancy is an exciting yet transformative journey. By comprehending the conversion from weeks to months, expectant mothers gain a greater sense of control and understanding. It allows them to confidently navigate the pregnancy duration, plan for the future, and cherish the miraculous process of bringing new life into the world.
Weeks Months (Approximation)
1-4 weeks 1 month
5-8 weeks 2 months
9-13 weeks 3 months
14-17 weeks 4 months
18-21 weeks 5 months
## Weeks to Months Conversion for Project Timelines
When it comes to project management and scheduling, converting weeks to months is a valuable tool for setting realistic timelines, assigning tasks, and monitoring progress. Whether you’re working on personal projects or professional endeavors, understanding the conversion from weeks to months enables project managers and teams to have a better overview of the project’s duration and plan accordingly, ensuring efficient completion.
By converting weeks to months, you can:
• Effectively plan and allocate resources
• Break down the project into manageable phases
• Assign tasks and set milestones
• Monitor progress and make adjustments
Creating a well-defined project timeline is essential for project success. Converting weeks to months provides a clear and concise way to communicate the project’s duration and milestones to stakeholders, team members, and clients. It helps in setting realistic expectations and ensuring that everyone is on the same page.
### Example:
Let’s consider a hypothetical project that involves building a website. The estimated timeline for the project is 21 weeks. Converting this timeline to months gives us a better understanding of the project’s duration.
Weeks Months
21 weeks Approximately 4.83 months
With the conversion, we can see that the project is expected to take approximately 4.83 months to complete. This information allows project managers to structure the timeline, allocate resources, delegate tasks, and track progress effectively. It provides clarity and enables a more realistic approach to project management.
Converting weeks to months is a valuable technique that helps project managers and teams plan and execute projects with greater efficiency and accuracy. By utilizing this conversion, project timelines can be better organized, leading to successful project completion.
## Advantage of Weeks to Months Conversion Calculator
To simplify the process of converting weeks to months, online conversion tools and calculators are available. These convenient tools provide quick and accurate results, saving you time and effort. By entering the desired number of weeks into the calculator, you can instantly obtain the corresponding number of months.
Utilizing a weeks to months conversion calculator offers several advantages. It eliminates the need for manual calculations, reducing the risk of errors. Whether you’re calculating a single conversion or dealing with multiple conversions, this tool streamlines the process and ensures accuracy.
The convenience of an online conversion tool cannot be overstated. With just a few clicks, you can obtain the precise conversion you need. Whether you’re a pregnant woman tracking your pregnancy timeline or a project manager scheduling tasks, having a reliable conversion calculator at your fingertips makes the process seamless and efficient.
By utilizing an online weeks to months conversion calculator, you can easily determine the equivalent number of months for any given number of weeks. This eliminates the need for manual calculations, providing you with the answers you need in an instant.
Experience the convenience and efficiency of online conversion tools and calculators. Try using a weeks to months conversion calculator today and simplify your conversion process.
## Conclusion
In conclusion, converting 21 weeks to months provides an approximate duration of 4.83 months. This conversion is particularly important for pregnant women who want to understand the timeline of their pregnancy. By considering the slight variation in the length of each month, accurate conversions can be achieved, allowing expectant mothers to track their progress and communicate due dates effectively.
Additionally, weeks to months conversion is also valuable in project management and scheduling. By converting weeks to months, project managers can establish realistic timelines and monitor progress more efficiently. Understanding the relationship between weeks and months helps teams plan and allocate tasks effectively, leading to successful project completion.
Conducting weeks to months conversion is made easier with online conversion tools. These tools offer convenience and accuracy, enabling quick calculations and eliminating the need for manual conversions. Whether it’s for pregnancy or project management, online conversion tools ensure precise results, saving time and effort.
In summary, understanding the conversion from weeks to months is beneficial for both pregnant women and project managers. By converting 21 weeks to approximately 4.83 months, individuals can gain clarity and better manage timelines and scheduling. Whether using conversion charts or online tools, accurate conversions help in tracking progress and achieving desired outcomes.
## FAQ
### How many months is 21 weeks?
21 weeks is approximately 4 and 3/4 (four and three quarters) months.
### How can I convert weeks to months?
You can use a conversion chart or a simple conversion formula to convert weeks to months. One month is approximately equal to 4.35 weeks.
### Why is it important to consider the slight variation in the length of each month when converting weeks to months?
A month is not exactly four weeks, except for February in a non-leap year. The slight variation in the length of each month should be taken into account for accurate conversions.
### What are some other units of time that can be used to convert weeks?
21 weeks is equal to 147 days, 3,528 hours, or 211,680 minutes.
### What is the exact conversion of 21 weeks to months?
The exact conversion of 21 weeks to months is approximately 4.83 months, considering the slight variation in the length of each month.
### Why is weeks to months conversion significant during pregnancy?
Weeks to months conversion helps expectant mothers understand the duration of their pregnancy and track their progress. It enables better communication of due dates and pregnancy timeline to healthcare providers and others.
### How can weeks to months conversion be helpful for project timelines?
Weeks to months conversion assists in setting realistic timelines, assigning tasks, and monitoring progress in project management. It provides a better overview of the project’s duration and facilitates efficient completion.
### Are there any convenient tools available for weeks to months conversion?
Yes, there are online conversion calculators that can quickly and accurately convert weeks to months. These tools save time and effort, especially when dealing with multiple conversions or when accuracy is crucial.
### Is there a conclusion or summary of weeks to months conversion?
While there is no specific conclusion, understanding the conversion from weeks to months provides clarity and helps in managing timelines and scheduling effectively.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 09, 2010 8:05 pm
Joined: Thu Jul 23, 2009 5:06 pm
Location: Berkeley, CA, USA
Julian wrote:
Stefan Schwalbe wrote:
Gelatinbrain: the 3.10.3 has a piece-type with 192 members. You need great endurance to solve that. I used: F, B,U4,B', R,F',R',F, B, F',R,F,R', U'4, R,F',R',F, B', F',R,F,R', F', to solve the 192 pieces one by one.
That is a clever algo. I couldn't find a pure algo to cycle smaller pieces on 3 different faces but I have found a pure commutated commutated commutator where 2 pieces are on the same face: (((L3, F), R'), U2). Setups are going to be tricky.
Interesting, my algo is
F2, R,B,U4, L',R,/*setup moves*/
L,[B,R,B',R,B,R'2,B'],L',[B,R2,B',R',B,R',B'],/*commutator*/
R',L,U'4,B',R',F'2,/*reverse setup moves*/
The core is only the second line. The first and third lines are setup moves. Among the three locations, two are on the back face in a single quadrant, which are very compact. The third is on the front face, far away from the other two. Every time I can use any combination of F, U, R to setup. I found setups pretty easy. But I can solve only one piece at a time.
Top
Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Fri Sep 10, 2010 12:58 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Julian wrote:
That is a clever algo. I couldn't find a pure algo to cycle smaller pieces on 3 different faces but I have found a pure commutated commutated commutator where 2 pieces are on the same face: (((L3, F), R'), U2). Setups are going to be tricky.
My algo (F, [B,U4,B'], [[R,F',R',F], B, [F',R,F,R']], U'4, [R,F',R',F], B', [F',R,F,R'], F',) is a commutated commutated commutator too :
-I started with (R,F',R',F): that cycles 3 center-pieces on the down face (DRB->DFR->DLF->) and the 3 center-pieces (FUR->URF->RFU->).
-Now I needed a good modificaton-move and than I would do the reverse of the first 4 moves.
-B seemed to be a good modification move. So I tried: (R,F',R',F),(B),(F',R,F,R'),(B').
-It gave a good result: The center pieces were all back. It cycled 6 corner-pieces on the down-face and 1 corner-piece on the right-face.
-That 1 corner piece must now be changed, without changing the other 6 corner-pieces. That does (B,U,B').
-The ready move sequence with (A and B swaped)looks now: [B,U,B',] [(R,F',R',F,) B, (F',R,F,R',), B'] [B, U', B'] [B, (R,F',R',F,) B', (F',R,F,R',)] (26 moves).
-Accidental some B moves in the middle combine and make it 4 moves shorter. It's not so easy, to see the commutator structure in it.
-I finally changed [B,U,B'] into [B,U4,B'] and enclosed all with F [...] F' to make it more convenient.
-For the setup moves I can use U, F, L without changing position1 or position2. Especially L at the end of the setup is usefull.
-My move-count was 4671. For a better move-count I could solve two pieces at once.
How would my move-sequence look in your kind of writing? Maybe: F,[([B,U4,B'],((R,F'),B))],F'
schuma wrote:
F2, R,B,U4, L',R,/*setup moves*/
L,[B,R,B',R,B,R'2,B'],L',[B,R2,B',R',B,R',B'],/*commutator*/
R',L,U'4,B',R',F'2,/*reverse setup moves*/
...
Among the three locations, two are on the back face in a single quadrant, which are very compact. The third is on the front face, far away from the other two. Every time I can use any combination of F, U, R to setup. I found setups pretty easy. But I can solve only one piece at a time.
The way of it is interesting. 28 moves is okay. My first approach was 50 moves. I forgot it. I agree: setups are easy. Maybe I can think into your solving way. I would like to know, how this alg. was build.
bmenrigh wrote:
I want to hear about your 2.1.1 solve strategy and solve order. Are your routines more efficient than mine posted here:
Julian wrote:
Ok, I will write that into separate posts as soon as possible. Julian, can you in return tell me your Pentultimate-approach wich got the fascinating move-count of 123. I suppose you're using a layer-method.
That's it for now, Stefan.
Top
Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Fri Sep 10, 2010 3:15 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
bmenrigh wrote:
I want to hear about your 2.1.1 solve strategy and solve order. Are your routines more efficient than mine posted here: http://www.twistypuzzles.com/forum/view ... 8&p=211229?
That is: center diamonds ((1,1), 3), center triangles ((1,1),1) and small two-color triangles ((1,1),3), where the (1,1) can sometimes be shortened to just 3 moves? Those 60 tiny 2-color triangles take me forever. Sometimes I have to use 8-10 setup moves to get them in the right position and orientation.
Here it is:
"Gelatinbrain 2.1.2" - Solution (single cut, face turning icosahedron)
Attachment:
File comment: corner-letter-notation
GB2.1.2.jpg [ 27.96 KiB | Viewed 6555 times ]
2.1.2 has all piece-types from 2.1.1 as well as 2.1.3. So you can solve all three puzzles with a 2.1.2 strategy.
The piece-types are:
-1. edge (30)
-2. face-corner (60) "kite"
-3. face (20)
-4. corner (12) with 5 orientations
-5. corner-edge (60) "butterfly"
The solving order is like given. Solve all pieces of one type, and than continue with the next type.
-prepare a color scheme, and a letter-notation scheme.
Algs:
-1. edge (30):
commute any pair of level-1-relation face-moves.
-use move-sequences like: [ABC, FBA, ABC', FBA',]. Do that from all angles.
That cycles 3 edges and scrambles a lot of other pieces wich does'nt matter.
-2. face-corner (60) "kite":
commute any pair of level-2-relation face-moves.
-use move-sequences like: [ABC, CDG, ABC', CDG',]. Do that from all angles.
Additional move-sequences to get the last pieces solved:
-a ring-cycler: [KEG', LJK, KEG, LJK, KEG', LJK, KEG,]
-two locations on one face: [[FBA', DCB, FBA], AIF, [FBA', DCB', FBA], AIF',]. Do that from all angles.
That cycles 3 face-corners, does'nt change any edge, and scrambles 3 other piece-types, which does'nt matter.
-3. face (20):
-move-sequence: [[LGD, KEG', LGD'], BHD, [LGD, KEG, LGD'], BHD',], use this move-sequence with appropriate setup-moves.
(You can vary BHD to BHD' or KEG' to KEG on need.
This move-sequence cycles 3 face-pieces, and scrambles no other piece.
-4. corner (12):
-move-sequence: [[LJK', HJL, LJK], KEG', [LJK', HJL', LJK], KEG], use this move-sequence with appropriate setup moves.
Solve the orientation too!
-for twisting a target use [CGE, EAC,] or [CGE, EAC,]2 or [EAC',CGE',] or [EAC',CGE',]2 or similar moves.
This cycles 3 corner-pieces and scrambles some corner-edges, wich does'nt matter.
-5. corner-edge (60) "butterfly":
-move-sequence: [[GLK', LGD, [LDH,] LGD', GLK,] KIE, [GLK', LGD, [LDH',] LGD', GLK,] KIE',], use this move sequence or bacic variations (for instance: KIE' instead of KIE), with appropriate setup moves.
This cycles 3 corner-edges and doesn't change other pieces.
A hint for finding setup moves:
-Find all moves, that change the target location.
-Remember all locations that you got.
-All that locations are 1 move afar from the target location.
-Draw a scheme with a "1" on all that locations.
-You can repeaet that and find all locations that are 2 moves afar and so on.
-Use only moves, that do not change the other 2 locations of the 3-cycle.
That is my solution for the 2.1.2.
I hope I have nothing forgotten. If not, please tell me.
Stefan.
P.S.
bmenrigh wrote:
Sometimes I have to use 8-10 setup moves to get them in the right position and orientation.
Note that the corner-edges have only one orientation!
Last edited by Stefan Schwalbe on Sun Sep 12, 2010 6:35 am, edited 1 time in total.
Top
Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 6:18 am
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Julian wrote:
Here is my approach for the 4.3.1:
I used a one letter per face notation. A, B, C, D, a, b, c, d.
I used a color scheme:
Attachment:
File comment: color and face-letter scheme
GB4.3.1a-.gif [ 6.55 KiB | Viewed 6531 times ]
The piece-types are:
1. face-corner (4 groups of 6 pieces)
2. face (8) (odd permutation parity possible)
3. corner-edge (24)
4. corner (6) (odd permutation parity possible)
Solve piece-type by piece-type in the given order.
1. face-corners (4 groups of 6 pieces)
Attachment:
File comment: groups
GB4.3.1b-.gif [ 1.42 KiB | Viewed 6524 times ]
The picture shows the 4 groups of face-corners.
Start with red. Look for the 3 red face-corners. Remember the 3 groups they are in. Look into the scheme for a face with that 3 groups. Solve them into that face.
Rotate the whole puzzle to get the red face on A (face: left view, up).
Look for the yellow pieces in the same way and solve them to the right face, adjacent to the red face. Now get red and yellow where they belong on your color scheme, in rotating the whole puzzle.
Moves like (CB,AB)2 cycle 3 face-corners of one group, don't scramble any other face-corners, scramble some other pieces wich doesn't matter.
1b. face-corners (4 groups of 6 pieces) parity fix
You can solve all of them, or end up with two odd groups.
In that case, find an edge with that two groups, and twist that edge. Now the groups are all even, and you can solve the face-corners.
2a. faces (8) parity-fix
Check the parity of the face-pieces without doing one move. If it is odd you can get it even with the following:
Do a single move on a "1"-"4"-edge. Groups 1 and 4 are now odd and the faces are even. Take a third group "3" into. Do a move on a "3"-"1"-edge, and a move on a "3"-"4"-edge. All groups are now even. Than you must solve some face-corners again.
That parity fix would look so: [Bd,AB,BC],(Bd,BC)2,(BC,AB)2,(Bd,AB)2.
Enhancement: You may solve the parity of the face-corners and the faces together, and save some moves.
2b. faces (8)
move-sequence: (DA,AB,BC,AB)2 cycles 3 face-pieces and doesn't change the allready solved face-corner-pieces.
So the face-corners and the faces are now all solved.
3: corner-edge (24)
use the following move-sequences with appropriate setup moves.
forward: ((AB,DC,AB)Ac,)2
backward:(Ac,(AB,DC,AB))2
left->right-reflection, forward: ((DA,BC,DA)Ac,)2
left->right-reflection, backward: (Ac,(DA,BC,DA))2
Solve one piece at once.
Set-up:
Attachment:
File comment: setup
GB4.3.1c.gif [ 1.84 KiB | Viewed 6531 times ]
The picture shows the depth of the locations. 0 are the target locations. x are the loc.1 and loc.2 for the 3-cycles. location-1,2,3 - pieces you can get into the targets without moving any x piece. Not so with 4 and 5- depth locations.
depth 5 pieces you must first get into depth 4 locations with one move.
To get depth 4 pieces into the targets you must first save one of the x - pieces with a setup, wich looks so:
For DA -> Bd : (moves:) dc, [AB,Bd], dc
For AB -> bD : (moves:) cb, [DA,bD], cb
The depth-4 pieces on the right view are set into the targets in a similar way.
4. corners (6)
The corners must now be even in permutation-parity, because of the solved face-corners and faces.
For the corners I use that (moves): [(AB, DC)2, (BC,DA)2, (DC, AB)2, (DA,BC)2]
That does a clean 3-cycle of corner-pieces: (ABCD->aCBd->BAcd->)
I solve the permutation and orientation at once.
Twist target BAcd similar.
finished,
Stefan.
P.S.
Julian can you tell me wich part of it has helped you enhancing your own solve for I don't believe, I had to tell you all that.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 11:45 am
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
4.5.2 solve:
Julian wrote:
my notation + color scheme
Attachment:
File comment: My notation and the color scheme
4.5.2a.gif [ 5.1 KiB | Viewed 6523 times ]
piece-types:
1. edge-wing (24)
2. left corner-face (24)
3. right corner-face (24)
4. corner (24)
solve order:
1. pair up edge wings
The piece in Ca we bring with one move to its other part. This move is Db. We need to setup the other part to Ac. We use only face-moves for that. If the other part is on cA, because we can not get it with face-moves to Ac, we do one edge-move (Ac). This breaks the edge-pairs Db and Bd. So bevore we do Ac we must setup unsolved edge pieces on both Db and Bd. At the end we may have no unsolved pieces to setup at Db and Bd. This is, when only two unsolved edge-pairs are left, and they are wrong twisted.
Here is a fix for that:
Bring one edge-pair to AB and the other to DA. Now do:
gelatinbrain-moves-notation: [DR, DBL, DB, UBR, URF, UBR', URF', DF, URF, UBR, URF', UBR', DF, DB, DBL', DR,]
Now all edges are paired.
2. solve all edge-pairs (12)
Use only face moves for that, because the edge- moves break edge-pairs
You may need to twist edges:
do (gelatinbrain-moves-notation): UFL', ULB, BL, ULB', UFL, URF', UFL', ULB, BL, ULB', UFL, URF,
This twists both the DA and AB edge-pairs.
You may need a swap for fixing the odd parity of edge-pairs.
do (gelatinbrain-moves-notation):
[DR, DBL, DB, [DF, [UBR, URF, UBR', URF'], DF, [URF, UBR, URF', UBR']], DB, DBL', DR],
[DF, DBL', DL, [[UFL', URF', UFL, URF], DR, [URF', UFL', URF, UFL], DR], DL, DBL, DF],
This swaps DA with AB.
3. pair up the corner-faces (24)
We bring a left corner-face to a right corner-face of the same color.
Do (gelatinbrain-moves-notation): [FL, UBR, FL, UBR'], DBL', [UBR, FL, UBR', FL], DBL,
backward : DBL', [FL, UBR, FL, UBR'], DBL, [UBR, FL, UBR', FL],
This cycles three left corner-faces. Use the location on the b-face for the target. This can be accesed in several ways without changing loc. 1 and loc. 2.
The left and right corner-faces should now be paired.
4. solve the corner-face-pairs (24):
Use (gelatinbrain-moves-notation): [UBR, UFL', UBR', UFL], BL, [UFL', UBR, UFL, UBR'], BL,
and the reverse of it. That cycles 3 corner-face-pairs.
5. corner-quarters (24)
Use (gelatinbrain-moves-notation):
FL, BRU, FL, BRU',
UL, BDR, UL, BDR',
BRU, FL, BRU', FL,
BDR, UL, BDR', UL,
and the backward of it:
UL, BDR, UL, BDR',
FL, BRU, FL, BRU',
BDR, UL, BDR', UL,
BRU, FL, BRU', FL,
That cycles 3 corner-quaters. Best-target is cA.
(With cA I mean the left corner-quater-piece on the left view. The right piece I would call Ac)
That's it
Stefan.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 1:01 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Stefan Schwalbe wrote:
4.5.2 solve:
...snip...
5. corner-quarters (24)
Use (gelatinbrain-moves-notation):
FL, BRU, FL, BRU',
UL, BDR, UL, BDR',
BRU, FL, BRU', FL,
BDR, UL, BDR', UL,
I haven't played with 4.5.x enough to comment on your solution outline other than to say your routines are very complicated! Interestingly, the above routine seems to trigger a bug in the Applet with the Undo button. If I paste in what you've listed above, I do get a corner piece 3-cycle but when I press the Undo button moves aren't undone correctly and even removing all 16 moves leaves the 3-cycle in place. Did you experience trouble with the Undo on this puzzle too?
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 1:57 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
bmenrigh wrote:
Stefan Schwalbe wrote:
4.5.2 solve:
...snip...
5. corner-quarters (24)
Use (gelatinbrain-moves-notation):
FL, BRU, FL, BRU',
UL, BDR, UL, BDR',
BRU, FL, BRU', FL,
BDR, UL, BDR', UL,
I haven't played with 4.5.x enough to comment on your solution outline other than to say your routines are very complicated! Interestingly, the above routine seems to trigger a bug in the Applet with the Undo button. If I paste in what you've listed above, I do get a corner piece 3-cycle but when I press the Undo button moves aren't undone correctly and even removing all 16 moves leaves the 3-cycle in place. Did you experience trouble with the Undo on this puzzle too?
Yes, I know that bug. That happens only when you use the Input-Text-Area on some puzzles. You can't undo the "Input"-Moves.
Do you know how to link to a special post?
Stefan.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 3:49 pm
Joined: Thu May 31, 2007 7:13 pm
Location: Bruxelles, Belgium
Stefan Schwalbe wrote:
Yes, I know that bug. That happens only when you use the Input-Text-Area on some puzzles. You can't undo the "Input"
Oops, je m'en occupe...
I updated the standalone version
You can now send your scores from the standalone program.
When you quit the program the current session is automatically saved and resumed when you restart.
So you don't have to keep your computer on during the night(more ecologic ).
It's only for Windows. Sorry for mac and linux users(not that I particularly like Windows ).
P.S.
Stefan, is there any signification in your stage name?
So far, I know that Agamemnon is a Greek mithological figure and Harmagedon a biblical term for the end of the world. But the combination of the two...???
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 4:55 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
I finally solved the Circle FTO but I'm not particularly happy with my understanding of the puzzle and move count. It seems my attitude towards solving puzzles has changed a lot in my several month solving hiatus because previously I would have been happy that I solved it and that I didn't get the worst move count.
I recognized that the middle circle pieces show the orientation of the centers. I also observed that circle pieces under the corners were paired. Unfortunately I don't understand what those paired pieces represent. I was able to come up with a 3-cycle for them but I couldn't relate them to pieces from any other puzzle I'm familiar with. With 3-cycles to move the paired pieces and the virtual centers I solved all of the circle pieces in about 150 moves. From there, I solved like a super-FTO which took me about 400 moves to get to here:
Attachment:
circle_fto_center_twist.png [ 15.67 KiB | Viewed 6649 times ]
I spent about an hour trying dozens of things before I figured out how to put an excess twist in the center without also putting an excess 3-cycling in the edges. My resulting fix that that case is about 100 moves.
As far as I can tell, the only way to sub-500 this puzzle is if you have a better understanding of the circle pieces. What am I missing? Is there some trick to approaching this puzzle?
_________________
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 5:33 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Stefan Schwalbe wrote:
Do you know how to link to a special post?
Yeah but phpBB doesn't make it easy.
If you mouse-over the post's quote button or the report ! button you'll get the post number from the link:
Code:
http://twistypuzzles.com/forum/posting.php?mode=quote&f=8&p=228893
The to link to that post you'd do:
http://twistypuzzles.com/forum/viewtopic.php?f=8&p=228893#p228893
_________________
Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 5:49 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
3.10.3
schuma wrote:
L,[B,R,B',R,B,R'2,B'],L',[B,R2,B',R',B,R',B'],/*commutator*/
I found setups pretty easy. But I can solve only one piece at a time.
I really like that algo -- "sliding Sune" for a pure cycle of just (7,1). Hmmm, I'm going to have to go with more difficult setups but solving 2-3 pieces every time. My setup method will be "choose homeboy and homeboy's neighbor" -- then get the neighbor into a circle, spin him around until he is 45 degrees twisted away from homeboy, then slot him into position. Then setup the other piece. I was going to set myself a target of 2500 with my (10,1) cycle but now I'll set myself a target of <=2000 moves, when I get around to solving it.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 6:18 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan Schwalbe wrote:
P.S. Julian can you tell me wich part of it has helped you enhancing your own solve for I don't believe, I had to tell you all that.
Here is my original 4.3.1 solution. Because I solved the corners first, my later algos were much less efficient. I also finished with a (7,1) pure algo where I see now that there is a (5,1). I knew about the orbitals and parity issues, but it has been very helpful to see your solving order and algos. I have just used your method (I solved most of the face-corners intuitively before cycling) to do a solve in 296 moves: face-corners in 40, centers in 24, edges in 162, and corners in 70. Thanks!
Your 2.1.2 solution is very interesting too. Some of us found a (9,1) pure algo for the corner-edges, then later we were pleased to find (6,1), but you found (5,1)! I was starting with the edges but then I did the corners before the kites (not sure why now), which was costing extra moves. I'll read your 4.5.2 solution tomorrow, and I'll also reply about my Pentultimate method.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sun Sep 12, 2010 9:21 pm
Joined: Thu Jul 23, 2009 5:06 pm
Location: Berkeley, CA, USA
Stefan Schwalbe wrote:
schuma wrote:
F2, R,B,U4, L',R,/*setup moves*/
L,[B,R,B',R,B,R'2,B'],L',[B,R2,B',R',B,R',B'],/*commutator*/
R',L,U'4,B',R',F'2,/*reverse setup moves*/
I would like to know, how this alg. was build.
In order to construct a clean 3-cycle, I tried some routines. I tried this: [R, U, R', U, R, U'2, R'], which is a 3x3x3 last layer algorithm. Of course, the definition for R is turning by 45deg rather than 90deg like in 3x3x3. But I found this routine and a D-turn form a great commutator. But the 3 positions in the 3-cycle turn out to be too close to each other, so that the setup is a little bit hard. Therefore I intentionally added some setup moves to separate one location away from the other two, so that I have more freedom to setup. I also re-oriented the cube to let me easier to think.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 1:07 am
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
bmenrigh wrote:
Stefan Schwalbe wrote:
Do you know how to link to a special post?
Yeah but phpBB doesn't make it easy.
If you mouse-over the post's quote button or the report ! button you'll get the post number from the link:
Code:
http://twistypuzzles.com/forum/posting.php?mode=quote&f=8&p=228893
The to link to that post you'd do:
http://twistypuzzles.com/forum/viewtopic.php?f=8&p=228893#p228893
There's a quicker/better way. See that white/orange page next to the text "Posted: Sun Sep 12, 2010 3:18 pm?" It has a link to that post.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 2:56 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Seeing Drewseph's amazing Choposaurus made me go back to Gelatinbrain 1.2.9, to try to find a shorter pure algorithm for the triangles. Here is a symmetrical (8,1) algo:
DCG,IJL, /* Setup moves */
HIL,ACD,HIL',ACD', /* Central (1,1) sequence using cutting planes as far apart as possible while still intersecting */
IJL',DCG', /* Undo moves */
KGL', /*Replace the isolated swapped piece with another */
DCG,IJL,ACD,HIL,ACD',HIL',IJL',DCG',
KGL
--------------------------------------------------------------
Then I noticed that the 6th and 7th move of the above algo are exactly opposite each other, so I fused them into a single move with a puzzle rotation, and here is the resulting (7,1) algo:
CGD,IJL,HIL,ACD,HIL',IJL,DEA',
EJF',
DEA,IJL',HIL,ACD',HIL',IJL',CGD',
GLK
This algorithm also cycles three edge pieces of the dual puzzle 2.1.5 pure. Now I wonder if it is possible to solve these puzzles in under 1000 moves?
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 3:27 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Julian wrote:
Stefan Schwalbe wrote:
P.S. Julian can you tell me wich part of it has helped you enhancing your own solve for I don't believe, I had to tell you all that.
Here is my original 4.3.1 solution. Because I solved the corners first, my later algos were much less efficient. I also finished with a (7,1) pure algo where I see now that there is a (5,1). I knew about the orbitals and parity issues, but it has been very helpful to see your solving order and algos. I have just used your method (I solved most of the face-corners intuitively before cycling) to do a solve in 296 moves: face-corners in 40, centers in 24, edges in 162, and corners in 70. Thanks!
...
Thank you! 296 is better than my move-count. I like that.
Thank you for showing me your own 4.3.1 solution.
gelatinbrain wrote:
Oops, je m'en occupe...
I am sorry. Your programm is still the greatest for me.
gelatinbrain wrote:
Stefan, is there any signification in your stage name?
So far, I know that Agamemnon is a Greek mithological figure and Harmagedon a biblical term for the end of the world. But the combination of the two...???
Don't worry about that. I found it joky, when it came into my mind.
bmenrigh wrote:
As far as I can tell, the only way to sub-500 this puzzle is if you have a better understanding of the circle pieces. What am I missing? Is there some trick to approaching this puzzle?
As I remember, I have solved all the middle circle pieces first.
schuma wrote:
In order to construct a clean 3-cycle, I tried some routines. I tried this: [R, U, R', U, R, U'2, R'], which is a 3x3x3 last layer algorithm. Of course, the definition for R is turning by 45deg rather than 90deg like in 3x3x3. But I found this routine and a D-turn form a great commutator. But the 3 positions in the 3-cycle turn out to be too close to each other, so that the setup is a little bit hard. Therefore I intentionally added some setup moves to separate one location away from the other two, so that I have more freedom to setup. I also re-oriented the cube to let me easier to think.
Thank you for the reply. I was amazed that [R, U, R', U, R, U'2, R'], which I also know from the cube works on the 3.10.3 .
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 4:30 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan, thanks for sharing your clever 4.5.2 method. Almost fully reducing to a Face-Turning Octahedron, with just broken corners left at the end. (Apart from the first few edge pieces, I cycled everything: edges then corners then center pieces.)
I like the way you find efficient non-commutator algos. For example, the best I can do solving the corners pure with a commutator is (8,1):
RF,DLF', /* Setup moves */
UBR',FL,UBR,FL, /* Face and edge back and forth to cycle groups of 4 pieces */
DLF,RF, /* Undo moves */
URF, /* Replace swapped corner piece with another */
RF,DLF',FL,UBR',FL,UBR,DLF,RF,
URF'
Which is of course 2 moves longer than your "4x4":
FL, BRU, FL, BRU',
UL, BDR, UL, BDR',
BRU, FL, BRU', FL,
BDR, UL, BDR', UL
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 5:09 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan Schwalbe wrote:
Julian, can you in return tell me your Pentultimate-approach wich got the fascinating move-count of 123. I suppose you're using a layer-method.
Here is my Pentultimate solution from around a year ago, which I have improved slightly since then. You are right, it is essentially a layer-by-layer method. Nowadays I try to completely finish one half before solving the last 6 centers. Of corners 6-10, sometimes the first one or two can be solved in 4 moves (not counting a rotation of the unsolved half to set up the corner), otherwise in 6 or 7 moves, if the corners are not in the adjacent layer. If the corners are in the ring adjacent to where they need to go, they are much harder to solve. To solve a corner adjacently takes me 9 moves and can only be done by pushing a neighboring corner out of position. (I have recently found that by adding a setup move to an algo Brandon found, a corner can be solved adjacently from 2 out of 3 orientations in 10 moves.)
My record of 123 moves was very lucky. At the time I said:
"Crazy luck solving the Pentultimate (1.1.7)! I solved one half of the puzzle in 70 moves, which is nothing special. But, when I checked the other half:
* The other 6 centers were already solved
* 2 of the last 10 corners were already solved
* 3 of the other 8 corners were sitting in a perfect 3-cycle (including orientation)
I calculate the odds of all these happening at 1 in 1,944.
So I had 5 corners left to solve after 88 moves; 3 corners left after 106 moves; and finished at 123 moves."
Brandon has written a program that has proved the last 6 centers can be solved from any permutation with no effect on the solved half, in a maximum of 10 moves, including a twist to roll them into place relative to the solved half. He has also found pure corner cycles of 10 and 11 moves. This is so efficient that there is only point in solving corners 6-10 one by one as long as they are in convenient positions, then we switch to solving the last 6 centers and cycling the remaining corners. I think we could soon be looking at an average solution length of around 140 moves!
Note to Brandon: Sorry, I will try to get that list of Pentultimate corner cycle possibilities to you later this week.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 5:24 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan -- copying your approach but making a slight change, I have found that we can cycle three corners of 4.3.1 pure in 14 moves:
FL,BL,FL,
UF,DF,UF,DF,
FL,BL,FL,
DF,UF,DF,UF
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 6:05 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Julian wrote:
Note to Brandon: Sorry, I will try to get that list of Pentultimate corner cycle possibilities to you later this week.
No problem at all! I'm really enjoying the conversation about algorithms between you and Stefan. I wish I had more to contribute; the puzzles Stefan is best at I've hardly touched. New thoughts and approaches are great to read about though.
I understand how long it takes to classify the algs I've already posted -- it took me a long time just to pop them in and spot their pattern -- looking at the orientations of the corners is going to take twice that time.
When I realized I wasn't using all my CPUs efficiently with my 11-move searching I started up my 12-move searching at a lower priority to keep the CPUs saturated. The 12-move search is already 1/16th completed and 56 algs have been found. I think it will be done in only two months. Several new patterns have come up already. I'm only going to post the new patterns until you've had a chance to check the orientations on the patterns I've already posted that don't cover all of the possible corner orientations. That way I don't flood the forum with dozens of routines that are already covered by the 10 or 11 moves routines.
As for moves, I think the first center, the 5 corners around it, and then the other 5 centers around that can be placed in 45 moves or so. After that the remaining centers can be done in 9 moves for a total of 54 moves. Then, supposing 2.5 corners are cycled into place in an average of 15 moves for each cycle, were looking at another 90 moves. That comes to 144 moves total A lucky solve like your 123 moves should drop into the sub-100 range. Once we've cataloged all the routines I'd love to see what you can do with them.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 13, 2010 11:14 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
This post is meant to catalog all of the useful pure Pentultimate/1.1.7 corner 3-cycles of length 12.
I have already posted all length 10 cycles and all length 11 cycles. Rather than post all length 12 routines I will only post the patterns that aren't covered by the 10 or 11 move routines. Once the corner orientations of those routines have been cataloged I will update this post with routines that fill in gaps.
These sequences will take my computer 2-3 months to finish finding. I will keep updating this post even if it is pushed several pages back in this thread.
Skinny Tall Triangle:
Attachment:
skinny_tall_triangle.png [ 56.84 KiB | Viewed 6535 times ]
[A1, B1, C3, B4, F1, C3, B4, E2, F1, B4, E2, A4]
[A1, B1, D3, B4, C1, D3, E4, A2, E1, C4, D2, A4]
[A1, B2, F4, E1, B2, A4, B3, A1, E4, B3, F1, A4]
[A1, B2, F4, E1, B2, E4, B3, F1, C4, A3, C1, A4]
Attachment:
left_askew.png [ 55.58 KiB | Viewed 6535 times ]
[A1, B1, D2, A4, D3, A1, B4, D3, C1, D2, C4, A4]
[A1, B1, D3, B4, F2, B1, C4, F2, B1, D3, C4, A4]
[A1, B1, C3, B4, C2, F1, A4, C2, A1, C3, F4, A4]
[A1, B1, C2, F4, E3, B1, F4, E3, F1, C2, F4, A4]
[A1, B2, A4, B3, A1, E4, B3, F1, B2, F4, E1, A4]
[A1, B2, D4, B3, A1, E4, B3, E1, B2, A4, D1, A4]
Counter-Clockwise Ninja Star:
Attachment:
ccw_ninja_star.png [ 55.96 KiB | Viewed 6535 times ]
[A1, B1, D2, B4, D3, C1, E4, A3, E1, D2, C4, A4]
[A1, B1, C2, B4, E3, B1, F4, E3, B1, C2, F4, A4]
[A1, B1, C3, D4, A2, D1, B4, C2, F1, C3, F4, A4]
[A1, B1, D3, C4, F2, B1, C4, F2, C1, D3, C4, A4]
[A1, B2, F4, B3, F1, C4, A3, F1, A2, F4, C1, A4]
[A1, B2, F4, D3, F1, E4, D3, F1, B2, E4, F1, A4]
StarTrek Communicator:
Attachment:
startrek_communicator.png [ 56.68 KiB | Viewed 6276 times ]
[A1, B3, F4, B2, E1, A4, B2, A1, B3, E4, F1, A4]
[A1, B3, D4, B2, D1, C4, B2, F1, A3, F4, C1, A4]
Length 13 move sequences are out of reach. It would take a distribute effort and many months to find them all. If somebody is willing to donate "nice -n 19" CPU time on a cluster, it would need about 920,000 CPU hours. On 256 nodes each with 8 cores that's less than 3 weeks.
_________________
Prior to using my real name I posted under the account named bmenrigh.
Last edited by Brandon Enright on Sat Sep 18, 2010 8:46 pm, edited 1 time in total.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 12:12 pm
Joined: Tue Feb 16, 2010 12:15 pm
Location: Sandnes, Norway
It seems I'm not able to open 3.1.24. I only get this when I try:
Attachment:
3.1.24.jpg [ 26.61 KiB | Viewed 6522 times ]
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 2:12 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Katten wrote:
It seems I'm not able to open 3.1.24. I only get this when I try: [...snip...]
3.1.21-3.1.24 aren't linked from the main page. You can only get to them via the "File" menu inside of the applet.
I did notice the scores page is truncated though so maybe Gelatinbrain ran out of disk space?
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 4:05 pm
Joined: Thu May 31, 2007 7:13 pm
Location: Bruxelles, Belgium
bmenrigh wrote:
Katten wrote:
It seems I'm not able to open 3.1.24. I only get this when I try: [...snip...]
3.1.21-3.1.24 aren't linked from the main page. You can only get to them via the "File" menu inside of the applet.
I did notice the scores page is truncated though so maybe Gelatinbrain ran out of disk space?
Looks OK. Maybe you just accessed at very critical moment...
3.1.24 was once removed, because I noticed it is functionally equivalent to simple 2x2x2. All who solved this should know it. But as it looks so popular, I restored it.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 4:37 pm
Joined: Tue Feb 16, 2010 12:15 pm
Location: Sandnes, Norway
Yes, I noticed when I just solved it. Not hard at all, though I'm not feeling too secure with circle puzzles just yet.
gelatinbrain wrote:
I really think that's a fascinating puzzle, thanks!
Last edited by Katja on Wed Sep 15, 2010 1:36 am, edited 1 time in total.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 6:14 pm
Joined: Thu Jul 23, 2009 5:06 pm
Location: Berkeley, CA, USA
gelatinbrain wrote:
(3.3.12) is an interesting puzzle. I have to confess that I don't deserve the record of 2:54 and 102 moves. It seems that it was caused by a bug.
Here is the reason. The first time I scramble it, the second layers were not moved at all. I didn't realize the second layers could be moved. It was exactly a helicopter cube for me, and I solved it like a helicopter cube. I was just wondering what was the difference between this thing and a helicopter cube.
The second time I played it, it works properly. I would like to solve it again. Gelatinbrain, can you remove my record with 2:54 and 102 moves? It shouldn't be that easy.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 6:25 pm
Joined: Fri Jul 16, 2010 8:25 pm
Location: Israel
I got the same problam, but to me it didnt get fixed..
It still acts like a helicopter cube.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 14, 2010 6:35 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Sharon Avidor wrote:
I got the same problam, but to me it didnt get fixed..
It still acts like a helicopter cube.
Java probably cached your applet. You should clear your Java "Temporary Internet Files" and try again.
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Wed Sep 15, 2010 1:40 am
Joined: Tue Feb 16, 2010 12:15 pm
Location: Sandnes, Norway
Sharon Avidor wrote:
The second time I played it, it works properly. I would like to solve it again. Gelatinbrain, can you remove my record with 2:54 and 102 moves? It shouldn't be that easy
This happened to me as well, so it's only fair that mine be removed too.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Wed Sep 15, 2010 1:11 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Hello,
I solved 1.1.4 and 1.1.6. For the face-corners I used an alg., I used too, when I first solved 1.1.5.
It is (14 moves):
F, C', F', C,
A'2, E'2&2, A2,
C', F, C, F',
A'2, E2&2, A2,
The 1.1.6 corners I solved with my old Pentultimate corner-cycler (18 moves):
F', C, F, C',
J', G, L, G', J,
C, F', C', F,
J', G, L', G', J,
This was the best alg. I found for the Pentultimate-corners. I tried first one of Brandons 10 move-algorithms, which worked on the Pentultimate, but on the 1.1.6 it scrambled a lot of pieces.
I will continue with the 1.1.x-puzzles.
I have found a mechanic version of the 1.1.5. It's called Starminx.
gelatinbrain wrote:
gelatinbrain: I really would like to know, how you can achieve this - creating so many puzzles. This must be a lot of hard work. Especially how you construct your models. I expect, you're using splitting algorithms.
Julian wrote:
Stefan -- copying your approach but making a slight change, I have found that we can cycle three corners of 4.3.1 pure in 14 moves:
FL,BL,FL,
UF,DF,UF,DF,
FL,BL,FL,
DF,UF,DF,UF
Yeah, thats good, thank you.
Julian wrote:
Here is my Pentultimate solution ...
Thank you, for explaining your Pentultimate solution.
I tried the old one, but I couldn't do some moves. For instance "f". I would like to enhance my own Pentultimate-solution - I will try again.
Julian wrote:
Stefan, thanks for sharing your clever 4.5.2 method. Almost fully reducing to a Face-Turning Octahedron, with just broken corners left at the end. (Apart from the first few edge pieces, I cycled everything: edges then corners then center pieces.)
...
I took pleasure in it. Ask for more!!
GuiltyBystander wrote:
bmenrigh wrote:
Stefan Schwalbe wrote:
Do you know how to link to a special post?
Yeah but phpBB doesn't make it easy.
If you mouse-over the post's quote button or the report ! button you'll get the post number from the link:
Code:
http://twistypuzzles.com/forum/posting.php?mode=quote&f=8&p=228893
The to link to that post you'd do:
http://twistypuzzles.com/forum/viewtopic.php?f=8&p=228893#p228893
There's a quicker/better way. See that white/orange page next to the text "Posted: Sun Sep 12, 2010 3:18 pm?" It has a link to that post.
Thank you! Bye.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Wed Sep 15, 2010 4:52 pm
Joined: Thu May 31, 2007 7:13 pm
Location: Bruxelles, Belgium
Stefan Schwalbe wrote:
gelatinbrain: I really would like to know, how you can achieve this - creating so many puzzles. This must be a lot of hard work. Especially how you construct your models. I expect, you're using splitting algorithms.
I think I found a good algorithm to cover a fairly wide range of twisty puzzles. So adding a new puzzle is just a question of a few lines of data.
So, 3.2.8 Circle Dino, 3.4.22 Circle Super X
Someday I will make my source public...
BTW, how is it called the puzzle in your new avatar?
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Wed Sep 15, 2010 6:28 pm
Joined: Thu Jul 23, 2009 5:06 pm
Location: Berkeley, CA, USA
Gelatinbrain, I sincerely appreciate your hard work. You are updating puzzles so frequently. Only because of you, I have the opportunity to play the various complicated puzzles created by Drewseph, TomZ etc, for FREE. And I am pretty sure that a great portion of puzzles they will create in the near future has been covered by your applets. Thank you very much.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Wed Sep 15, 2010 6:33 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
schuma wrote:
Gelatinbrain, I sincerely appreciate your hard work. You are updating puzzles so frequently. Only because of you, I have the opportunity to play the various complicated puzzles created by Drewseph, TomZ etc, for FREE. And I am pretty sure that a great portion of puzzles they will create in the near future has been covered by your applets. Thank you very much.
Well said. I love seeing new puzzles being made that I have either already solved or have the opportunity to play with on your applet. Twisting your applet feels as real to me as owning the puzzle. Your hard work to bring us new puzzles is very appreciated!
_________________
Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 16, 2010 2:23 am
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan Schwalbe wrote:
I tried first one of Brandons 10 move-algorithms, which worked on the Pentultimate, but on the 1.1.6 it scrambled a lot of pieces.
The (6,1) algo in my Pentultimate solution works with 1.1.6 too, so you can leave the corners of 1.1.6 to the end if you like. But there is also a (3,1) algo to cycle the center-corner pieces of 1.1.4, 1.1.5 and 1.1.6 pure. Hint: The "3" are face moves and the "1" is a slice (shift-click) move.
Stefan Schwalbe wrote:
Thank you, for explaining your Pentultimate solution. I tried the old one, but I couldn't do some moves. For instance "f".
The moves in lower case are puzzle rotations. For example, with "f" you rotate the whole puzzle clockwise 72 degrees around the F face. I did this so I could leave the main part of the algorithm the same for every orientation case.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 16, 2010 2:34 am
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
bmenrigh wrote:
schuma wrote:
Gelatinbrain, I sincerely appreciate your hard work. You are updating puzzles so frequently. Only because of you, I have the opportunity to play the various complicated puzzles created by Drewseph, TomZ etc, for FREE. And I am pretty sure that a great portion of puzzles they will create in the near future has been covered by your applets. Thank you very much.
Well said. I love seeing new puzzles being made that I have either already solved or have the opportunity to play with on your applet. Twisting your applet feels as real to me as owning the puzzle. Your hard work to bring us new puzzles is very appreciated!
I'll third that -- your prog has brought great pleasure to my life.
Gelatinbrain -- I know you have already added an amazing variety of puzzles, but I have remembered a category of puzzles that is missing! I was going to say something a long time ago, then I forgot, and TomZ's recent physical Helicopter Skewb has reminded me.
Could you please add a category for vertex + edge turning cubes? Some combinations of Dino Cube/Master Skewb/Skewb + 3.3.1/3.3.3/3.3.5/3.3.7 would probably make good puzzles.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 16, 2010 3:04 am
Joined: Tue Sep 08, 2009 8:41 am
Location: The Blue Mountains, Australia
Julian wrote:
But there is also a (3,1) algo to cycle the center-corner pieces of 1.1.4, 1.1.5 and 1.1.6 pure. Hint: The "3" are face moves and the "1" is a slice (shift-click) move.
The (3,1) i use is different, the three is face, slice, face and the 1 is a face move. I posted it back on page 36, as i said there i always thought you used the same one hahaha apparently not.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 16, 2010 3:08 am
Joined: Tue Feb 16, 2010 12:15 pm
Location: Sandnes, Norway
Julian wrote:
Could you please add a category for vertex + edge turning cubes? Some combinations of Dino Cube/Master Skewb/Skewb + 3.3.1/3.3.3/3.3.5/3.3.7 would probably make good puzzles.
I agree. Then we won't have to use 3.2.3 to simulate a Dino Cube solve anymore (like I did last night ) etc. I believe these puzzles would be greatly appreciated in your vast collection of virtual puzzles.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 16, 2010 2:39 pm
Joined: Sat Jul 11, 2009 1:09 pm
Location: My House
I haven't been on GB for a while and now I'd like to go back on it, but I'm just getting a blank area where the puzzle should be; does anyone know what's going on?
Alex
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Fri Sep 17, 2010 4:28 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Elwyn wrote:
Julian wrote:
But there is also a (3,1) algo to cycle the center-corner pieces of 1.1.4, 1.1.5 and 1.1.6 pure. Hint: The "3" are face moves and the "1" is a slice (shift-click) move.
The (3,1) i use is different, the three is face, slice, face and the 1 is a face move. I posted it back on page 36, as i said there i always thought you used the same one hahaha apparently not.
Thank you Elwyn for the (3,1). I didn't even think, that this is possible. . Julian, can you tell me the (6,1), for the Pentultimate-corners or at least another hint?
gelatinbrain wrote:
Stefan Schwalbe wrote:
gelatinbrain: I really would like to know, how you can achieve this - creating so many puzzles. This must be a lot of hard work. Especially how you construct your models. I expect, you're using splitting algorithms.
I think I found a good algorithm to cover a fairly wide range of twisty puzzles. So adding a new puzzle is just a question of a few lines of data.
So, 3.2.8 Circle Dino, 3.4.22 Circle Super X
Someday I will make my source public...
BTW, how is it called the puzzle in your new avatar?
Thank you. I expected something like this. Great thing. Don't stop.
bmenrigh wrote:
schuma wrote:
Gelatinbrain, I sincerely appreciate your hard work. You are updating puzzles so frequently. Only because of you, I have the opportunity to play the various complicated puzzles created by Drewseph, TomZ etc, for FREE. And I am pretty sure that a great portion of puzzles they will create in the near future has been covered by your applets. Thank you very much.
Well said. I love seeing new puzzles being made that I have either already solved or have the opportunity to play with on your applet. Twisting your applet feels as real to me as owning the puzzle. Your hard work to bring us new puzzles is very appreciated!
Your words really moved me. I feel the same.
gelatinbrain wrote:
BTW, how is it called the puzzle in your new avatar?
It is your words, that made it a puzzle - funny Idea. For me it was just a visualization of the commutator principle.
P.S. I recently solved the 1.1.8 and the 1.1.9 "gigaminx". For the 1.1.8 corner-faces i used a (3,1) and for the edge-halfs i used a (7,1). I guess there is a better alg. With my 1.1.9 solve i will not appear in the rankings, for it was too bad.
Bye, Stefan.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Fri Sep 17, 2010 4:48 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Stefan Schwalbe wrote:
P.S. I recently solved the 1.1.8 and the 1.1.9 "gigaminx". For the 1.1.8 corner-faces i used a (3,1) and for the edge-halfs i used a (7,1). I guess there is a better alg.
It seems like most of us had a hard time with 1.1.8 so good job.
For the "edge-halves" I too used a (7,1) when I first solved it. I just revisited it though and this (5,1) popped out at me as easy and kinda obvious. I wonder why I didn't spot it before
[F'2,C',B'2,C,F2], [E&2], [F'2,C',B2,C,F2], [E'&2]
Note that by modifying the commutated E&2 slice you can vary what gets cycled rather easily.
Elwyn came up with a brilliant reduction to Megaminx trick for this puzzle. He wrote about it here. See the posts after that one too for more discussion about 1.1.8.
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sat Sep 18, 2010 5:36 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
1.2.9 (physical version = Drewseph's Chopasaurus)
New solution outline
Stage 1 - Solve 8 corners using intuitive sequences: 5 corners around a face, then 3 adjacent corners in the next layer/ring. This stage should only take around 30 moves. With the last 3 of these corners, the idea is to get the corner in a place to spin it around to the correct orientation without disturbing other pieces, then set it up, and finally put it in place with a simple 2-gen sequence.
Stage 2 - Solve the remaining corners, permuting and orienting at the same time, mostly a pair at a time, using (4,1) commutators.
Stage 3 - Two thirds of the time you will end up with a single twisted corner. If you do, twist it to its correct orientation and then re-solve the other 9 corners in 3 cycles.
Stage 4 - Solve the centers using (6,1) commutators.
Stage 5 - Solve the triangles using (7,1) pure commutators.
Algos for stages 4 and 5 can be found by experimenting with (1,1) commutators using two corners that are far apart. The stage 4 algo has a single setup and undo move either side of a (1,1), and the stage 5 algo has two setup and undo moves, but a move overlap (consecutive moves on the same axis) allows the sequence to be reduced from 8 to 7 moves.
Solved today in 752 moves using this method.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Sat Sep 18, 2010 5:46 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
Stefan Schwalbe wrote:
Julian, can you tell me the (6,1), for the Pentultimate-corners or at least another hint?
The 6 moves are the sequence called P (or Pm for the mirror image) in my solution. Another way of looking at it is to take three faces that all touch each other and call them F, U, and R, like a Rubik's cube. F U R U' R' F' isolates a single swapped corner in half of the Pentultimate.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 20, 2010 2:50 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
I was laying in bed, unable to sleep thinking about the center-twist "parity" I got on the Circle FTO and how all the fixes I tried didn't come out pure. I got to thinking about what is required for a orientation parity twist to be pure and I realized that it is easy to find twists on 1.2.2 that meet those requirements:
Attachment:
1.2.2_parity_pure.png [ 92.48 KiB | Viewed 6330 times ]
The routine is:
...and it's pure (24 moves).
I suppose this isn't such an amazing thing for 1.2.2 but I think I'm pretty close to adapting it to 1.2.3 and deeper cut vertex turners such as 1.2.9. The principal works for all modulo-3 puzzles but finding short sequences that meet the right conditions seems non-trivial for the deeper cut puzzles.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 20, 2010 1:10 pm
Joined: Mon Jul 21, 2008 4:52 am
Location: Brighton, UK
bmenrigh wrote:
I was laying in bed, unable to sleep thinking about the center-twist "parity" I got on the Circle FTO and how all the fixes I tried didn't come out pure. I got to thinking about what is required for a orientation parity twist to be pure and I realized that it is easy to find twists on 1.2.2 that meet those requirements. The routine is:
...and it's pure (24 moves).
I suppose this isn't such an amazing thing for 1.2.2 but I think I'm pretty close to adapting it to 1.2.3 and deeper cut vertex turners such as 1.2.9. The principal works for all modulo-3 puzzles but finding short sequences that meet the right conditions seems non-trivial for the deeper cut puzzles.
Genius! Believe it or not, you have solved annoying corner orientation problems for 1.2.2-1.2.9.
With 1.2.3, the most efficient solving order I know is wide triangles, corners, centers, thin triangles, so the algo doesn't need to be pure, just not to affect wide triangles. It does the job perfectly for 1.2.3.
It is natural to solve 1.2.4 and 1.2.5 corners first so no problem there anyway, just a single twist if necessary before solving the centers.
1.2.6 has no corners so that leaves 1.2.7-1.2.9, where again it is natural to solve them corners first, so all we care about is that we can twist a single corner without affecting any other corners. I tried the same pattern as your algo but spacing the 4 turning points further apart, and here we are:
[DCG,JEK,DCG',
DCG,JEK',DCG',
You have just made solving 1.2.2-1.2.9 a lot more fun -- thanks! (Now I need to take some time to understand exactly how your magic algo works...)
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 20, 2010 3:39 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Hello again, it's great to be here
bmenrigh wrote:
The routine is:
...and it's pure (24 moves).
That is an amazing approach. It's great.
My moves for it were (30):
[AEF', AFB', AEF,]
[ACD, EDK, ACD'], AEF, [ACD, EDK', ACD'], AEF',
Julian wrote:
Stefan Schwalbe wrote:
Julian, can you tell me the (6,1), for the Pentultimate-corners or at least another hint?
The 6 moves are the sequence called P (or Pm for the mirror image) in my solution. Another way of looking at it is to take three faces that all touch each other and call them F, U, and R, like a Rubik's cube. F U R U' R' F' isolates a single swapped corner in half of the Pentultimate.
Thank you, Julian . Your 1.2.9 solve has impressed me very much.
bmenrigh wrote:
It seems like most of us had a hard time with 1.1.8 so good job.
For the "edge-halves" I too used a (7,1) when I first solved it. I just revisited it though and this (5,1) popped out at me as easy and kinda obvious. I wonder why I didn't spot it before
[F'2,C',B'2,C,F2], [E&2], [F'2,C',B2,C,F2], [E'&2]
Note that by modifying the commutated E&2 slice you can vary what gets cycled rather easily.
Elwyn came up with a brilliant reduction to Megaminx trick for this puzzle. He wrote about it here. See the posts after that one too for more discussion about 1.1.8.
Thank you Brandon. The (5,1) is good. Elwyn's solve of the 1.1.8 has also impressed me as it saves many moves. It seems to be hard to me, to pair the edges in his way. I tried the idea to write down the 2 colors of the 30 edge-pairs on a sheet of paper and tick off the solved edges but there must be a better way and I didn't really do that. Elwyn, if you want, you can tell me how you did that.
Stefan.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Mon Sep 20, 2010 7:56 pm
Joined: Fri Feb 06, 2009 2:57 pm
Location: Pittsburgh
1.2.9 is too hard for me. One day I WILL SOLVE IT.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 21, 2010 12:38 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Julian wrote:
Believe it or not, you have solved annoying corner orientation problems for 1.2.2-1.2.9.
With 1.2.3, the most efficient solving order I know is wide triangles, corners, centers, thin triangles, so the algo doesn't need to be pure, just not to affect wide triangles. It does the job perfectly for 1.2.3.
It is quite a privilege to be able to work with, and sometimes even contribute to such talented solvers. I spent another hour today trying to do better that what I came up with last night in my head in 30 minutes and no computer. No success though... I can't figure out how to mask of the extra thin triangle on 1.2.3 to make the routine pure. Even if I do come up with a solution, it will probably be 18 x 2 moves which hardly seems worth it.
But, much to my amazement, I don't need to adapt my routine to 1.2.9 because you already have . Individually we come up with good routines and solutions. Collectively our solutions are brilliant. It really feels great to interact with such good solvers.
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Tue Sep 21, 2010 1:54 am
Joined: Tue Sep 08, 2009 8:41 am
Location: The Blue Mountains, Australia
Interesting, gelatinbrain has done something new. I went to look at the records for 1.2.9 and found an odd "help" link above it which lead right to this thread, specifically to Julian's recent solutions outline.
Stefan Schwalbe wrote:
I tried the idea to write down the 2 colors of the 30 edge-pairs on a sheet of paper and tick off the solved edges but there must be a better way and I didn't really do that. Elwyn, if you want, you can tell me how you did that.
You did just as i did, wrote them down and crossed them off. I think i might be able to do it without writing them down now because of how well i know the colour scheme but it would be hard.
Stefan Schwalbe wrote:
With my 1.1.9 solve i will not appear in the rankings, for it was too bad.
Did you use reduction? If you want you should try solving it again like This I've always liked the gigaminx
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 23, 2010 12:09 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Julian wrote:
1.2.9 (physical version = Drewseph's Chopasaurus)
...
Stage 3 - Two thirds of the time you will end up with a single twisted corner. If you do, twist it to its correct orientation and then re-solve the other 9 corners in 3 cycles
Here is a way to fix the orientation-parity early. Check the parity without doing one move. Than fix it with one move if it is 1 or 2.
The orientation parity of all corners is computed in adding the orientation-state of each of the 20 corners, wich can be 0: correct, 1: clockwise twisted, 2: counterclockwise twisted. The summ is than taken modulo 3. 1 is fixed with a single counterclockwise move. 2 is fixed with a single clockwise move. 0 is allready correct.
The corners are divided up into 4 groups as indicated in the picture.
Attachment:
1.2.9-o-parity.jpg [ 20.09 KiB | Viewed 6207 times ]
1: corners with a white facelet
2: corners with a yellow facelet
3: corners with one left color and two right colors
4: corners with one right color and two left colors
So each corner-piece has a 0-facelet, and each corner-location has a 0-facelet-location. Go through the 20 corner-locations and look where the 0-facelet of the corner-piece is in relation to the 0-facelet-location. Add the numbers that you get together. When you are through take it modulo 3 and fix the corner-orientation-parity with one single move.
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Post subject: Re: Gelatin Brain's Applet Solutions Discussion ThreadPosted: Thu Sep 23, 2010 2:02 pm
Joined: Sun Aug 29, 2010 1:56 pm
Location: Berlin, Germany
Julian: Today I checked your new 1.2.9 outline (posted on this page), and I solved it in 900 moves, thanks to this brilliant method. Especcially the (7,1) for the face-edges is half as short as my alg. would have been. The solve of the corners at the beginning is very good, except I would fix the parity earlier as I have explained in my last post. I need not to say that the 1.2.9 method will also solve the 2.1.5. The only difference will be the orientation of the corner pieces and the face pieces. I'm always happy, when I solve a puzzle for the first time.
Elwyn: I solved the gigaminx again with your method, and this time I reached into the fewest move rankings. Your method is clever in saving moves, and it is a fun to do it in your way.
Stefan.
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Calcite | Level 5
## How to plot certain values in SAS
Hi,
How to plot the graph for only 50 and 75 and rest of the values should not be ploted.And 50 and 75 are different colors.
Please find the attached document for the query and data like
0.249377 5.17532 0.623441 10.3494 1.1222 15.3048 1.74564 20.1719 2.53533 25.0603 3.53283 30.1184 4.65503 35.0611 5.98504 40.0206 7.56442 45.0366 9.47631 50.0665 11.7207 55.0245 14.4223 60.0602 17.5811 65.0406 21.2801 70.034 25.6858 75.0302 30.9643 80.0342 37.739 85.0182 47.0075 90.0126 61.0141 95.0102 100 100
Thanks and regards
4 REPLIES 4
Super User
## Re: sas
Use either SGPLOT with a WHERE clause and filter it for values 50/75
Your values don't have whole values of 50 or 75 so in the context of Your question it doesn't make sense.
Calcite | Level 5
## Re: sas
for example i have two variables
1) percentages_agents
i have the values like
0.249377 5.17532 0.623441 10.3494 1.1222 15.3048 1.74564 20.1719 2.53533 25.0603 3.53283 30.1184 4.65503 35.0611 5.98504 40.0206 7.56442 45.0366 9.47631 50.0665 11.7207 55.0245 14.4223 60.0602 17.5811 65.0406 21.2801 70.034 25.6858 75.0302 30.9643 80.0342 37.739 85.0182 47.0075 90.0126 61.0141 95.0102 100 100
then i have to highlite only 50,75 values of percentages_premium rest of the values should not be plotted.
i have done
symbol value=squarefilled color=vibg cv=vibg interpol=join;
proc gplot data=annexus.annexus_2016a_group_final1_per;
plot cpp*cap/
haxis=0 to 100 by 10
vaxis=0 to 100 by 25
vminor=0
hminor=0
regeqn;
run;
quit;
Diamond | Level 26
## Re: How to plot certain values in SAS
Post test data in the form of a datastep. Don't post attachments like docx/xlsx as these are dangerous and wont be donwloaded. As for your problem,
where int(value) in (50,70);
Can be added to code to have only those values whose whole part is 50 or 70.
SAS Super FREQ
## Re: How to plot certain values in SAS
hi
As re Reeza suggested I would go for Proc SGPLOT, find below some sample code that can get you started.
``````data have;
infile cards dlm=",";
input x y;
cards;
0.249377,5.175325
0.623441,10.3494
1.122195,15.30481
1.745636,20.17191
2.535328,25.06026
3.532835,30.11836
4.655029,35.0611
5.985037,40.02055
7.564422,45.03658
9.476309,50.06652
11.7207,55.02452
14.42228,60.06016
17.58105,65.04058
21.28013,70.03401
25.68579,75.03019
30.96426,80.03419
37.73899,85.0182
47.00748,90.01258
61.01413,95.01025
100,100
;
data have2;
set have;
if int(y) = 50 then do;
x2 = x;
y2 = y;
group=1;
end;
if int(y) = 75 then do;
x2 = x;
y2 = y;
group=2;
end;
if 50 <= int(y) <= 75 then do;
x3 = x;
y3 = y;
end;
run;
proc sgplot data=have2 noautolegend;
styleattrs
datasymbols=(DiamondFilled)
datacontrastcolors=(red green)
;
series x=x y=y ;
scatter x=x2 y=y2 / group=group markerattrs=(size=20) ;
run;
proc sgplot data=have2 noautolegend;
series x=x y=y /;
series x=x3 y=y3 / lineattrs=(THICKNESS=5);
refline 50 75;
run;``````
Bruno
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Algebra: A Combined Approach (4th Edition)
False, $\dfrac{\sqrt[3] {12}}{\sqrt[3] 4}=\sqrt[3] {\dfrac{12}{4}}=\sqrt[3] {3}$
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Calculus Calculus Math Forum
February 16th, 2009, 01:23 PM #1 Member Joined: Feb 2009 Posts: 76 Thanks: 0 Quadratic surfaces Reduce the equation to one of the standard forms, classify the surface and sketch it. 4x^2 - 9y^2 + z^2 + 36 = 0 so... my work--> (4x^2 - 9y^2 + z^2 + 36 = 0) / 36 x^2 /9 - y^2 / 4 + z^2 /36 = -1 -x^2 /9 + y^2 / 4 - z^2 /36 = 1 I am not sure what to classify this surface. I think its an hyperboloid of Two sheets. but I am not sure how to sketch this
February 17th, 2009, 04:40 AM #2 Senior Member Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Quadratic surfaces I just tried sketching it. It might be easier to scale the $x$ and $z$ axes down a bit.
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0
2.1kviews
Determine the hardness of water sample.
The hardness of 10,000 Litres of hard water sample was completely removed by passing it through a zeolite softener. The zeolite softener required 5000 litres of NaCl solution containing 1170 mg of NaCl/litre. Determine the hardness of water sample.
0
85views
1170 mg of NaCl → 1 litre
∴ 5000 litres solutions contains → 5000 X 1170 = 5850000 mg of NaCl
Now, CaCO3 equivalent $= 5850000 X \frac{100}{2 X 58.5} \\ = 5 \times 10^6 \ \ mg \ \ of \ \ CaCO3 \ \ equivalent$
$1 litre → 5 X {10}^6 \ \ ppm \\ 10^4 litre → x \\ ∴ x = 5 X 10^6 X 10^4$
Hardness of water sample = 5 X $10^{10}$ ppm
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👉 Try now NerdPal! Our new math app on iOS and Android
# Find the integral $\int x^2\cos\left(x\right)dx$
Go!
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Text mode
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## Final answer to the problem
$x^2\sin\left(x\right)+2x\cos\left(x\right)-2\sin\left(x\right)+C_0$
Got another answer? Verify it here!
## Step-by-step Solution
How should I solve this problem?
• Choose an option
• Integrate by partial fractions
• Integrate by substitution
• Integrate by parts
• Integrate using tabular integration
• Integrate by trigonometric substitution
• Weierstrass Substitution
• Integrate using trigonometric identities
• Integrate using basic integrals
• Product of Binomials with Common Term
Can't find a method? Tell us so we can add it.
1
We can solve the integral $\int x^2\cos\left(x\right)dx$ by applying the method of tabular integration by parts, which allows us to perform successive integrations by parts on integrals of the form $\int P(x)T(x) dx$. $P(x)$ is typically a polynomial function and $T(x)$ is a transcendent function such as $\sin(x)$, $\cos(x)$ and $e^x$. The first step is to choose functions $P(x)$ and $T(x)$
$\begin{matrix}P(x)=x^2 \\ T(x)=\cos\left(x\right)\end{matrix}$
Learn how to solve integral calculus problems step by step online.
$\begin{matrix}P(x)=x^2 \\ T(x)=\cos\left(x\right)\end{matrix}$
Learn how to solve integral calculus problems step by step online. Find the integral int(x^2cos(x))dx. We can solve the integral \int x^2\cos\left(x\right)dx by applying the method of tabular integration by parts, which allows us to perform successive integrations by parts on integrals of the form \int P(x)T(x) dx. P(x) is typically a polynomial function and T(x) is a transcendent function such as \sin(x), \cos(x) and e^x. The first step is to choose functions P(x) and T(x). Derive P(x) until it becomes 0. Integrate T(x) as many times as we have had to derive P(x), so we must integrate \cos\left(x\right) a total of 3 times. With the derivatives and integrals of both functions we build the following table.
## Final answer to the problem
$x^2\sin\left(x\right)+2x\cos\left(x\right)-2\sin\left(x\right)+C_0$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
SnapXam A2
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### Main Topic: Integral Calculus
Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.
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#### The mole fraction of 20% (mass/mass) Kl (aq) is (density of kl(aq)= 1.202gml-1) Option 1) 0.0463 Option 2) 0.05 Option 3) 0.0263 Option 4) 0.03
As we learned in concept
Mole Fraction -
-
20 Mass / mass KI(aq) means 20g of KI in 100g of solution.
20g of KI in 80g of water
Option 1)
0.0463
Option is incorrect
Option 2)
0.05
Option is incorrect
Option 3)
0.0263
Option is correct
Option 4)
0.03
Option is incorrect
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Mike, I have a lens/optics question...
# Thread: Mike, I have a lens/optics question...
1. ## Mike, I have a lens/optics question...
Okay, after overhearing a conversation this weekend, I've done a bit of research and I have a couple of questions for the optically inclined (I am not.. lol)
I understand the relationship between diopter, focal length and power, but... when dealing with an archery lens the focal length would be the distance from your eye to the lens, I assume... does that differ if you use a straight peep or a corrected peep (clarifier installed) (do you now measure the distance to the peep from the lens, or how do you factor that into the equation)??
Next question is, how does the clarifier diopter influence the ultimate power of the lens based on it's diopter and the focal length? Is there a formula or a nomograph that can calculate this? I know the formula for calculating the ultimate power is 1/(1-D x ESI/39.37) (ESI = eye to scope/lens distance in inches(focal length))
Just trying to understand fully, so I can educate someone who is misguided..
2.
3. Join Date
Jan 2011
Location
Virginia
Posts
1,581
Give Chuck call or hit him up on FB Sticky. He will tell you all you need to know. Might as well get the info from the "Eye Dr." and #1 lens guy out there
4.
5. Originally Posted by IGluIt4U
Okay, after overhearing a conversation this weekend, I've done a bit of research and I have a couple of questions for the optically inclined (I am not.. lol)
I understand the relationship between diopter, focal length and power, but... when dealing with an archery lens the focal length would be the distance from your eye to the lens, I assume... does that differ if you use a straight peep or a corrected peep (clarifier installed) (do you now measure the distance to the peep from the lens, or how do you factor that into the equation)??
Next question is, how does the clarifier diopter influence the ultimate power of the lens based on it's diopter and the focal length? Is there a formula or a nomograph that can calculate this? I know the formula for calculating the ultimate power is 1/(1-D x ESI/39.37) (ESI = eye to scope/lens distance in inches(focal length))
Just trying to understand fully, so I can educate someone who is misguided..
Umm....I was told there would be no math involved on this forum!?!?
6.
7. Originally Posted by Mitchhunt
Umm....I was told there would be no math involved on this forum!?!?
They lied?
Give Chuck call or hit him up on FB Sticky. He will tell you all you need to know. Might as well get the info from the "Eye Dr." and #1 lens guy out there
Well, I put Mike in the title, but didn't mean to limit it to him.. thought that perhaps others may benefit from the responses/discussion, that's why I posted it in here.. (guess I need to send a message to Eye Dr. and get him to post over here, eh? lol)
9. Join Date
Jan 2011
Location
Virginia
Posts
1,581
Probably. That's way more technical then I get. I buy a lens in the power I want and put it in my scope. Put a peep in and shoot.
Sent from my iPhone using Tapatalk
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Two Plus Two Older Archives The Quizshow
FAQ Members List Calendar Search Today's Posts Mark Forums Read
#1
08-05-2002, 08:04 AM
Guest Posts: n/a
The Quizshow
Hi there...
Here is an old but good one. (please excuse my spelling and choice of words - english is not my first language)
You are the last person left in a quizshow. You now have the possibility to win a car. The quizmaster shows you 3 doors. Behind 1 of them, there is a car. Behind the other 2 there is nothing. He tells you to pick a door, and you do. The door remains closed. Now he says: "I'll help you - I'll show you a door with nothing behind. He opens 1 of the doors that you didn't pick. Now he says: "Would like to change your mind and pick the other closed door, or would you like the one you've picked in the first place?"
Now you have 3 options:
1) Change your mind, and pick the other door
2) Stick to the door you've picked first
3) It doesn't make any difference
Well - what's it gonna be (And why!)
Yours
Mik
#2
08-05-2002, 08:33 AM
Guest Posts: n/a
switch
I'll let other's elaborate.
#3
08-05-2002, 12:37 PM
Guest Posts: n/a
Re: switch
Ditto as to Bruces' response.
Jimbo
#4
08-05-2002, 03:08 PM
Guest Posts: n/a
Re: switch
Originally, was 2/3's chance of NOT being the door you selected... so it's 66.7% that it's behind the other door
#5
08-05-2002, 07:24 PM
Guest Posts: n/a
Re: The Quizshow
It doesn't make any difference.
They are going to show you a door with nothing behind it anyway, whether you pick the right one or not. The 33.3% chance that the door they opened will split evenly between the two doors remaining.
If in a three-handed no ante HE game all three of you are dealt AKs, preflop your chance of winning is proportional to the number of players in the hand. One of them folding doesn't suddenly give either one of you 2/3 of the equity.
#6
08-05-2002, 09:19 PM
Guest Posts: n/a
Re: The Quizshow
Ok Monte let's play. I'll switch doors and give you even money odds. Since you think I'm an 2-1 dog you should be eager to play.
#7
08-06-2002, 03:19 PM
Guest Posts: n/a
Re: The Quizshow
BruceZ is right. Here's how it works:
Let's say that you don't change your mind. The probability that you pick the door with the car behind is 1/3. So if you don't change your mind you will win the car 1 out of 3 times.
Now, let's say that you change your mind every time. The probability that you picked the door with the car behind is still 1/3. Let's say the car is behind door #1:
1) You pick door 1 (Quizmaster opens door 2 or 3)You switch = No car
2) You pick door 2 (Quizmaster opens door 3)You switch = Win the car
3) You pick door 3 (Quizmaster opens door 2) You switch = Win the car
The correct answer is to change your mind and switch. The strange thing is, that 4/5 people that you give this riddle to, will stick to their intuition and hang on to the door they picked first.
Has anyone got any ideas on how to use this at the pokertable? How about you BruceZ?
Yours
Mik
#8
08-06-2002, 07:34 PM
Guest Posts: n/a
Re: The Quizshow
There are many poker applications of Bayes' therorem which is what this is based on. Sklansky gave an example recently from jacks or better draw poker where the odds that your opponent has a set go from 2% to 10% once the player opens since you now know that he has jacks or better.
In fact, much of reading hands involves modifying the probabilities of what hands your opponent has based on additional information you gain on each round by the way he bets. You in turn attempt to keep your opponent from doing the same thing to you by making him miscalculate the probabilites of certain hands that you might hold. You do this either by playing in a way that changes the probability of the hand you are holding relative to other hands, or by sometimes playing hands in a way that is different from how you normally play them so that your opponent will miscalculate on future hands.
Revealing information which causes others to calculate whatever probabilities you want them to calculate is also used between nations in economic matters.
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CONV2
Two dimensional convolution.
C = CONV2(A, B) performs the 2-D convolution of matrices
A and B. If [ma,na] = size(A) and [mb,nb] = size(B), then
size(C) = [ma+mb-1,na+nb-1].
C = CONV2(A,B,'shape') returns a subsection of the 2-D
convolution with size specified by 'shape':
'full' - (default) returns the full 2-D convolution,
'same' - returns the central part of the convolution
that is the same size as A.
'valid' - returns only those parts of the convolution
that are computed without the zero-padded
edges, size(C) = [ma-mb+1,na-nb+1] when
size(A) > size(B).
CONV2 is fastest when size(A) > size(B).
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# Lab_1.xlsx
Description
141 1 Mark ID # Submission Date Problem 1 Problem 2 Problem 3 Problem 4 Total Student Name Problem # King Fahd University of Petroleum & Minerals Civil & Environmental Engineering Department CE 318 Numerical & Statistical Methods in Civil Engineering Academic Term Assignment # Instructor: Dr. Husain Jubran Al-Gahtani How to use Excel built-in functions Exercise 1 Fill-in the third column by computing the respective function given in the second column x Formula Result -1 Abs(x) 3.14 Cos(x) 3.14
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1411 Mark ID #Submission DateProblem 1Problem 2Problem 3Problem 4TotalStudent NameProblem # King Fahd University of Petroleum & Minerals ivil & Environmental Engineering Departme CE 318 Numerical & Statistical Methods in Civil Engineering Academic Term Assignment # Instructor: Dr. Husain Jubran Al-Gahtani How to use Excel built-in functionsExercise 1 Fill-in the third column by computing the respective function given in the second columnxFormulaResult-1Abs(x)3.14Cos(x)3.14Sin(x) π pi()0.1Exp(x)2.5Even(x)2.5Odd(x)2.35Floor(x,a)2.35Ceiling(x,a)16Sqrt(x)0.5Atan(x)180RadiansAs an example, to use the SIN formula just write =(equal to) SIN followed by the number in brackets as shown in the figure. As soon as you close the brackets and hit enter the formula is evaluated and displayed in the same cell as shown in the figure. Note that you can use capital, small or mixed letters (Excel is not case-sensitive) Try more functions (many of them stored under Formulas menue Repeating Calculations Create a table in which the first column is x, the second column is f(x) = x 2 +sin x +e x Repeat the caculations for x = 0.1, 0.2 to 1.2, then plot the data of x versus f(x) Add a third column containing the function x*f(x) to the data and add it to the plot xf(x)x*f(x)0.11.21500430.0012150.21.46007210.2920140.31.7353790.5206140.42.0412430.8164970.52.37814681.1890730.62.74676131.6480570.73.14797042.2035790.83.5828972.8663180.94.052933.64763714.55975284.5597531.15.10537345.6159111.25.6921566.830587 Exercise 2 -80 < x < 4V(x)=126 < x< 842-5x8 < x< 10-8x0 < x < 4M(x)12x-806 < x< 8 -2.5x^2+42x-170 8 < x< 10x (m)V (N)M (N.m)0-804-8-324.00112-31.988612-8771.58269-35.510-80The first column in the following table represents distance measured from the left end of a beam. The second column represents the shear force and the third represents the bending moment. Add another two columns that display the results of V and M, then plot them in one figure (x-axis the distance x and y-axis for shear V & moment M). Can you add a y-axis for the moment so that it has a more suitable scale. (Hint:right mouse-click on the moment curve -> Format data series -> select Soon (in this lab) you will learn a more efficient way of formulating the shear and moment using IF statement 01234567800.511.5f(x)x*f(x) You should get the following results -35-30-25-20-15-10-5051015051015Shear (N)Bending Moment(N.m)-35-30-25-20-15-10-50510-10-5051015051015Shear (N)Bending Moment(N.m)
Jul 24, 2017
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https://nerdutella.com/q1338-C-plus-plus-program-to-print-all-possible-subset-of-a-set
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Q:
# C++ program to print all possible subset of a set
need an explanation for this answer? contact us directly to get an explanation for this answer
Program:
``````#include <bits/stdc++.h>
using namespace std;
void allPossibleSubset(int arr[], int n)
{
int count = pow(2, n);
// The outer for loop will run 2^n times to print all subset .
// Here variable i will act as a binary counter
for (int i = 0; i < count; i++) {
// The inner for loop will run n times ,
// As the maximum number of elements a set can have is n
// This loop will generate a subset
for (int j = 0; j < n; j++) {
// This if condition will check if jth bit in
// binary representation of i is set or not
// if the value of (i & (1 << j)) is not 0 ,
// include arr[j] in the current subset
// otherwise exclude arr[j]
if ((i & (1 << j)) != 0)
cout << arr[j] << " ";
}
cout << "\n";
}
}
int main()
{
int n;
cout << "Enter size of the set\n";
cin >> n;
int arr[n];
cout << "Enter Elements of the set\n";
for (int i = 0; i < n; i++)
cin >> arr[i];
allPossibleSubset(arr, n);
return 0;
}``````
Output
```Enter size of the set
4
Enter Elements of the set
1 2 3 4
1
2
1 2
3
1 3
2 3
1 2 3
4
1 4
2 4
1 2 4
3 4
1 3 4
2 3 4
1 2 3 4```
need an explanation for this answer? contact us directly to get an explanation for this answer
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https://rkenmi.com/posts/reversing-words-in-a-string-the-double-reverse?lang=en
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@rkenmi - Reversing words in a string : the double reverse
# Reversing words in a string : the double reverse
## Problem
Reverse all words (separated by space) in a string
## Input
• $$s$$ - a string
## Approach
Nice case and point for really examining the examples.
First, a few examples of reversing words in a string.
"Wu Tang Clan" => "Clan Tang Wu"
"r R r" => "r R r"
"C a c" => "c a C"
"Winter is coming" => "coming is Winter"
It might not come across one to consider completely reversing a string and studying the examples.
If you reverse a sentence, say - "smell u l8r".
You'll have the output: r8l u llems.
At first sight, this doesn't appear to show much. A key visualization here however, is that after reversing the original string, the characters are lined up in the correct positions if we were to reverse all words in a string.
That is, if the final output we are looking for is smell u l8r => l8r u smell, then r8l u llems has the spaces in the correct positions and the alphanumeric characters in relative (but not quite correct) positions.
The trick from here is to reverse only the words in this generated output : r8l => l8r, u => u, llems => smell.
Then the final output becomes: l8r u smell.
Try it yourself!
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https://community.smartsheet.com/discussion/74204/formula-help-to-return-differnt-conditions
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# Formula help to return differnt conditions
Options
Hi
I am after help in developing a formula to look for certain conditions. I require the highlighted cell in blue perimeter to record certain condition if a condition is true from the 5 cells below.
If all cells return Completed then return "Completed"
If all cells return Not Started, then Return "Not Started"
If all cells return mixed conditions, then return "In Progress"
I have the following fields within the column as dropdowns.
Not Started
In Progress
Waiting For Checks
Checks in Progress
Completed
• ✭✭✭✭✭✭
Options
You may need to adjust this based on where you're placing the formula and how many rows are below it, but it will give you the general idea:
=IF(COUNTIF(staus2:status6, "Complete")=5, "Complete", IF(COUNTIF(staus2:status6, "Not Started")=5, "Not Started", "In Progress"))
Work?
Mark
I'm grateful for your "Vote Up" or "Insightful". Thank you for contributing to the Community.
• ✭✭✭✭✭✭
Options
You may need to adjust this based on where you're placing the formula and how many rows are below it, but it will give you the general idea:
=IF(COUNTIF(staus2:status6, "Complete")=5, "Complete", IF(COUNTIF(staus2:status6, "Not Started")=5, "Not Started", "In Progress"))
Work?
Mark
I'm grateful for your "Vote Up" or "Insightful". Thank you for contributing to the Community.
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template!
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https://www.turito.com/ask-a-doubt/state-the-gradient-and-y-intercept-of-the-line-y-3x-8-q52a0b555
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Maths-
General
Easy
Question
State the gradient and y-intercept of the line y = -3x + 8.
Hint:
The correct answer is: we get Gradient = -3 y-intercept = 8
Step by step solution:The given equation of the line isy = - 3x + 8Comparing the above equation withy = mx + cWe getm = -3; c = 8Thus, we getGradient = -3y-intercept = 8
The student needs to remember all the different forms of equation of a line and what each term and notation signifies in the equation.
We can find the slope and y-intercept directly from the general form of the equation too; slope = and y-intercept =, where the general form of equation of a line is ax + by + c = 0.
General
Maths-
In the standard form of a polynomial, the terms are written in ______order of their degree.
Maths-General
General
Maths-
Identify a binomial.
Maths-General
General
Maths-
What must be added to 5y2+3y + 1 to get 3y3 -7y2+8
Maths-General
General
Maths-
Find the value of the y-intercept in the equation 3x-y+5=0.
Maths-General
General
Maths-
𝐴𝑃 is perpendicular bisector of 𝐵𝐶. 1. What segment lengths are equal? Explain your reasoning. 2. Find AB.3. Explain why D is on 𝐴𝑃.
Maths-General
General
Maths-
Identify a monomial.
Maths-General
General
Maths-
Convert the equation 5x + 4y = 12 into y = mx + c and find its y-intercept.
Maths-General
General
Maths-
What must be subtracted from 3x2-5xy -2y2-3 to get 5x2-7xy -3y2+3x
Maths-General
General
Maths-
𝐴𝑃 is the perpendicular bisector of 𝐵𝐶. a. What segment lengths in the diagram are equal? b. Is A on 𝐴𝑃?
Maths-General
General
Maths-
Use the square of a binomial to find the value. 562
Maths-General
General
Maths-
Is the given conjecture correct? Provide arguments to support your answer.The product of any two consecutive odd numbers is 1 less than a perfect square.
Maths-General
General
Maths-
Name the polynomial based on its degree and number of terms.
Maths-General
General
Maths-
In an academic contest correct answers earn 12 points and incorrect answers lose 5points. In the final round, school A starts with 165 points and gives the same numberof correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.ii)How many answers did each school get correct in the final round?
A linear equation in one variable is an equation that has only one solution and is expressed in the form ax+b = 0, where a and b are two integers and x is a variable. 2x+3=8, for example, is a linear equation with a single variable. As a result, this equation has only one solution, x = 5/2.
The standard form of a linear equation in one variable is: ax + b = 0
Where,
The letters 'a' and 'b' are real numbers.
'a' and 'b' are both greater than zero.
In an academic contest correct answers earn 12 points and incorrect answers lose 5points. In the final round, school A starts with 165 points and gives the same numberof correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.ii)How many answers did each school get correct in the final round?
Maths-General
A linear equation in one variable is an equation that has only one solution and is expressed in the form ax+b = 0, where a and b are two integers and x is a variable. 2x+3=8, for example, is a linear equation with a single variable. As a result, this equation has only one solution, x = 5/2.
The standard form of a linear equation in one variable is: ax + b = 0
Where,
The letters 'a' and 'b' are real numbers.
'a' and 'b' are both greater than zero.
General
Maths-
Determine the equation of the line that passes through
Maths-General
General
Maths-
Is the given conjecture correct? Provide arguments to support your answer.The product of any two consecutive even numbers is 1 less than a perfect square
Maths-General
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https://www.ozgrid.com/freebies/excel-get-maximum-number-2-numbers
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# OzGrid
Excel: Get Maximum Number Between 2 Numbers
Category: [Excel] Demo Available
# Excel: Get Maximum Number Between 2 Numbers
Got any Excel/VBA Questions? Free Excel Help
This Custom Function will extract the maximum number between any two specified numbers. If used on large ranges of data, it will slow down Excel's calculation.
The Code
```Function GetMaxBetween(rCells As Range, MinNum, MaxNum)
Dim rRange As Range
Dim vMax
Dim aryNums()
Dim i As Integer
ReDim aryNums(rCells.Count)
For Each rRange In rCells
vMax = rRange
Select Case vMax
Case MinNum + 0.01 To MaxNum - 0.01
aryNums(i) = vMax
i = i + 1
Case Else
GetMaxBetween = 0
End Select
Next rRange
GetMaxBetween = WorksheetFunction.Max(aryNums)
End Function```
To use this UDF push Alt+F11 and go Insert>Module and paste in the code. Push Alt+Q and save. The Function will appear under "User Defined" in the Paste Function dialog box (Shift+F3). Use the Function in any cell as shown below.
=GetMaxBetween(A1:A100,9,10)
This would return the highest number in the range A1:A100 that is less than 10, but greater than 9. Only accurate up to 0.01
Click here to visit our Free 24/7 Excel/VBA Help Forum where there are thousands of posts you can get information from, or you can join the Forum and post your own questions.
### Gallery
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https://www.folkstalk.com/2022/10/get-current-week-2-with-code-examples.html
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Get Current Week 2 With Code Examples
Get Current Week 2 With Code Examples
Hello everyone, In this post, we will examine how to solve the Get Current Week 2 problem using the computer language.
```function getCurrentWeek() {
var currentDate = moment();
var weekStart = currentDate.clone().startOf('isoWeek');
var weekEnd = currentDate.clone().endOf('isoWeek');
var days = [];
for (var i = 0; i <= 6; i++) {
}
console.log(days);
}
getCurrentWeek();```
We were able to demonstrate how to correct the Get Current Week 2 bug by looking at a variety of examples taken from the real world.
How do YOu calculate current week?
Add the number of days to the current weekday using getDay() and divide it by 7. We will get the current week's number.24-May-2022
Is it week 1 or week 2?
View the week numbers of 2022
How do I get the current week in Python?
YOu can use the isocalender function from the datetime module to get the current week. Create the object of the date first, then call isocalender() on this object. This returns a 3-tuple of Year, Week number and Day of Week.12-Jun-2020
What is the current week in 2022?
It is currently week 39 in 2022. There are 13 weeks remaining.
How do I get the current week number in Excel?
Excel WEEKNUM Function
• Summary. The Excel WEEKNUM function takes a date and returns a week number (1-54) that corresponds to the week of year.
• Get the week number for a given date.
• A number between 1 and 54.
• =WEEKNUM (serial_num, [return_type])
• serial_num – A valid Excel date in serial number format.
What is the current week number 2021?
Week Numbers for 2021
What is first week of the month?
First week of a month is the the first seven days of the month. It may start as a Sunday or a Monday or any day as a matter of fact and ends on the 7th day of the month. Second week like wise are the next 7 days after the first week and so on for the 3rd and 4th week.
What date is week 22 in 2022?
The date of the calendar Week Week 22 and other years
week 39
How do I get this week in JavaScript?
JavaScript: Get the week end date
• Sample Solution:-
• HTML Code: <!
• JavaScript Code: function endOfWeek(date) { var lastday = date.getDate() – (date.getDay() – 1) + 6; return new Date(date.setDate(lastday)); } dt = new Date(); console.log(endOfWeek(dt).toString());
• Flowchart:
• Live Demo:
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Trending
# Pick A Card, Any Card!
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.
## Riddler Express
From Mike Strong, a climate change problem:
In each of the last three years — 2014, 2015 and 2016 — a new global temperature record has been set. Assuming that accurate temperature records exist since 1880, what is the probability of this happening at random?
## Riddler Classic
From Mont Chris Hubbard, a prediction puzzle inspired by his habit of trying to identify the funniest name as early as possible in a movie’s scrolling end credits:
From a shuffled deck of 100 cards that are numbered 1 to 100, you are dealt 10 cards face down. You turn the cards over one by one. After each card, you must decide whether to end the game. If you end the game on the highest card in the hand you were dealt, you win; otherwise, you lose.
What is the strategy that optimizes your chances of winning? How does the strategy change as the sizes of the deck and the hand are changed?
## Solution to last week’s Riddler Express
Congratulations to 👏 Bob Rietz 👏 of Asheville, North Carolina, winner of last week’s Express puzzle!
In standard American bingo, a bingo card is a five-by-five grid of squares. The columns are labeled B, I, N, G and O, in that order. The five squares in the B column can be filled with the numbers 1 through 15, those in the I column with the numbers 16 through 30, those in the N column 31 through 45, and so on. The square in the very center of the grid is a “free space” on every card. How many different possible bingo cards are there?
There are 552,446,474,061,128,648,601,600,000.
For the four columns B, I, G and O — those without the free space — there are 15 ways to choose the first number in the column, then 14 ways to choose the second, 13 to choose the third, 12 to choose the fourth and 11 to choose the fifth. (The number of options descend because once a number is selected for one square, it can’t appear in another.) For the N column — with the free space in the middle — we need to pick only four numbers: There are 15 ways to choose the first, 14 ways to choose the second, 13 to choose the third and 12 to choose the fourth. Now, we multiply all those possibilities together!
((15cdot 14cdot 13cdot 12cdot 11)^4cdot (15cdot 14cdot 13cdot 12)approx 5.52cdot 10^{26})
## Solution to last week’s Riddler Classic
Congratulations to 👏 Andrew Zwicky 👏 of Cedar Falls, Iowa, winner of last week’s Classic puzzle!
Imagine that it’s the beginning of time, and the Supreme Court’s…
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# Edge operators¶
Edge operators are used in image processing within edge detection algorithms. They are discrete differentiation operators, computing an approximation of the gradient of the image intensity function.
```import numpy as np
import matplotlib.pyplot as plt
from skimage import filters
from skimage.data import camera
from skimage.util import compare_images
image = camera()
edge_roberts = filters.roberts(image)
edge_sobel = filters.sobel(image)
fig, axes = plt.subplots(ncols=2, sharex=True, sharey=True,
figsize=(8, 4))
axes[0].imshow(edge_roberts, cmap=plt.cm.gray)
axes[0].set_title('Roberts Edge Detection')
axes[1].imshow(edge_sobel, cmap=plt.cm.gray)
axes[1].set_title('Sobel Edge Detection')
for ax in axes:
ax.axis('off')
plt.tight_layout()
plt.show()
```
Different operators compute different finite-difference approximations of the gradient. For example, the Scharr filter results in a less rotational variance than the Sobel filter that is in turn better than the Prewitt filter [1] [2] [3]. The difference between the Prewitt and Sobel filters and the Scharr filter is illustrated below with an image that is the discretization of a rotation- invariant continuous function. The discrepancy between the Prewitt and Sobel filters, and the Scharr filter is stronger for regions of the image where the direction of the gradient is close to diagonal, and for regions with high spatial frequencies. For the example image the differences between the filter results are very small and the filter results are visually almost indistinguishable.
```x, y = np.ogrid[:100, :100]
# Creating a rotation-invariant image with different spatial frequencies.
image_rot = np.exp(1j * np.hypot(x, y) ** 1.3 / 20.).real
edge_sobel = filters.sobel(image_rot)
edge_scharr = filters.scharr(image_rot)
edge_prewitt = filters.prewitt(image_rot)
diff_scharr_prewitt = compare_images(edge_scharr, edge_prewitt)
diff_scharr_sobel = compare_images(edge_scharr, edge_sobel)
max_diff = np.max(np.maximum(diff_scharr_prewitt, diff_scharr_sobel))
fig, axes = plt.subplots(nrows=2, ncols=2, sharex=True, sharey=True,
figsize=(8, 8))
axes = axes.ravel()
axes[0].imshow(image_rot, cmap=plt.cm.gray)
axes[0].set_title('Original image')
axes[1].imshow(edge_scharr, cmap=plt.cm.gray)
axes[1].set_title('Scharr Edge Detection')
axes[2].imshow(diff_scharr_prewitt, cmap=plt.cm.gray, vmax=max_diff)
axes[2].set_title('Scharr - Prewitt')
axes[3].imshow(diff_scharr_sobel, cmap=plt.cm.gray, vmax=max_diff)
axes[3].set_title('Scharr - Sobel')
for ax in axes:
ax.axis('off')
plt.tight_layout()
plt.show()
```
As in the previous example, here we illustrate the rotational invariance of the filters. The top row shows a rotationally invariant image along with the angle of its analytical gradient. The other two rows contain the difference between the different gradient approximations (Sobel, Prewitt, Scharr & Farid) and analytical gradient.
The Farid & Simoncelli derivative filters [4], [5] are the most rotationally invariant, but require a 5x5 kernel, which is computationally more intensive than a 3x3 kernel.
```x, y = np.mgrid[-10:10:255j, -10:10:255j]
image_rotinv = np.sin(x ** 2 + y ** 2)
image_x = 2 * x * np.cos(x ** 2 + y ** 2)
image_y = 2 * y * np.cos(x ** 2 + y ** 2)
def angle(dx, dy):
"""Calculate the angles between horizontal and vertical operators."""
return np.mod(np.arctan2(dy, dx), np.pi)
true_angle = angle(image_x, image_y)
angle_farid = angle(filters.farid_h(image_rotinv),
filters.farid_v(image_rotinv))
angle_sobel = angle(filters.sobel_h(image_rotinv),
filters.sobel_v(image_rotinv))
angle_scharr = angle(filters.scharr_h(image_rotinv),
filters.scharr_v(image_rotinv))
angle_prewitt = angle(filters.prewitt_h(image_rotinv),
filters.prewitt_v(image_rotinv))
def diff_angle(angle_1, angle_2):
"""Calculate the differences between two angles."""
return np.minimum(np.pi - np.abs(angle_1 - angle_2),
np.abs(angle_1 - angle_2))
diff_farid = diff_angle(true_angle, angle_farid)
diff_sobel = diff_angle(true_angle, angle_sobel)
diff_scharr = diff_angle(true_angle, angle_scharr)
diff_prewitt = diff_angle(true_angle, angle_prewitt)
fig, axes = plt.subplots(nrows=3, ncols=2, sharex=True, sharey=True,
figsize=(8, 8))
axes = axes.ravel()
axes[0].imshow(image_rotinv, cmap=plt.cm.gray)
axes[0].set_title('Original image')
axes[1].imshow(true_angle, cmap=plt.cm.hsv)
axes[2].imshow(diff_sobel, cmap=plt.cm.inferno, vmin=0, vmax=0.02)
axes[2].set_title('Sobel error')
axes[3].imshow(diff_prewitt, cmap=plt.cm.inferno, vmin=0, vmax=0.02)
axes[3].set_title('Prewitt error')
axes[4].imshow(diff_scharr, cmap=plt.cm.inferno, vmin=0, vmax=0.02)
axes[4].set_title('Scharr error')
color_ax = axes[5].imshow(diff_farid, cmap=plt.cm.inferno, vmin=0, vmax=0.02)
axes[5].set_title('Farid error')
colorbar_ax = fig.add_axes([0.90, 0.10, 0.02, 0.50])
fig.colorbar(color_ax, cax=colorbar_ax, ticks=[0, 0.01, 0.02])
for ax in axes:
ax.axis('off')
plt.show()
```
Total running time of the script: ( 0 minutes 0.407 seconds)
Gallery generated by Sphinx-Gallery
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Lec16_Structural_Reliability_Example
# Lec16_Structural_Reliability_Example - R – 0.5 Var[ln(R =...
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Lecture 16; page 12 Structural Reliability - Revisited S = load effect (lognormal) lognormal) R = resistance of structure to load lognormal) V = R/S = safety factor for structure < 1 failure > 1 no failure P[ V < 1 ] = P[ R S < 1]
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Lecture 16; page 13 SOLUTION STRATEGY: Resistance Load Safety Factor Real Space variables R S R/S = exp[ U ] | | /\ \/ \/ | Log Space variables lnR lnS —> U = lnR – lnS ~ N
Lecture 16; page 14 Let R and S be independent and lognormal where E[R] = 120 E[S] = 75 Var[R] = (10) 2 Var[S] = (15) 2 Find mean and variance of ln(R) Var[ ln(R) ] = ln[ 1 + σ R 2 R 2 ] = ln[ 1 + (10) 2 /120 2 ] = 0.00692 = (0.0832) 2 E[ ln(R) ] = ln[μ
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Unformatted text preview: R ] – 0.5 Var[ln(R)] = ln[120] – 0.5 (0.0832) 2 = 4.78 Doing the same for ln[S] Var[ ln(S) ] = ln[ 1 + σ S 2 /μ S 2 ] = ln[ 1 + (15) 2 /75 2 ] = 0.0392 = (0.2) 2 E[ ln(S) ] = ln[μ S ] – 0.5 Var[ln(S)] = ln[75] – 0.5 (0.2) 2 = 4.30 Lecture 16; page 15 The variable of interest is U = ln(R) – ln(S) ~ N(μ lnR – μ lnS , σ lnR 2 + σ lnS 2 ) = N( 4.78 – 4.30, (0.0832) 2 + (0.2) 2 ) = N( 0.486, 0.046 ) = N( 0.486, (0.21) 2 ) P[ R/S < 1] = P[ ln(R) – ln(S) < 0 ] = P[ (U-μ U )/ σ U < –0.486/0.21 ] = P[ Z < –0.486/0.21 ] = P[ Z < – 2.26 ] = 0.012 = 1.2% (using tables) Using bivariate normal model obtained failure probability of 0.6% —> model choice matters....
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## This note was uploaded on 12/05/2011 for the course CEE 3040 at Cornell.
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# 3 Digit Multiplication Worksheet Hard
In Free Printable Worksheets219 views
4.07 / 5 ( 204votes )
Top Suggestions 3 Digit Multiplication Worksheet Hard :
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Author: Olga Riske
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
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Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest Research Report
## Summary
As stock data is characterized by highly noisy and non-stationary, stock price prediction is regarded as a knotty problem. In this paper, we propose new two-stage ensemble models by combining empirical mode decomposition (EMD) (or variational mode decomposition (VMD)), extreme learning machine (ELM) and improved harmony search (IHS) algorithm for stock price prediction, which are respectively named EMD–ELM–IHS and VMD–ELM–IHS. We evaluate Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest prediction models with Deductive Inference (ML) and Statistical Hypothesis Testing1,2,3,4 and conclude that the PTA stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Buy PTA stock.
## Key Points
1. How do you decide buy or sell a stock?
2. Technical Analysis with Algorithmic Trading
3. What is statistical models in machine learning?
## PTA Target Price Prediction Modeling Methodology
We consider Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest Stock Decision Process with Deductive Inference (ML) where A is the set of discrete actions of PTA stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Statistical Hypothesis Testing)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Deductive Inference (ML)) X S(n):→ (n+16 weeks) $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of PTA stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## PTA Stock Forecast (Buy or Sell) for (n+16 weeks)
Sample Set: Neural Network
Stock/Index: PTA Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest
Time series to forecast n: 22 Nov 2022 for (n+16 weeks)
According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Buy PTA stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Adjusted IFRS* Prediction Methods for Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest
1. When a group of items that constitute a net position is designated as a hedged item, an entity shall designate the overall group of items that includes the items that can make up the net position. An entity is not permitted to designate a non-specific abstract amount of a net position. For example, an entity has a group of firm sale commitments in nine months' time for FC100 and a group of firm purchase commitments in 18 months' time for FC120. The entity cannot designate an abstract amount of a net position up to FC20. Instead, it must designate a gross amount of purchases and a gross amount of sales that together give rise to the hedged net position. An entity shall designate gross positions that give rise to the net position so that the entity is able to comply with the requirements for the accounting for qualifying hedging relationships.
2. Compared to a business model whose objective is to hold financial assets to collect contractual cash flows, this business model will typically involve greater frequency and value of sales. This is because selling financial assets is integral to achieving the business model's objective instead of being only incidental to it. However, there is no threshold for the frequency or value of sales that must occur in this business model because both collecting contractual cash flows and selling financial assets are integral to achieving its objective.
3. For some types of fair value hedges, the objective of the hedge is not primarily to offset the fair value change of the hedged item but instead to transform the cash flows of the hedged item. For example, an entity hedges the fair value interest rate risk of a fixed-rate debt instrument using an interest rate swap. The entity's hedge objective is to transform the fixed-interest cash flows into floating interest cash flows. This objective is reflected in the accounting for the hedging relationship by accruing the net interest accrual on the interest rate swap in profit or loss. In the case of a hedge of a net position (for example, a net position of a fixed-rate asset and a fixed-rate liability), this net interest accrual must be presented in a separate line item in the statement of profit or loss and other comprehensive income. This is to avoid the grossing up of a single instrument's net gains or losses into offsetting gross amounts and recognising them in different line items (for example, this avoids grossing up a net interest receipt on a single interest rate swap into gross interest revenue and gross interest expense).
4. When designating a hedging relationship and on an ongoing basis, an entity shall analyse the sources of hedge ineffectiveness that are expected to affect the hedging relationship during its term. This analysis (including any updates in accordance with paragraph B6.5.21 arising from rebalancing a hedging relationship) is the basis for the entity's assessment of meeting the hedge effectiveness requirements.
*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.
## Conclusions
Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest assigned short-term B2 & long-term Baa2 forecasted stock rating. We evaluate the prediction models Deductive Inference (ML) with Statistical Hypothesis Testing1,2,3,4 and conclude that the PTA stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Buy PTA stock.
### Financial State Forecast for PTA Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B2Baa2
Operational Risk 8467
Market Risk6885
Technical Analysis3375
Fundamental Analysis4374
Risk Unsystematic5277
### Prediction Confidence Score
Trust metric by Neural Network: 75 out of 100 with 778 signals.
## References
1. J. N. Foerster, Y. M. Assael, N. de Freitas, and S. Whiteson. Learning to communicate with deep multi-agent reinforcement learning. In Advances in Neural Information Processing Systems 29: Annual Conference on Neural Information Processing Systems 2016, December 5-10, 2016, Barcelona, Spain, pages 2137–2145, 2016.
2. Semenova V, Goldman M, Chernozhukov V, Taddy M. 2018. Orthogonal ML for demand estimation: high dimensional causal inference in dynamic panels. arXiv:1712.09988 [stat.ML]
3. M. Babes, E. M. de Cote, and M. L. Littman. Social reward shaping in the prisoner's dilemma. In 7th International Joint Conference on Autonomous Agents and Multiagent Systems (AAMAS 2008), Estoril, Portugal, May 12-16, 2008, Volume 3, pages 1389–1392, 2008.
4. Knox SW. 2018. Machine Learning: A Concise Introduction. Hoboken, NJ: Wiley
5. S. J. Russell and P. Norvig. Artificial Intelligence: A Modern Approach. Prentice Hall, Englewood Cliffs, NJ, 3nd edition, 2010
6. Breiman L. 1996. Bagging predictors. Mach. Learn. 24:123–40
7. Bastani H, Bayati M. 2015. Online decision-making with high-dimensional covariates. Work. Pap., Univ. Penn./ Stanford Grad. School Bus., Philadelphia/Stanford, CA
Frequently Asked QuestionsQ: What is the prediction methodology for PTA stock?
A: PTA stock prediction methodology: We evaluate the prediction models Deductive Inference (ML) and Statistical Hypothesis Testing
Q: Is PTA stock a buy or sell?
A: The dominant strategy among neural network is to Buy PTA Stock.
Q: Is Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest stock a good investment?
A: The consensus rating for Cohen & Steers Tax-Advantaged Preferred Securities and Income Fund Common Shares of Beneficial Interest is Buy and assigned short-term B2 & long-term Baa2 forecasted stock rating.
Q: What is the consensus rating of PTA stock?
A: The consensus rating for PTA is Buy.
Q: What is the prediction period for PTA stock?
A: The prediction period for PTA is (n+16 weeks)
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The very first computers existed long, long before the microchip-powered devices we use every day here in the future. And over the centuries, there have been various different kinds of mechanical calculators and other computing devices that use physical means—unalterable cogs and gears—to turn one number into a different one. That this is possible is obvious, but how is it possible, exactly?
This 1953 training film has you covered. Forty minutes long (all of them good, I promise), it provides a slow, coherent, and exhaustive explanation of exactly how gears can do math, all framed in the context of the Navy's now-ancient but still mind-blowingly impressive fire control computers. These geared behemoths carefully computed the angles needed to hit moving targets from other moving targets with arcing shells fired from giant naval cannons. No small feat.
The tutorial starts simple, with how just two gears can do multiplication. From there it's just a few more steps to calculus. It makes more, elegant sense than you would think. Just watch:
###### More From Popular Mechanics
This content is imported from youTube. You may be able to find the same content in another format, or you may be able to find more information, at their web site.
Mechanical Computer (All Parts) - Basic Mechanisms In Fire Control Computers
Watch on
This isn't the only old video out there that's proven to be as instructive as any new-fangled YouTubery ever could be. You may recall this lovely 1937 lesson on the operation of a car differential, which is every bit as useful today. Or this 1942 Disney-animated film about how to fire a Boys Anti-tank rifle, which has maybe outlived its usefulness a little bit. And while they heyday of the giant, sophisticated mechanical computer may be long gone, the logic behind its many interlocking gears and pegs and cams is as mesmerizing as ever.
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Parallel Brute Force
Parallel Brute Force
You are encouraged to solve this task according to the task description, using any language you may know.
Find, through brute force, the five-letter passwords corresponding with the following SHA-256 hashes:
```1. 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
2. 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
3. 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
```
Your program should naively iterate through all possible passwords consisting only of five lower-case ASCII English letters. It should use concurrent or parallel processing, if your language supports that feature. You may calculate SHA-256 hashes by calling a library or through a custom implementation. Print each matching password, along with its SHA-256 hash.
BaCon
`PRAGMA INCLUDE <openssl/sha.h>PRAGMA LDFLAGS -lcrypto OPTION MEMTYPE unsigned char LOCAL buffer[32], passwd[5] TYPE unsigned charLOCAL result TYPE unsigned char*LOCAL a,b,c,d,e TYPE INT DATA "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f", "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad" WHILE TRUE READ secret\$ IF NOT(LEN(secret\$)) THEN BREAK FOR i = 0 TO 31 buffer[i] = DEC(MID\$(secret\$, i*2+1, 2)) NEXT FOR a = 97 TO 122 FOR b = 97 TO 122 FOR c = 97 TO 122 FOR d = 97 TO 122 FOR e = 97 TO 122 passwd[0] = a passwd[1] = b passwd[2] = c passwd[3] = d passwd[4] = e result = SHA256(passwd, 5, 0) FOR i = 0 TO SHA256_DIGEST_LENGTH-1 IF PEEK(result+i) != buffer[i] THEN BREAK NEXT IF i = SHA256_DIGEST_LENGTH THEN PRINT a,b,c,d,e,secret\$ FORMAT "%c%c%c%c%c:%s\n" BREAK 5 END IF NEXT NEXT NEXT NEXT NEXTWEND`
Output:
```apple:3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm:74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
```
C
Translation of: C#
`// \$ gcc -o parabrutfor parabrutfor.c -fopenmp -lssl -lcrypto// \$ export OMP_NUM_THREADS=4// \$ ./parabrutfor #include <stdio.h>#include <stdlib.h>#include <string.h>#include <omp.h>#include <openssl/sha.h> typedef unsigned char byte; int matches(byte *a, byte* b) { for (int i = 0; i < 32; i++) if (a[i] != b[i]) return 0; return 1;} byte* StringHashToByteArray(const char* s) { byte* hash = (byte*) malloc(32); char two[3]; two[2] = 0; for (int i = 0; i < 32; i++) { two[0] = s[i * 2]; two[1] = s[i * 2 + 1]; hash[i] = (byte)strtol(two, 0, 16); } return hash;} void printResult(byte* password, byte* hash) { char sPass[6]; memcpy(sPass, password, 5); sPass[5] = 0; printf("%s => ", sPass); for (int i = 0; i < SHA256_DIGEST_LENGTH; i++) printf("%02x", hash[i]); printf("\n");} int main(int argc, char **argv){ #pragma omp parallel { #pragma omp for for (int a = 0; a < 26; a++) { byte password[5] = { 97 + a }; byte* one = StringHashToByteArray("1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad"); byte* two = StringHashToByteArray("3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b"); byte* three = StringHashToByteArray("74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"); for (password[1] = 97; password[1] < 123; password[1]++) for (password[2] = 97; password[2] < 123; password[2]++) for (password[3] = 97; password[3] < 123; password[3]++) for (password[4] = 97; password[4] < 123; password[4]++) { byte *hash = SHA256(password, 5, 0); if (matches(one, hash) || matches(two, hash) || matches(three, hash)) printResult(password, hash); } free(one); free(two); free(three); } } return 0;}`
Output:
```apple => 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm => 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
C#
`using System;using System.Linq;using System.Text;using System.Threading.Tasks; class Program{ static void Main(string[] args) { Parallel.For(0, 26, a => { byte[] password = new byte[5]; byte[] hash; byte[] one = StringHashToByteArray("1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad"); byte[] two = StringHashToByteArray("3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b"); byte[] three = StringHashToByteArray("74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"); password[0] = (byte)(97 + a); var sha = System.Security.Cryptography.SHA256.Create(); for (password[1] = 97; password[1] < 123; password[1]++) for (password[2] = 97; password[2] < 123; password[2]++) for (password[3] = 97; password[3] < 123; password[3]++) for (password[4] = 97; password[4] < 123; password[4]++) { hash = sha.ComputeHash(password); if (matches(one, hash) || matches(two, hash) || matches(three, hash)) Console.WriteLine(Encoding.ASCII.GetString(password) + " => " + BitConverter.ToString(hash).ToLower().Replace("-", "")); } }); } static byte[] StringHashToByteArray(string s) { return Enumerable.Range(0, s.Length / 2).Select(i => (byte)Convert.ToInt16(s.Substring(i * 2, 2), 16)).ToArray(); } static bool matches(byte[] a, byte[] b) { for (int i = 0; i < 32; i++) if (a[i] != b[i]) return false; return true; }}`
Output:
```apple => 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm => 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Clojure
`(ns rosetta.brute-force (:require [clojure.math.combinatorics :refer [selections]]) ;; https://github.com/clojure/math.combinatorics (:import [java.util Arrays] [java.security MessageDigest])) ;;https://rosettacode.org/wiki/Parallel_Brute_Force (def targets ;; length = 5 ["1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad" "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b" "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"]) ;; HELPER/UTIL fns;;================= (defn digest "Given a byte-array <bs> returns its hash (also a byte-array)." ^bytes [^MessageDigest md ^bytes bs] (.digest md bs)) (defn char-range "Helper fn for easily producing character ranges." [start end] (map char (range (int start) (inc (int end))))) (def low-case-eng-bytes "Our search-space (all lower case english characters converted to bytes)." (map byte (char-range \a \z))) (defn hex->bytes "Converts a hex string to a byte-array." ^bytes [^String hex] (let [len (.length hex) ret (byte-array (/ len 2))] (run! (fn [i] (aset ret (/ i 2) ^byte (unchecked-add-int (bit-shift-left (Character/digit (.charAt hex i) 16) 4) (Character/digit (.charAt hex (inc i)) 16)))) (range 0 len 2)) ret)) (defn bytes->hex "Converts a byte-array to a hex string." [^bytes bs] (.toString ^StringBuilder (areduce bs idx ret (StringBuilder.) (doto ret (.append (format "%02x" (aget bs idx))))))) ;; MAIN LOGIC;;=========== (defn check-candidate "Checks whether the SHA256 hash of <candidate> (a list of 5 bytes), matches <target>. If it does, returns that hash as a hex-encoded String. Otherwise returns nil." [^bytes target sha256 candidate] (let [candidate-bytes (byte-array candidate) ^bytes candidate-hash (sha256 candidate-bytes)] (when (Arrays/equals target candidate-hash) (let [answer (String. candidate-bytes)] (println "Answer found for:" (bytes->hex candidate-hash) "=>" answer) answer)))) (defn sha256-brute-force "Top level function. Returns a list with the 3 answers." [space hex-hashes] (->> hex-hashes (map hex->bytes) ;; convert the hex strings to bytes (pmap ;; parallel map the checker-fn (fn [target-bytes] (let [message-digest (MessageDigest/getInstance "SHA-256") ;; new digest instance per thread sha256 (partial digest message-digest)] (some (partial check-candidate target-bytes sha256) (selections space 5))))))) `
Output:
```Answer found for: 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b => apple
Answer found for: 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f => mmmmm
Common Lisp
Library: lparallel
`(defpackage #:parallel-brute-force (:use #:cl #:lparallel)) (in-package #:parallel-brute-force) (defparameter *alphabet* "abcdefghijklmnopqrstuvwxyz")(defparameter *hash0* "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad")(defparameter *hash1* "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b")(defparameter *hash2* "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f")(defparameter *kernel-size* 7) (defun sha-256 (input) (ironclad:byte-array-to-hex-string (ironclad:digest-sequence :sha256 (ironclad:ascii-string-to-byte-array input)))) (defun call-with-5-char-string (fun first-char) (loop with str = (make-array 5 :element-type 'character :initial-element first-char) for c1 across *alphabet* do (setf (char str 1) c1) (loop for c2 across *alphabet* do (setf (char str 2) c2) (loop for c3 across *alphabet* do (setf (char str 3) c3) (loop for c4 across *alphabet* do (setf (char str 4) c4) (funcall fun (copy-seq str))))))) (defmacro with-5-char-string ((str first-char) &body body) `(call-with-5-char-string (lambda (,str) ,@body) ,first-char)) (defun find-passwords-with (first-char) (let (results) (with-5-char-string (str first-char) (let ((hash (sha-256 str))) (when (or (string= hash *hash0*) (string= hash *hash1*) (string= hash *hash2*)) (push (list str hash) results)))) (nreverse results))) (defun find-passwords () (setf *kernel* (make-kernel *kernel-size*)) (let ((results (unwind-protect (pmapcan #'find-passwords-with *alphabet*) (end-kernel)))) (dolist (r results) (format t "~A: ~A~%" (first r) (second r)))))`
Output:
```apple: 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm: 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
D
There is at least one more method not shown for doing the task in parallel, which uses the std.concurrency module instead.
`import std.digest.sha;import std.parallelism;import std.range;import std.stdio; // Find the five lower-case letter strings representing the following sha256 hashesimmutable p1 = cast(ubyte[32]) x"1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad";immutable p2 = cast(ubyte[32]) x"3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b";immutable p3 = cast(ubyte[32]) x"74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"; void main() { import std.datetime.stopwatch; auto sw = StopWatch(AutoStart.yes); // Switch these top loops to toggle between non-parallel and parrallel solutions. // foreach(char a; 'a'..'z'+1) { foreach(i, a; taskPool.parallel(iota('a', 'z'+1))) { char[5] psw; psw[0] = cast(char) a; foreach(char b; 'a'..'z'+1) { psw[1] = b; foreach(char c; 'a'..'z'+1) { psw[2] = c; foreach(char d; 'a'..'z'+1) { psw[3] = d; foreach(char e; 'a'..'z'+1) { psw[4] = e; auto hash = psw.sha256Of; if (equal(hash, p1) || equal(hash, p2) || equal(hash, p3)) { writefln("%s <=> %(%x%)", psw, hash); } } } } } } sw.stop; writeln(sw.peek);} //Specialization that supports static arrays toobool equal(T)(const T[] p, const T[] q) { if (p.length != q.length) { return false; } for(int i=0; i<p.length; i++) { if (p[i] != q[i]) { return false; } } return true;}`
Output:
Parallel run time: 9 secs, 684 ms, 678 ╬╝s, and 6 hnsecs
Sequential run time: 29 secs, 298 ms, and 837 ╬╝s
```apple <=> 3a7bd3e236a3d29eea436fcfb7e44c735d117c42d1c183542b6b9942dd4f1b
mmmmm <=> 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
F#
` (*Nigel Galloway February 21st., 2017*)let N n i g e l = let G = function |"3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b"->Some(string n+string i+string g+string e+string l) |"74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"->Some(string n+string i+string g+string e+string l) |"1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad"->Some(string n+string i+string g+string e+string l) |_->None G ([|byte n;byte i;byte g;byte e;byte l|]|>System.Security.Cryptography.SHA256.Create().ComputeHash|>Array.map(fun (x:byte)->System.String.Format("{0:x2}",x))|>String.concat "")open System.Threading.Taskslet n1 = Task.Factory.StartNew(fun ()->['a'..'m']|>List.collect(fun n->['a'..'m']|>List.collect(fun i->['a'..'m']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n2 = Task.Factory.StartNew(fun ()->['a'..'m']|>List.collect(fun n->['a'..'m']|>List.collect(fun i->['n'..'z']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n3 = Task.Factory.StartNew(fun ()->['a'..'m']|>List.collect(fun n->['n'..'z']|>List.collect(fun i->['a'..'m']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n4 = Task.Factory.StartNew(fun ()->['a'..'m']|>List.collect(fun n->['n'..'z']|>List.collect(fun i->['n'..'z']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n5 = Task.Factory.StartNew(fun ()->['n'..'z']|>List.collect(fun n->['a'..'m']|>List.collect(fun i->['a'..'m']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n6 = Task.Factory.StartNew(fun ()->['n'..'z']|>List.collect(fun n->['a'..'m']|>List.collect(fun i->['n'..'z']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n7 = Task.Factory.StartNew(fun ()->['n'..'z']|>List.collect(fun n->['n'..'z']|>List.collect(fun i->['a'..'m']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l))))))let n8 = Task.Factory.StartNew(fun ()->['n'..'z']|>List.collect(fun n->['n'..'z']|>List.collect(fun i->['n'..'z']|>List.collect(fun g->['a'..'z']|>List.collect(fun e->['a'..'z']|>List.choose(fun l->N n i g e l)))))) for r in n1.Result@n2.Result@n3.Result@n4.Result@n5.Result@n6.Result@n7.Result@n8.Result do printfn "%s" r `
Output:
```mmmmm
apple
zyzzx
```
Go
This solution runs 26 goroutines, one for each possible password first letter. Goroutines run in parallel on a multicore system.
`package main import ( "crypto/sha256" "encoding/hex" "log" "sync") var hh = []string{ "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad", "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f",} func main() { log.SetFlags(0) hd := make([][sha256.Size]byte, len(hh)) for i, h := range hh { hex.Decode(hd[i][:], []byte(h)) } var wg sync.WaitGroup wg.Add(26) for c := byte('a'); c <= 'z'; c++ { go bf4(c, hd, &wg) } wg.Wait()} func bf4(c byte, hd [][sha256.Size]byte, wg *sync.WaitGroup) { p := []byte("aaaaa") p[0] = c p1 := p[1:]p: for { ph := sha256.Sum256(p) for i, h := range hd { if h == ph { log.Println(string(p), hh[i]) } } for i, v := range p1 { if v < 'z' { p1[i]++ continue p } p1[i] = 'a' } wg.Done() return }}`
Output:
```zyzzx 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
apple 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
```
Java
Works with: Java version 1.5
This example uses Java's multithreading capabilities. Note that the Java Virtual Machine will use native Threads if the underlying platform supprts them. If there is no native thread support, the Java VM will emulate threads. This implementation runs 3 threads (one per hash to crack), and short-stops when a match for a hash is found.
`import javax.xml.bind.DatatypeConverter;import java.security.MessageDigest;import java.security.NoSuchAlgorithmException;import java.util.Arrays;import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors; /** * "Main Program" that does the parallel processing */public class ParallelBruteForce { public static void main(String[] args) throws NoSuchAlgorithmException { //the hashes to be cracked String[] hashes = {"1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad", "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"}; //An ExecutorService is a high-level parallel programming facility, that can execute a number of tasks //the FixedThreadPool is an ExecutorService that uses a configurable number of parallel threads ExecutorService executorService = Executors.newFixedThreadPool(3); //Submit one Task per hash to the thread po for (String hash : hashes) { executorService.submit(new Forcer(hash)); } //An ExecutorSerice must be shut down properly (this also causes the program to await termination of // all pending tasks in the thread pool) executorService.shutdown(); }} /** * The Class that contains the actual brute-forcing task. * <p> * It implements the build-in Interface "Runnable", so it can be run on a Thread or a Thread-Execution-Facility * (such as an ExecutorService). */class Forcer implements Runnable { private static final int LENGTH = 5; //These will sore the hash to be cracked in both bytes (required for comparison) and String representation // (required for output) private final byte[] crackMe; private final String crackMeString; //The MessageDigest does the SHA-256 caclulation. Note that this may throw a NoSuchAlgorithmException when there // is no SHA-256 implementation in the local standard libraries (but that algorithm is mandatory, so this code // probably will never throw that Excpetion private final MessageDigest digest = MessageDigest.getInstance("SHA-256"); public Forcer(String crackMe) throws NoSuchAlgorithmException { this.crackMeString = crackMe; this.crackMe = DatatypeConverter.parseHexBinary(crackMe); } @Override public void run() { String match = ""; //all loops use this array for their counters. This is very dirty and should never be done in production! char[] chars = new char[LENGTH]; //used for short-stopping when a match is found - one could abuse the match-variable for this, but this is // much clearer boolean done = false; for (chars[0] = 'a'; chars[0] <= 'z' && !done; chars[0]++) { for (chars[1] = 'a'; chars[1] <= 'z' && !done; chars[1]++) { for (chars[2] = 'a'; chars[2] <= 'z' && !done; chars[2]++) { for (chars[3] = 'a'; chars[3] <= 'z' && !done; chars[3]++) { for (chars[4] = 'a'; chars[4] <= 'z' && !done; chars[4]++) { //the String creation is necessary to get the encoding right String canidate = new String(chars); //genenrate SHA-256 hash using Java's standard facilities byte[] hash = digest.digest(canidate.getBytes()); if (Arrays.equals(hash, crackMe)) { match = canidate; done = true; } } } } } } System.out.println(String.format("Hash %s has the following match : %s", crackMeString, match)); }} `
Output:
```Hash 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b has the following match : apple
Hash 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f has the following match : mmmmm
Hash 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad has the following match : zyzzx```
Julia
`@everywhere using SHA @everywhere function bruteForceRange(startSerial, numberToDo) targets = ["1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad", "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"] targets = map(hex2bytes, targets) for count = 1 : numberToDo password = [UInt8(97 + x) for x in digits(UInt8, startSerial + count, 26, 5)] hashbytes = sha256(password) if (hashbytes[1] == 0x11 || hashbytes[1] == 0x3a || hashbytes[1] == 0x74) && findfirst(targets, hashbytes) > 0 hexstring = join(hex(x,2) for x in hashbytes) passwordstring = join(map(Char, password)) println("\$passwordstring --> \$hexstring") end end return 0end @everywhere perThread = div(26^5, Sys.CPU_CORES)pmap(x -> bruteForceRange(x * perThread, perThread), 0:Sys.CPU_CORES-1) `
Output:
```From worker 2: apple --> 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
From worker 3: zyzzx --> 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
From worker 4: mmmmm --> 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f```
Kotlin
Translation of: C#
`// version 1.1.51 import java.security.MessageDigest fun stringHashToByteHash(hash: String): ByteArray { val ba = ByteArray(32) for (i in 0 until 64 step 2) ba[i / 2] = hash.substring(i, i + 2).toInt(16).toByte() return ba} fun ByteArray.matches(other: ByteArray): Boolean { for (i in 0 until 32) { if (this[i] != other[i]) return false } return true} fun main(args: Array<String>) { val stringHashes = listOf( "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad", "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f" ) val byteHashes = List(3) { stringHashToByteHash(stringHashes[it]) } val letters = List(26) { (97 + it).toByte() } letters.stream().parallel().forEach { val md = MessageDigest.getInstance("SHA-256") val range = 97..122 val pwd = ByteArray(5) pwd[0] = it for (i1 in range) { pwd[1] = i1.toByte() for (i2 in range) { pwd[2] = i2.toByte() for (i3 in range) { pwd[3] = i3.toByte() for (i4 in range) { pwd[4] = i4.toByte() val ba = md.digest(pwd) for (j in 0..2) { if (ba.matches(byteHashes[j])) { val password = pwd.toString(Charsets.US_ASCII) println("\$password => \${stringHashes[j]}") break } } } } } } }}`
Output:
```mmmmm => 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
apple => 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
```
Mathematica
`testPassword[pass_String] := If[MemberQ[{16^^1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad, 16^^3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b, 16^^74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f}, Hash[pass, "SHA256"]], Print[pass]]; chars=CharacterRange["a","z"]; ParallelDo[ testPassword[StringJoin[a, b, c, d, e]], {a, chars}, {b, chars}, {c, chars}, {d, chars}, {e, chars}]`
Modula-2
`MODULE PBF;FROM FormatString IMPORT FormatString;FROM SHA256 IMPORT SHA256,Create,Destroy,HashBytes,Finalize,GetHash;FROM SYSTEM IMPORT ADR,ADDRESS,BYTE;FROM Terminal IMPORT Write,WriteString,WriteLn,ReadChar;FROM Threads IMPORT Thread,CreateThread,WaitForThreadTermination; PROCEDURE PrintHexBytes(str : ARRAY OF BYTE; limit : INTEGER);VAR buf : ARRAY[0..7] OF CHAR; i,v : INTEGER;BEGIN i := 0; WHILE i<limit DO v := ORD(str[i]); IF v < 16 THEN WriteString("0") END; FormatString("%h", buf, v); WriteString(buf); INC(i); ENDEND PrintHexBytes; PROCEDURE Check(str : ARRAY OF CHAR);TYPE HA = ARRAY[0..31] OF BYTE;CONST h1 = HA{3aH, 7bH, 0d3H, 0e2H, 36H, 0aH, 3dH, 29H, 0eeH, 0a4H, 36H, 0fcH, 0fbH, 7eH, 44H, 0c7H, 35H, 0d1H, 17H, 0c4H, 2dH, 1cH, 18H, 35H, 42H, 0bH, 6bH, 99H, 42H, 0ddH, 4fH, 1bH}; h2 = HA{74H, 0e1H, 0bbH, 62H, 0f8H, 0daH, 0bbH, 81H, 25H, 0a5H, 88H, 52H, 0b6H, 3bH, 0dfH, 6eH, 0aeH, 0f6H, 67H, 0cbH, 56H, 0acH, 7fH, 7cH, 0dbH, 0a6H, 0d7H, 30H, 5cH, 50H, 0a2H, 2fH}; h3 = HA{11H, 15H, 0ddH, 80H, 0fH, 0eaH, 0acH, 0efH, 0dfH, 48H, 1fH, 1fH, 90H, 70H, 37H, 4aH, 2aH, 81H, 0e2H, 78H, 80H, 0f1H, 87H, 39H, 6dH, 0b6H, 79H, 58H, 0b2H, 07H, 0cbH, 0adH};VAR hash : SHA256; out : ARRAY[0..31] OF BYTE; i : CARDINAL; match : BOOLEAN;BEGIN hash := Create(); HashBytes(hash, ADR(str), HIGH(str)+1); Finalize(hash); GetHash(hash, out); Destroy(hash); match := TRUE; FOR i:=0 TO HIGH(out) DO IF out[i] # h1[i] THEN match := FALSE; BREAK END END; IF match THEN WriteString(str); WriteString(" "); PrintHexBytes(out, 32); WriteLn; RETURN END; match := TRUE; FOR i:=0 TO HIGH(out) DO IF out[i] # h2[i] THEN match := FALSE; BREAK END END; IF match THEN WriteString(str); WriteString(" "); PrintHexBytes(out, 32); WriteLn; RETURN END; match := TRUE; FOR i:=0 TO HIGH(out) DO IF out[i] # h3[i] THEN match := FALSE; BREAK END END; IF match THEN WriteString(str); WriteString(" "); PrintHexBytes(out, 32); WriteLn ENDEND Check; PROCEDURE CheckWords(a : CHAR);VAR word : ARRAY[0..4] OF CHAR; b,c,d,e : CHAR;BEGIN word[0] := a; FOR b:='a' TO 'z' DO word[1] := b; FOR c:='a' TO 'z' DO word[2] := c; FOR d:='a' TO 'z' DO word[3] := d; FOR e:='a' TO 'z' DO word[4] := e; Check(word) END END END ENDEND CheckWords; PROCEDURE CheckAF(ptr : ADDRESS) : CARDINAL;VAR a : CHAR;BEGIN FOR a:='a' TO 'f' DO CheckWords(a) END; RETURN 0END CheckAF; PROCEDURE CheckGM(ptr : ADDRESS) : CARDINAL;VAR a : CHAR;BEGIN FOR a:='g' TO 'm' DO CheckWords(a) END; RETURN 0END CheckGM; PROCEDURE CheckNS(ptr : ADDRESS) : CARDINAL;VAR a : CHAR;BEGIN FOR a:='n' TO 's' DO CheckWords(a) END; RETURN 0END CheckNS; PROCEDURE CheckTZ(ptr : ADDRESS) : CARDINAL;VAR a : CHAR;BEGIN FOR a:='t' TO 'z' DO CheckWords(a) END; RETURN 0END CheckTZ; VAR t1,t2,t3,t4 : Thread; s1,s2,s3,s4 : CARDINAL;BEGIN CreateThread(t1,CheckAF,NIL,0,TRUE); CreateThread(t2,CheckGM,NIL,0,TRUE); CreateThread(t3,CheckNS,NIL,0,TRUE); CreateThread(t4,CheckTZ,NIL,0,TRUE); WaitForThreadTermination(t1,-1,s1); WaitForThreadTermination(t2,-1,s2); WaitForThreadTermination(t3,-1,s3); WaitForThreadTermination(t4,-1,s4); WriteString("Done"); WriteLn; ReadCharEND PBF.`
Output:
```apple 3A7BD3E2360A3D29EEA436FCFB7E44C735D117C42D1C1835420B6B9942DD4F1B
mmmmm 74E1BB62F8DABB8125A58852B63BDF6EAEF667CB56AC7F7CDBA6D7305C50A22F
Done```
Perl
Uses threads library to do naive search using 26 threads ("aaaaa" .. "azzzz", "baaaa" .. "bzzzz", etc.). No effort is made to early exit.
`use Digest::SHA qw/sha256_hex/;use threads;use threads::shared;my @results :shared; print "\$_ : ",join(" ",search(\$_)), "\n" for (qw/ 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f/); sub search { my \$hash = shift; @results = (); \$_->join() for map { threads->create('tsearch', \$_, \$hash) } 0..25; return @results;} sub tsearch { my(\$tnum, \$hash) = @_; my \$s = chr(ord("a")+\$tnum) . "aaaa"; for (1..456976) { # 26^4 push @results, \$s if sha256_hex(\$s) eq \$hash; \$s++; }}`
Output:
```1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad : zyzzx
3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b : apple
74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f : mmmmm
```
Perl 6
This solution can be changed from parallel to serial by removing the `.race` method.
`use Digest::SHA;constant @alpha2 = [X~] <a m p y z> xx 2;constant @alpha3 = [X~] <e l m p x z> xx 3; my %WANTED = set < 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad>; sub find_it ( \$first_two ) { return gather for \$first_two «~« @alpha3 -> \$password { my \$digest_hex = sha256(\$password).list.fmt('%02x', ''); take "\$password => \$digest_hex" if %WANTED{\$digest_hex}; }} .say for flat @alpha2.race(:1batch).map: {.&find_it.cache};`
Output:
```apple => 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm => 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Testers can adjust the run speed by replacing the @alpha constants with any of the below:
` # True to actual RC task, but slowestconstant @alpha2 = 'aa' .. 'zz';constant @alpha3 = 'aaa' .. 'zzz';# Reduced alphabets for speed during developmentconstant @alpha2 = [X~] <a m p y z> xx 2;constant @alpha3 = [X~] <e l m p x z> xx 3;# Alphabets reduced by position for even more speedconstant @alpha2 = [X~] <a m z>, <p m y>;constant @alpha3 = [X~] <m p z>, <l m z>, <e m x>;# Completely cheatingconstant @alpha2 = <ap mm zy>;constant @alpha3 = <ple mmm zzx>;`
Python
`import multiprocessingfrom hashlib import sha256 def HashFromSerial(serial): divisor = 456976 letters = [] for i in range(5): letter, serial = divmod(serial, divisor) letters.append( 97 + int(letter) ) divisor /= 26 return (letters, sha256(bytes(letters)).digest()) def main(): h1 = bytes().fromhex("1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad") h2 = bytes().fromhex("3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b") h3 = bytes().fromhex("74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f") numpasswords = int(26 ** 5) chunksize = int(numpasswords / multiprocessing.cpu_count()) with multiprocessing.Pool() as p: for (letters, digest) in p.imap_unordered(HashFromSerial, range(numpasswords), chunksize): if digest == h1 or digest == h2 or digest == h3: password = "".join(chr(x) for x in letters) print(password + " => " + digest.hex()) if __name__ == "__main__": main()`
Output:
```apple => 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm => 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Racket
Tests are included firstly to check it works, but they also provide an opportunity to time the single threaded version.
`#lang racket/base(require racket/place racket/list racket/match ;; requires sha package. install it in DrRacket's "File/Install Package..." ;; or with raco: ;; % raco pkg install sha sha (only-in openssl/sha1 hex-string->bytes)) (define (brute css targs) (define (sub-work i) (let ((cs (list-ref css i))) (in-range (car cs) (cdr cs)))) (define-values (as bs cs ds es) (apply values (map sub-work (range 5)))) (define s (make-bytes 5)) (for*/list ((a as) #:when (bytes-set! s 0 a) (b bs) #:when (bytes-set! s 1 b) (c cs) #:when (bytes-set! s 2 c) (d ds) #:when (bytes-set! s 3 d) (e es) #:when (bytes-set! s 4 e) (h (in-value (sha256 s))) (t (in-list targs)) #:when (bytes=? t h)) (eprintf "found ~s -> ~s~%" t s) (cons (bytes-copy s) t))) ;; ---------------------------------------------------------------------------------------------------(unless (place-enabled?) (error "We're using places... they're not enabled!")) (define target-list (map hex-string->bytes (list "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad" "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b" "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"))) (define (run-place/assign-task sub-task) (define there (place here (match-define (cons work targs) (place-channel-get here)) (place-channel-put here (brute work targs)))) (place-channel-put there (cons sub-task target-list)) there) (define (task->subtasks css n-tasks) (match css [(list (and initial-range (cons A Z+)) common-tail ...) (define step (quotient (+ n-tasks (- Z+ A)) n-tasks)) (for/list ((a (in-range A Z+ step))) ;; replace the head with a sub-task head (cons (cons a (min (+ a step) Z+)) common-tail))])) (define readable-pair (match-lambda [(cons x (app bytes->hex-string s)) (cons x s)])) (define (parallel-brute css (n-tasks (processor-count))) (define the-places (map run-place/assign-task (task->subtasks css n-tasks))) (define collected-results (append* (map place-channel-get the-places))) (map readable-pair collected-results)) (define 5-char-lowercase-work (make-list 5 (cons (char->integer #\a) (add1 (char->integer #\z))))) ;; ---------------------------------------------------------------------------------------------------(module+ main (time (parallel-brute 5-char-lowercase-work))) ;; ---------------------------------------------------------------------------------------------------(module+ test (require rackunit) (check-equal? (bytes->hex-string (sha256 #"mmmmm")) "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f" "SHA-256 works as expected") (check-equal? (hex-string->bytes "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f") #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/" "This is the raw value we'll be hashing to") (define m-idx (char->integer #\m)) (define m-idx+ (add1 m-idx)) (check-equal? (brute (make-list 5 (cons m-idx m-idx+)) target-list) (list (cons #"mmmmm" #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/"))) ;; Brute works without parallelism ;; check when you have the time... it takes a minute (literally) (check-equal? (time (brute 5-char-lowercase-work target-list)) '((#"apple" . #":{\323\3426\n=)\356\2446\374\373~D\3075\321\27\304-\34\0305B\vk\231B\335O\e") (#"mmmmm" . #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/") (#"zyzzx" . #"\21\25\335\200\17\352\254\357\337H\37\37\220p7J*\201\342x\200\361\2079m\266yX\262\a\313\255")) "without parallelism, it works"))`
Output:
Test phase of run:
```found #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/" -> #"mmmmm"
found #":{\323\3426\n=)\356\2446\374\373~D\3075\321\27\304-\34\0305B\vk\231B\335O\e" -> #"apple"
found #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/" -> #"mmmmm"
found #"\21\25\335\200\17\352\254\357\337H\37\37\220p7J*\201\342x\200\361\2079m\266yX\262\a\313\255" -> #"zyzzx"
cpu time: 19593 real time: 19581 gc time: 2247
```
Main phase of run:
```found #"\21\25\335\200\17\352\254\357\337H\37\37\220p7J*\201\342x\200\361\2079m\266yX\262\a\313\255" -> #"zyzzx"
found #":{\323\3426\n=)\356\2446\374\373~D\3075\321\27\304-\34\0305B\vk\231B\335O\e" -> #"apple"
found #"t\341\273b\370\332\273\201%\245\210R\266;\337n\256\366g\313V\254\177|\333\246\3270\\P\242/" -> #"mmmmm"
cpu time: 30641 real time: 4681 gc time: 0
'((#"apple" . "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b")
(#"mmmmm" . "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f")
Rust
In this solution the number of threads is the number of logical processors on the machine. `distribute_work()` distributes the work more or less equally between the threads.
`// [dependencies]// rust-crypto = "0.2.36"// num_cpus = "1.7.0"// hex = "0.2.0" extern crate crypto;extern crate num_cpus;extern crate hex; use std::thread;use std::cmp::min;use crypto::sha2::Sha256;use crypto::digest::Digest;use hex::{FromHex, ToHex}; fn main() { let hashes = vec![ decode("1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad"), decode("3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b"), decode("74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"), ]; let mut threads = Vec::new(); let mut ranges = distribute_work(); while let Some(range) = ranges.pop() { let hashes = hashes.clone(); threads.push(thread::spawn( move || search(range.0, range.1, hashes.clone()), )); } while let Some(t) = threads.pop() { t.join().ok(); }} fn search(from: [u8; 5], to: [u8; 5], hashes: Vec<[u8; 256 / 8]>) { let mut password = from.clone(); while password <= to { let mut sha256 = Sha256::new(); sha256.input(&password); let mut result = [0u8; 256 / 8]; sha256.result(&mut result); for hash in hashes.iter() { if *hash == result { println!( "{}{}{}{}{} {}", password[0] as char, password[1] as char, password[2] as char, password[3] as char, password[4] as char, hash.to_hex() ); } } password = next(&password); } } fn next(password: &[u8; 5]) -> [u8; 5] { let mut result = password.clone(); for i in (0..result.len()).rev() { if result[i] == b'z' { if i == 0 { result[i] = b'z' + 1; } else { result[i] = b'a'; } } else { result[i] += 1; break; } } result.clone()} fn distribute_work() -> Vec<([u8; 5], [u8; 5])> { let mut ranges = Vec::new(); let num_cpus = min(num_cpus::get(), 26) as u8; let div = 25 / num_cpus; let mut remainder = 25 % num_cpus; let mut from = b'a'; while from < b'z' { let to = from + div + if remainder > 0 { remainder -= 1; 1 } else { 0 }; ranges.push(( [from, from, from, from, from + 1].clone(), [to, to, to, to, to].clone(), )); from = to; } ranges[0].0[4] = b'a'; ranges.clone()} fn decode(string: &str) -> [u8; 256 / 8] { let mut result = [0; 256 / 8]; let vec = Vec::from_hex(string).unwrap(); for i in 0..result.len() { result[i] = vec[i]; } result.clone()}`
Output:
```apple 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
mmmmm 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
```
zkl
The built in thread message passing object uses the OS to do the heavy lifting and, as a result, isn't suited to high through put (ie passing passwords to cracking threads, equally distributing passwords to each cracking thread). Instead, each thread gets a range of passwords to crack and use signals to coordinate. Way more code with the drawback that one thread may have to do all the work.
This was run on a Intel i7 4 core 8 thread Linux box.
Uses the message hashing extension library (DLL).
Translation of: C#
`var [const] MsgHash=Import.lib("zklMsgHash");var [const] gotEm=Atomic.Int(); // global signal for all threads const THREADS=9, // how we will split task, THREADS<=26 CHR_a="a".toAsc(); fcn crack(c,n,hashes){ // thread sha256:=MsgHash.SHA256; // the SHA-256 hash method, byte bucket bytes,hash := Data(),Data().howza(0); // byte buckets to reduce garbage production firstLtrs:=(c+CHR_a).walker(n); ltrs:=CHR_a.walker; // iterator starting at 97/"a" foreach a,b,c,d,e in (firstLtrs,ltrs(26),ltrs(26),ltrs(26),ltrs(26)){ if(not hashes2go) return(); // all cracked, stop, not really needed bytes.clear(a,b,c,d,e); // recycle Data, faster than creating Strings sha256(bytes,1,hash); // put hash in hash if(hashes.holds(hash)){ println(bytes.text," --> ",hash.pump(String,"%02x".fmt)); hashes2go.dec(); // I cracked one, let mom thread know } }}`
`hashes:=T("3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b", "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f", "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad");// convert hex strings to binary; cuts down conversions during crackfcn hex2binary(s){ s.pump(Data,Void.Read,fcn(a,b){ (a+b).toInt(16) }) }hashes:=hashes.apply(hex2binary); hashes2go.set(hashes.len()); // number of codes to cracknum,xtra := 26/THREADS, 26%THREADS; // try for the most even spread over threadss:=0; do(THREADS){ // start threads n:=num + ((xtra-=1)>=0); crack.launch(s.toInt(),n,hashes); s+=n;}hashes2go.waitFor(0); // wait until all cracked, just exit, OS kills threads`
```mmmmm --> 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
apple --> 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
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3. Q3e
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Video Solution
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Similar Question 1
<p> Find the location of the vertex for the following parabolas and state the maximum or minimum value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = 2(x - 6)^2 + 3 </code></p>
Similar Question 2
<p>Sketch a graph of each quadratic relation. Pick the graph that's closest to your graph. State the transformation mapping.</p> <ul> <li><code class='latex inline'>y=(x-5)^2+3</code></li> </ul>
Similar Question 3
<p> Write and sketch the curve that satisfies <strong>all</strong> of the following conditions.</p> <ul> <li>i)The parabola has vertical shift of <code class='latex inline'>-3</code> from the origin</li> <li>ii)The parabola has horizontal shift of <code class='latex inline'>5</code> from the origin</li> <li>iii) The parabola passes through <code class='latex inline'>(0, -8)</code></li> </ul>
Similar Questions
Learning Path
L1 Quick Intro to Factoring Trinomial with Leading a
L2 Introduction to Factoring ax^2+bx+c
L3 Factoring ax^2+bx+c, ex1
Now You Try
<p>Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = -2(6x - 12)^2 - 1 </code></p>
<p> Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = -2(4 - 3x)^2 + 3 </code></p>
<p> Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = -2(2 - 3x)^2 - 1 </code></p>
<p>Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = -\frac{1}{16}(4x + 16)^2 + 2 </code></p>
<p>State the vertex and the axis of symmetry of each parabola.</p><p><strong>a)</strong> <code class='latex inline'>y=x^2+5</code></p><p><strong>b)</strong> <code class='latex inline'>y=(x-3)^2</code></p><p><strong>c)</strong> <code class='latex inline'>y=-3x^2</code></p><p><strong>d)</strong> <code class='latex inline'>y=(x+7)^2</code></p><p><strong>e)</strong> <code class='latex inline'>\displaystyle{y=\frac{1}{2}x^2}</code></p><p><strong>f)</strong> <code class='latex inline'>y=(x+6)^2+12</code></p>
<p>Sketch a graph of each quadratic relation. Pick the graph that's closest to your graph. State the transformation mapping.</p> <ul> <li><code class='latex inline'>y=(x-5)^2+3</code></li> </ul>
<p>Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = 2(5x - 10)^2 </code></p>
<p>Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = (2x - 2)^2 + 1 </code></p>
<p> Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = (3- x)^2 + 2 </code></p>
<p> Find the location of the vertex for the following parabolas and state the maximum or minimum value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = -2(x - 3)^2 - 11 </code></p>
<p> Find the location of the vertex for the following parabolas and state the maximum or minimum value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = 2(x - 6)^2 + 3 </code></p>
<p> Write and sketch the curve that satisfies <strong>all</strong> of the following conditions.</p> <ul> <li>i)The parabola has vertical shift of <code class='latex inline'>-3</code> from the origin</li> <li>ii)The parabola has horizontal shift of <code class='latex inline'>5</code> from the origin</li> <li>iii) The parabola passes through <code class='latex inline'>(0, -8)</code></li> </ul>
<p>Sketch a graph of each quadratic relation. Pick the graph that's closest to your graph. State the transformation mapping.</p> <ul> <li><code class='latex inline'>y=(x+1)^2-2</code></li> </ul>
<p> Find the location of the vertex for the following parabolas and state the maximum or minimum value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = (x - 2)^2 + 2 </code></p>
<p> Find the location of the vertex for the following parabolas by converting it into proper vertex form. State the max or min value. Sketch the graph.</p><p><code class='latex inline'>\displaystyle y = 2(5 -x)^2 + 1 </code></p>
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# Find common elements in three linked lists
• Difficulty Level : Medium
• Last Updated : 09 Jan, 2023
Given three linked lists, find all common elements among the three linked lists.
Examples:
```Input :
10 15 20 25 12
10 12 13 15
10 12 15 24 25 26
Output : 10 12 15
Input :
1 2 3 4 5
1 2 3 4 6 9 8
1 2 4 5 10
Output : 1 2 4```
Method 1 : (Simple)
Use three-pointers to iterate the given three linked lists and if any element common print that element.
Time complexity of the above solution will be O(N*N*N)
Method 2 : (Use Merge Sort)
In this method, we first sort the three lists and then we traverse the sorted lists to get the intersection.
Following are the steps to be followed to get intersection of three lists:
1. Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
2. Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
3. Sort the third Linked List using merge sort. This step takes O(pLogp) time. Refer this post for details of this step.
4. Linearly scan three sorted lists to get the intersection. This step takes O(m + n + p) time. This step can be implemented using the same algorithm as sorted arrays algorithm discussed here.
Time complexity of this method is O(mLogm + nLogn + plogp) which is better than method 1’s time complexity.
Method 3 : (Hashing)
Following are the steps to be followed to get intersection of three lists using hashing:
1. Create an empty hash table. Iterate through the first linked list and mark all the element frequency as 1 in the hash table. This step takes O(m) time.
2. Iterate through the second linked list and if current element frequency is 1 in hash table mark it as 2. This step takes O(n) time.
3. Iterate the third linked list and if the current element frequency is 2 in hash table mark it as 3. This step takes O(p) time.
4. Now iterate first linked list again to check the frequency of elements. if an element with frequency three exist in hash table, it will be present in the intersection of three linked lists. This step takes O(m) time.
Time complexity of this method is O(m + n + p) which is better than time complexity of method 1 and 2.
Below is the implementation of the above idea.
## C++
`// C++ program to find common element``// in three unsorted linked list``#include ``#define max 1000000``using` `namespace` `std;` `/* Link list node */``struct` `Node {`` ``int` `data;`` ``struct` `Node* next;``};` `/* A utility function to insert a node at the``beginning of a linked list */``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{`` ``struct` `Node* new_node =`` ``(``struct` `Node *)``malloc``(``sizeof``(``struct` `Node));`` ``new_node->data = new_data;`` ``new_node->next = (*head_ref);`` ``(*head_ref) = new_node;``}` `/* print the common element in between``given three linked list*/``void` `Common(``struct` `Node* head1,`` ``struct` `Node* head2, ``struct` `Node* head3)``{`` ` ` ``// Creating empty hash table;`` ``unordered_map<``int``, ``int``> hash;`` ` ` ``struct` `Node* p = head1;`` ``while` `(p != NULL) {`` ` ` ``// set frequency by 1`` ``hash[p->data] = 1;`` ``p = p->next;`` ``}`` ` ` ``struct` `Node* q = head2;`` ``while` `(q != NULL) {`` ` ` ``// if the element is already exist in the`` ``// linked list set its frequency 2`` ``if` `(hash.find(q->data) != hash.end())`` ``hash[q->data] = 2;`` ``q = q->next;`` ``}`` ` ` ``struct` `Node* r = head3;`` ``while` `(r != NULL) {`` ``if` `(hash.find(r->data) != hash.end() &&`` ``hash[r->data] == 2)`` ` ` ``// if the element frequency is 2 it means`` ``// its present in both the first and second`` ``// linked list set its frequency 3`` ``hash[r->data] = 3;`` ``r = r->next;`` ``}` ` ` ` ``for` `(``auto` `x : hash) {`` ` ` ``// if current frequency is 3 its means`` ``// element is common in all the given`` ``// linked list`` ``if` `(x.second == 3)`` ``cout << x.first << ``" "``;`` ``}``}` `// Driver code``int` `main()``{` ` ``// first list`` ``struct` `Node* head1 = NULL;`` ``push(&head1, 20);`` ``push(&head1, 5);`` ``push(&head1, 15);`` ``push(&head1, 10);` ` ``// second list`` ``struct` `Node* head2 = NULL;`` ``push(&head2, 10);`` ``push(&head2, 20);`` ``push(&head2, 15);`` ``push(&head2, 8);`` ` ` ``// third list`` ``struct` `Node* head3 = NULL;`` ``push(&head3, 10);`` ``push(&head3, 2);`` ``push(&head3, 15);`` ``push(&head3, 20);` ` ``Common(head1, head2, head3);`` ` ` ``return` `0;``}`
## Java
`// Java program to find common element``// in three unsorted linked list``import` `java.util.*;` `class` `GFG``{``static` `int` `max = ``1000000``;` `/* Link list node */``static` `class` `Node``{`` ``int` `data;`` ``Node next;``};` `/* A utility function to insert a node``at the beginning of a linked list */``static` `Node push(Node head_ref,`` ``int` `new_data)``{`` ``Node new_node = ``new` `Node();`` ``new_node.data = new_data;`` ``new_node.next = head_ref;`` ``head_ref = new_node;`` ``return` `head_ref;``}` `/* print the common element in between``given three linked list*/``static` `void` `Common(Node head1,`` ``Node head2,`` ``Node head3)``{`` ` ` ``// Creating empty hash table;`` ``HashMap hash = ``new` `HashMap();`` ` ` ``Node p = head1;`` ``while` `(p != ``null``)`` ``{`` ` ` ``// set frequency by 1`` ``hash. put(p.data, ``1``);`` ``p = p.next;`` ``}`` ` ` ``Node q = head2;`` ``while` `(q != ``null``)`` ``{`` ` ` ``// if the element is already exist in the`` ``// linked list set its frequency 2`` ``if` `(hash.containsKey(q.data))`` ``hash. put(q.data, ``2``);`` ``q = q.next;`` ``}`` ` ` ``Node r = head3;`` ``while` `(r != ``null``)`` ``{`` ``if` `(hash.containsKey(r.data)&&`` ``hash.get(r.data) == ``2``)`` ` ` ``// if the element frequency is 2 it means`` ``// its present in both the first and second`` ``// linked list set its frequency 3`` ``hash. put(r.data, ``3``);`` ``r = r.next;`` ``}` ` ``for` `(Map.Entry x : hash.entrySet())`` ``{`` ` ` ``// if current frequency is 3 its means`` ``// element is common in all the given`` ``// linked list`` ``if` `(x.getValue() == ``3``)`` ``System.out.println(x.getKey() + ``" "``);`` ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ` ` ``// first list`` ``Node head1 = ``null``;`` ``head1 = push(head1, ``20``);`` ``head1 = push(head1, ``5``);`` ``head1 = push(head1, ``15``);`` ``head1 = push(head1, ``10``);` ` ``// second list`` ``Node head2 = ``null``;`` ``head2 = push(head2, ``10``);`` ``head2 = push(head2, ``20``);`` ``head2 = push(head2, ``15``);`` ``head2 = push(head2, ``8``);`` ` ` ``// third list`` ``Node head3 = ``null``;`` ``head3 = push(head3, ``10``);`` ``head3 = push(head3, ``2``);`` ``head3 = push(head3, ``15``);`` ``head3 = push(head3, ``20``);` ` ``Common(head1, head2, head3);``}``}` `// This code is contributed by 29AjayKumar`
## Python3
`# Python3 program to find common element``# in three unsorted linked list``max` `=` `1000000` `# Link list node``class` `Node:`` ` ` ``def` `__init__(``self``, data):`` ` ` ``self``.data ``=` `data`` ``self``.``next` `=` `None`` ` `# A utility function to insert a node at the``# beginning of a linked list``def` `push( head_ref, new_data):`` ` ` ``new_node ``=` `Node(new_data)`` ``new_node.``next` `=` `head_ref`` ``head_ref ``=` `new_node`` ``return` `head_ref` `# Print the common element in between``# given three linked list``def` `Common(head1, head2, head3):`` ` ` ``# Creating empty hash table;`` ``hash` `=` `dict``()`` ` ` ``p ``=` `head1`` ``while` `(p !``=` `None``):`` ` ` ``# Set frequency by 1`` ``hash``[p.data] ``=` `1`` ``p ``=` `p.``next`` ` ` ``q ``=` `head2`` ` ` ``while` `(q !``=` `None``):`` ` ` ``# If the element is already exist in the`` ``# linked list set its frequency 2`` ``if` `(q.data ``in` `hash``):`` ``hash``[q.data] ``=` `2`` ` ` ``q ``=` `q.``next` ` ``r ``=` `head3`` ` ` ``while` `(r !``=` `None``):`` ``if` `(r.data ``in` `hash``) ``and` `hash``[r.data] ``=``=` `2``:`` ` ` ``# If the element frequency is 2 it means`` ``# its present in both the first and second`` ``# linked list set its frequency 3`` ``hash``[r.data] ``=` `3`` ` ` ``r ``=` `r.``next`` ` ` ``for` `x ``in` `hash``.keys():`` ` ` ``# If current frequency is 3 its means`` ``# element is common in all the given`` ``# linked list`` ``if` `(``hash``[x] ``=``=` `3``):`` ``print``(x, end ``=` `' '``)`` ` `# Driver code``if` `__name__``=``=``'__main__'``:`` ` ` ``# First list`` ``head1 ``=` `None`` ``head1 ``=` `push(head1, ``20``)`` ``head1 ``=` `push(head1, ``5``)`` ``head1 ``=` `push(head1, ``15``)`` ``head1 ``=` `push(head1, ``10``)`` ` ` ``# Second list`` ``head2 ``=` `None`` ``head2 ``=` `push(head2, ``10``)`` ``head2 ``=` `push(head2, ``20``)`` ``head2 ``=` `push(head2, ``15``)`` ``head2 ``=` `push(head2, ``8``)`` ` ` ``# Third list`` ``head3 ``=` `None`` ``head3 ``=` `push(head3, ``10``)`` ``head3 ``=` `push(head3, ``2``)`` ``head3 ``=` `push(head3, ``15``)`` ``head3 ``=` `push(head3, ``20``)`` ` ` ``Common(head1, head2, head3)`` ` `# This code is contributed by rutvik_56`
## C#
`// C# program to find common element``// in three unsorted linked list``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``static` `int` `max = 1000000;` `/* Link list node */``public` `class` `Node``{`` ``public` `int` `data;`` ``public` `Node next;``};` `/* A utility function to insert a node``at the beginning of a linked list */``static` `Node push(Node head_ref,`` ``int` `new_data)``{`` ``Node new_node = ``new` `Node();`` ``new_node.data = new_data;`` ``new_node.next = head_ref;`` ``head_ref = new_node;`` ``return` `head_ref;``}` `/* print the common element in between``given three linked list*/``static` `void` `Common(Node head1,`` ``Node head2,`` ``Node head3)``{`` ` ` ``// Creating empty hash table;`` ``Dictionary<``int``, `` ``int``> hash = ``new` `Dictionary<``int``,`` ``int``>();`` ` ` ``Node p = head1;`` ``while` `(p != ``null``)`` ``{`` ` ` ``// set frequency by 1`` ``hash.Add(p.data, 1);`` ``p = p.next;`` ``}`` ` ` ``Node q = head2;`` ``while` `(q != ``null``)`` ``{`` ` ` ``// if the element is already exist in the`` ``// linked list set its frequency 2`` ``if` `(hash.ContainsKey(q.data))`` ``hash[q.data] = 2;`` ``q = q.next;`` ``}`` ` ` ``Node r = head3;`` ``while` `(r != ``null``)`` ``{`` ``if` `(hash.ContainsKey(r.data)&&`` ``hash[r.data] == 2)`` ` ` ``// if the element frequency is 2 it means`` ``// its present in both the first and`` ``// second linked list set its frequency 3`` ``hash[r.data] = 3;`` ``r = r.next;`` ``}` ` ``foreach``(KeyValuePair<``int``, ``int``> x ``in` `hash)`` ``{`` ` ` ``// if current frequency is 3 its means`` ``// element is common in all the given`` ``// linked list`` ``if` `(x.Value == 3)`` ``Console.Write(x.Key + ``" "``);`` ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{`` ` ` ``// first list`` ``Node head1 = ``null``;`` ``head1 = push(head1, 20);`` ``head1 = push(head1, 5);`` ``head1 = push(head1, 15);`` ``head1 = push(head1, 10);` ` ``// second list`` ``Node head2 = ``null``;`` ``head2 = push(head2, 10);`` ``head2 = push(head2, 20);`` ``head2 = push(head2, 15);`` ``head2 = push(head2, 8);`` ` ` ``// third list`` ``Node head3 = ``null``;`` ``head3 = push(head3, 10);`` ``head3 = push(head3, 2);`` ``head3 = push(head3, 15);`` ``head3 = push(head3, 20);` ` ``Common(head1, head2, head3);``}``}` `// This code is contributed by Princi Singh`
## Javascript
``
Output:
`10 15 20`
Time Complexity : O(m + n + p)
Auxiliary Space : O(m + n + p)
My Personal Notes arrow_drop_up
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# Limit of Maximum
• Nov 16th 2011, 01:40 AM
younhock
Limit of Maximum
Anyone can help me ?
QUESTION :
for all x in (0,1), n is Natural number
f_n (x) := max{1/x , n }
find lim f_n (x) when n -> infinity.
• Nov 16th 2011, 07:10 AM
HallsofIvy
Re: Limit of Maximum
Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?
If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.
• Nov 16th 2011, 07:31 AM
younhock
Re: Limit of Maximum
Quote:
Originally Posted by HallsofIvy
Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?
If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.
i had updated the question to make it clear.
I was interested to know what is the limit function which f_n approaching to .
• Nov 16th 2011, 12:37 PM
HallsofIvy
Re: Limit of Maximum
It's pretty straight forward if you write out a few terms in the sequence of functions.
For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So $f_n(x)= 1/x$ for x< 1/n and $f_n(x)= n$ for x> 1/n, In particular, $f_1(x)= 1/x$ for all x in (0, 1). $f_2(x)= 1/x$ for 0< x<= 1/2, $f_2(x)= 2$ for 1/2<= x< 1. $f_3(x)= 1/x$ for 0< x<= 1/3 and $f_3(x)= 3$ for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, $f_n(x)> N$.
• Nov 16th 2011, 06:30 PM
younhock
Re: Limit of Maximum
Quote:
Originally Posted by HallsofIvy
It's pretty straight forward if you write out a few terms in the sequence of functions.
For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So $f_n(x)= 1/x$ for x< 1/n and $f_n(x)= n$ for x> 1/n, In particular, $f_1(x)= 1/x$ for all x in (0, 1). $f_2(x)= 1/x$ for 0< x<= 1/2, $f_2(x)= 2$ for 1/2<= x< 1. $f_3(x)= 1/x$ for 0< x<= 1/3 and $f_3(x)= 3$ for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, $f_n(x)> N$.
i see, so we will get f_n(x) -> infinity eventually. Thanks yea.
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Cody
# Problem 10. Determine whether a vector is monotonically increasing
Solution 1877349
Submitted on 17 Jul 2019 by Ricky Hoang
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [0 1 2 3 4]; assert(isequal(mono_increase(x),true));
2 Pass
x = [0]; assert(isequal(mono_increase(x),true));
3 Pass
x = [0 0 0 0 0]; assert(isequal(mono_increase(x),false));
4 Pass
x = [0 1 2 3 -4]; assert(isequal(mono_increase(x),false));
5 Pass
x = [-3 -4 2 3 4]; assert(isequal(mono_increase(x),false));
6 Pass
x = 1:.1:10; assert(isequal(mono_increase(x),true));
7 Pass
x = cumsum(rand(1,100)); x(5) = -1; assert(isequal(mono_increase(x),false));
8 Pass
x = cumsum(rand(1,50)); assert(isequal(mono_increase(x),true));
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# Is Algebra 2 Hard? Let’s Find Out!
Is Algebra 2 hard? This is a question we get pretty often, so we’ve decided to write a comprehensive blog post about it. Algebra 2 is a relatively difficult class, but it doesn’t need to be. Truth be told, you shouldn’t have a major problem with this type of algebra if you have a firm grasp on the prerequisites: exponents, integer arithmetic, pre-algebra, fraction arithmetic, geometry, and Algebra 1.
However, keep in mind that there are many things hanging in the balance when it comes to Algebra 2. Regardless of their major, college students need to pass Algebra 2 to get their Bachelor’s degree. Some programs also require this for students looking to get an Associate’s Degree. If you are looking for science degrees, you will also need to pass Pre-calculus and Calculus. However, let’s focus on Algebra 2 for now.
## Is Algebra 2 Hard? Well, It Depends
Is algebra 2 easy? Why is algebra 2 so hard? These are two very important questions. Let’s take a look at some of the things that certainly make this course a difficult one for college students (or even high school students):
• The Algebra 2 course usually lasts for an entire year in high school, so it’s definitely a comprehensive course. Many colleges, however, have just a 6-month course. This is simply not enough to teach all students the algebraic manipulation skills and the arithmetic (exponents or fractions are two good examples) one needs to solve complex problems.
• Many professors tend to assume that their students already know some concepts introduced by Algebra 2. This makes solving hard algebra 2 equations very difficult for students.
• In many cases, students are not even proficient in Algebra 1, which also causes some problems.
• Things like finding the equation of the exponential or solving a quadratic equation with two or more complex solutions are pretty difficult to understand for most students. These exercises require a lot of time and effort to resolve. One also needs a lot of practice with these kinds of equations.
### Important Algebra 2 Concepts and Skills
Let’s talk about some Algebra 2 concepts and skills that students need to know before even considering passing the course. Basically, Algebra 2 will require you to master 12 major areas of mathematics:
1. Transformations of functions
2. Modeling
3. Complex numbers
4. Polynomial factorization
5. Equations
7. Polynomial division
8. Polynomial graphs
9. Exponential models
10. Polynomial arithmetic
11. Trigonometry
12. Logarithms
Each area has a number of different things you should master. For instance, polynomial arithmetic requires you to know how to add or subtract polynomials, how to multiply polynomials by polynomials, and even obtain the average rate of change of polynomials.
As for the skills required to effectively solve Algebra 2 problems, you will need to master Algebra 1. And of course, the most important equation you need to learn how to solve is the classic “y=ax2 + bx + c”. Another very important skill you will need to master is the ability to apply the Algebra II concepts you learn to real-world problems.
Finally, many students ask us “do you need geometry for Algebra 2?” Even though Geometry is a prerequisite for Algebra 2 (so is Algebra 1), there isn’t really much to be scared about. You will encounter the occasional need for a geometry formula, most often in word problems. However, this does not mean you shouldn’t learn geometry. Remember, you need to pass that class if you want to get enrolled in Algebra 2 (and not prep track math class).
### The Difference Between Algebra 1 and 2
So, what’s the difference between Algebra 1 and 2? Most students who managed to understand Algebra 1 are wondering what’s in store for them during the Algebra 2 class. However, before we show you the differences between the two, we want to assure you that you can pass the course. You just need to dedicate more time to the Algebra 2 course. You need to practice. After all, practice makes perfect. You will also need to learn the ropes of mathematical writing. This also takes practice to master. Now, let’s take a look at some of the differences between the two courses:
Algebra 1 Algebra 2 Overview You will learn the concepts behind algebra. This means learning numbers, functions, measurement, using rational expressions, exponents, inequalities, linear equations, coordinates, and so on. Algebra 2 is a more complex course that aims to teach you many other areas of mathematics. For instance, you will learn polynomials, logarithms and logarithmic expressions, polynomials, trigonometry, sequences and series, and so on. Level Algebra 1 is like an introduction to algebra, so you will have a basic understanding of the various concepts after you finish the course. Algebra 2 is more advanced and builds upon the things you’ve learned in Algebra 1 to help you understand difficult concepts. Life Application Algebra 1 is very important to master if you want to understand the concepts behind Algebra 2 problem solving. In Algebra 2, you will learn to apply the simple things you’ve learned in Algebra 1 to solve complex, real-life problems. What You Will Learn Equations and inequalities are two of the most important things you will learn in Algebra 1. Algebra 2 will teach you more advanced equations, like the logarithmic and exponential ones. You will also learn about polynomial calculus, as well as quadratic equations, matrices and graphs. Course Progression You can’t progress to the Algebra 2 course without passing the Algebra 1 class. You can’t be enrolled in the Calculus class without first passing the Algebra 2 class. Concepts Covered Algebra 1 works only with integers and rational numbers, which are not very difficult to understand and use. Algebra 2 introduces and discusses complex numbers and uses them to solve complex problems.
### How Hard Is Pre Calculus Compared to Algebra 2
How hard is Algebra 2? How hard is pre calculus compared to Algebra 2? These two questions are difficult to answer. As you can see from the information we have provided above, there is a big difference between Algebra 1 and 2. The latter is more difficult. However, how difficult it would be for you is not something we can predict. If you are relatively good at understanding mathematical concepts, you should do very well in Algebra 2.
Also, you shouldn’t be too worried about precalculus either. This is basically a more difficult version of Algebra 2 that aims to prepare you for the Calculus class. You will have to learn more trigonometry, as well as work with some pretty nasty decimals and fractions. You should also expect to do some complex factoring and division work. Other things students find difficult are reasoning and proofs, and sets and set operations. However, you shouldn’t worry too much about these if you are willing to dedicate the time and effort required by this class. And remember, if you run into any trouble, our expert writers (all are ENL writers, by the way) will help you solve any problem in no time.
### Solving Difficult Algebra 2 Problems Quickly
Unfortunately, there is no easy way to solve difficult Algebra 2 problems quickly. Otherwise, why would anyone consider math difficult? And no, there is no software that can solve any problem correctly. The best way to solve these problems quickly is to learn everything about basic algebraic concepts and operations, learn all the problem solving concepts used for quick solutions (like the principle of inverses), learn all the basic transformations, and perhaps learn a few useful patterns for solving specific equations.
### Learn How to Be Good at Algebra
Getting your professor to award you high grades in Algebra 2 is not as easy as you think. These professors usually have pretty high standards and expectations from their students. So you need to learn how to be good at algebra without spending all your weekends on it. Here are some of the best tips our experienced math problem solvers and technical writers can give you:
• Try to solve the problems that were discussed in class – after each and every class. This way, you will understand the solutions and will know how to apply them to solve more complex problems further down the line.
• If you don’t understand something, don’t be afraid to ask. Ask your classmates, your parents or even your professor. Also, you will find a lot of help (even guidelines on how to solve certain problems and equations) on the Internet.
• Get prep online. You will be amazed by how much you can learn when our math professional explains everything to you. Also, you can ask questions and get the perfect answers in minutes. What better way to learn math quickly?
• Study and then study some more. You will need a lot of practice to solve complex problems, so it is extremely important to commit the time and effort necessary to understand all the concepts. If you fall behind, you’ll find it very difficult to catch up.
• Practice solving problems. Don’t just real the theoretical framework. Make sure you work on real equations, trying to solve them.
• Organize your time and sleep well. Most students get desperate before a test and spend day and night trying to learn Algebra 2. This is not the way to go. You will just burn yourself out in record time. To be efficient, you need to organize your time and dedicate two or three hours every day to mathematics. You can take the weekend off.
• You should not rely on online calculators or equation-solving software. These will just give you the answer. Your professor wants to see how you solve the problem step by step. You won’t get a decent grade using math programs, guaranteed.
### Can Somebody Do My Math Homework?
I think I need somebody to do my math. Where could I get some affordable assistance from? College and university students sometimes need some math help. It’s not something to be ashamed about. Trust us, our math experts help hundreds of students every month. You can get top-notch assistance with your math homework from the best experts on the Internet in mere minutes right here. Also, we offer tutoring and prep services for people who want to excel at not only Algebra 2, but also precalculus and even calculus.
We have experts online day and night, so you can get the help you need as fast as possible. Speaking of our experts, did you know that they all have at least a PhD degree in a field in mathematics? You will get custom academic content from a team of genuine experts who know what they’re talking about. Many of our experts are also former teachers, so they know exactly what your professor wants to see. Get cheap help from our company and we’ll make learning Algebra 2 a piece of cake!
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THIS
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# THIS - PowerPoint PPT Presentation
THIS. IS. Jeopardy. Your. With. Host. Ms Purcell. Jeopardy. Cylinders. Area . Volume. Rectangles. Circles. Prisms . 100. 100. 100. 100. 100. 100. 200. 200. 200. 200. 200. 200. 300. 300. 300. 300. 300. 300. 400. 400. 400. 400. 400. 400. 500. 500. 500. 500.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## THIS
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Presentation Transcript
1. THIS IS Jeopardy
2. Your With Host... Ms Purcell
3. Jeopardy Cylinders Area Volume Rectangles Circles Prisms 100 100 100 100 100 100 200 200 200 200 200 200 300 300 300 300 300 300 400 400 400 400 400 400 500 500 500 500 500 500
4. What is Area? A 100
5. Length x Width A 200
6. Sides cubed A 300
7. Base x height A 400
8. What is Volume B 100
9. Length x width C 100
10. Sides cubed C 300
11. DAILY DOUBLE DAILY DOUBLE Place A Wager C 400
12. 60 C 400
13. 16 C 500
14. What is the radius? D 200
15. What is Pi? D 300
16. The relationship between the diameter and the circumference. There are 3.14 diameters in one circumference D 300
17. 3.14 D 400
18. 3.14 x 4 = 12.56 D 500
19. A cube. E 200
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The Babson task is a kind of chess problem of the form "white to move and mate black in N moves against any defence" with the following play:
#White makes his first move.
#Black defends by promoting a pawn to queen, rook, bishop or knight.
#White responds by promoting a pawn to queen, rook, bishop or knight respectively (if black promoted to rook, so does white, if black promoted to knight, so does white and so on). No other promotion (or any other move) leads to mate in the stipulated number of moves.
The task is named after the first person to speculate about the possibility of such a problem, Joseph Ney Babson. It is regarded as one of the greatest challenges for a composer of chess problems to devise a satisfying Babson task problem, and for around half a century the task was considered to be near-impossible in directmate form.
Technically, the task can be regarded as a form of Allumwandlung with corresponding promotions by black and white (an Allumwandlung is a problem which contains, at some point in the solution, promotions to each of the four possible pieces - such problems had already been composed before Babson devised his task).
Chess diagram|=
tleft
Wolfgang Pauly, 1912
= 8 | | | | | | |bl| |= 7 |kl| | |pd|bl|pl|pl| |= 6 | | | |pl| | | | |= 5 | | | | | |pl| | |= 4 | | | | | | |pl|pl|= 3 |pd| |pd| | | | |kd|= 2 |pd|pl|pd| | | |rl| |= 1 | |bd|rl| |nl| | | |= a b c d e f g h
This 1912 problem by Wolfgang Pauly is, as it were, a three-quarter Babson task - three of black's promotions are matched by white. White to move and mate in four:
The key is 1.b3, after which there are the following lines:
:1... a1Q 2. f8Q Qb2 3. Qa8 Qxc1 4. Qf3 mate:1... a1R 2. f8R a2 3. Rf6 Kxh4 4. Rh6 mate:1... a1N 2. f8N a2 3. Ng6 Nxb3 4. Nf4 mate
However, this is not a full Babson, because 1... a1B 2. f8B does not work - white must instead play 2. f8Q, with similar play to above.
elfmate Babsons
The earliest Babson tasks are all in the form of a selfmate - this is where white, moving first, must force black to mate him against his will within a specified number of moves. In 1914, Babson himself published a selfmate which achieved the task, although three different white pawns shared the promotions. The first problem in which a single black and single white pawn were involved in the promotions was by Henry Wald Bettmann, and won 1st prize in the Babson Task Tourney 1925-26.
Chess diagram|=
tleft
Henry Wald Bettmann. 1st Prize in the 1925-26 Babson task tourney.
= 8 | | | | | | | |nl|= 7 |pl|pd|pd| | |pl| | |= 6 |rl|rd|kd| | | | |pl|= 5 |kl| |pl| | | | | |= 4 | |pl| | |pl| | |pl|= 3 | | |pl| | | | |ql|= 2 | | | | | |pd| |rl|= 1 | | | | | |nl|bl| |= a b c d e f g h
A selfmate in three.
The key move in Bettmann's problem (left) is 1.a8=B, after which the play goes:: 1...fxg1=Q 2. f8=Q::2... Qxf1/Qxc5 3.b5 Qxb5#::2... Q-any 3.AnyxQ Rxa6#:1...fxg1=R 2. f8=R R-any 3.anyxR Rxa6#:1...fxg1=B 2. f8=B B-any 3.anyxB Rxa6#:1... fxg1=N 2. f8=N N-any 3.anyxN Rxa6#
A number of other selfmate Babson tasks with one pawn of each colour doing all the promotions followed this one.
Directmate Babsons
Composing a Babson task problem in directmate form (where white moves first, and must checkmate black against any defence within a stipulated number of moves) was thought so difficult that very little effort was put into solving it until the 1960s, when Pierre Drumare began his work on the problem which would occupy him for the next twenty years or so. He managed to compose a Babson task problem using nightriders (a Fairy piece which moves like a knight, but can make any number of knight-like moves in the same direction in one go) instead of knights, but found it hard to devise one using normal pieces - because of their limited range, it is difficult to justify white promoting to a knight because of black promoting to one way over the other side of the board.
When Drumare did eventually succeed using conventional pieces in 1980, the result was regarded as highly unsatisfactory, even by Drumare himself. It is a mate in five (first published "Memorial Seneca", 1980):Chess diagram|=
tleft
Pierre Drumare, "Memorial Camil Seneca", 1980
= 8 | | | |rl| | |qd|bl|= 7 | | | | | |pl| | |= 6 | |nl| | | | | |bl|= 5 | |kl| | | |pl| | |= 4 |rl|pl|pl| |pd|rl| | |= 3 | |bl| |pl|nd| | | |= 2 |rl|pd|pd|kd|nd| |pd| |= 1 |bd|ql|bd|rd|rd|bd|rd|bd|= a b c d e f g h
The key is 1.Rf2, after which black captures on b1 are answered by white captures on g8.
Efficiency in chess problems is considered a great boon, but Drumare's attempt is very inefficient - no less than 30 men are on the board. It also has six promoted pieces in the initial position (even a single promoted piece is considered something of a "cheat" in chess problems), which is in any case illegal - it could not be reached in the course of a game (one of the white f pawns must have made a capture, and the white and black b and c pawns must have made two captures between them, making three in total, yet only two units are missing from the board). Despite all these flaws, it is the first complete Babson task.
In 1982, two years after composing this problem, Drumare gave up, saying that the Babson task would never be satisfactorily solved.The following year, Leonid Yarosh, a football coach from Kazan then virtually unknown as a problem composer, came up with a much better Babson task problem than Drumare's – the position is legal, it is much simpler than Drumare's problem, and there are no promoted pieces on board. First published in March 1983 in the famous Russian chess magazine "Shakhmaty v SSSR", this is generally thought of as the first satisfactory solution of the Babson task. Drumare himself had high praise for the problem. It is a mate in four:Chess diagram|=
tleft
Leonid Yarosh, "Shakhmaty v SSSR", March 1983
= 8 | |qd| | | | | |rl|= 7 |pl| | |nl| | | | |= 6 | | | | | |pd| | |= 5 | | |pd| | |pd| | |= 4 | | |pl|kd| |bd| |nd|= 3 | | | | | | |pl| |= 2 |pl|nl|pd|pl| | | |kl|= 1 |bl|ql| | | |rl| |bl|= a b c d e f g h
The key is 1.Rxh4, and the main lines are:
:cxb1Q 2.axb8Q Qxb2 (2... Qe4 3.Qxf4 Qxf4 4.Rxf4 mate) 3.Qb3 Qc3 4.Qxc3#:cxb1R 2.axb8R Rxb2 3.Rb3 Kxc4 4.Rxf4 mate:cxb1B 2.axb8B Be4 3.Bxf4 Bxh1 4.Be3 mate:cxb1N 2.axb8N Nxd2 3.Nc6+ Kc3 4.Rc1 mateHowever, Yarosh's problem has a small flaw – the key is a capture, something which is generally frowned upon in problems. Also, when first presented the black piece at h4 was a pawn, but a computer discovered an additional solution by 1.axb8N in that construction which is not there when a knight is substituted at h4. Nevertheless, when Dutch author Tim Krabbé saw this version in the Soviet publication ´64´, he records that the realisation that somebody had at last solved the Babson Task had the effect upon him as if he had " ... opened a newspaper and seen the headline ´Purpose Of Life Discovered´."Yarosh worked on the problem, and in August 1983 an improved version of it with a non-capturing key appeared in "Shakhmaty v SSSR". It is generally considered one of the greatest chess problems ever composed. Again, mate in four:
Chess diagram|=
tleft
Leonid Yarosh, "Shakhmaty v SSSR", August 1983
= 8 |bl|qd| |bl| |kl| | |= 7 | | | |pl|pd|nl| | |= 6 |pl| | | |pl|pd| | |= 5 |pl| |pd| | |pl| | |= 4 | | |pl|kd| |bd| |rl|= 3 | |pd| | | | | | |= 2 |pd|nl| |pl| |pl| | |= 1 |ql|rl| | | | | | |= a b c d e f g h
The key here is non-capturing and also thematic (it is logically related to the rest of the solution): 1.a7. The variations are largely the same as in the original:
:axb1Q 2.axb8Q Qxb2 (2... Qe4 3.Qxf4 Qxf4 4.Rxf4 mate) 3.Qxb3 Qc3 4.Qbxc3 mate:axb1R 2.axb8R Rxb2 3.Rxb3 Kxc4 4.Qa4 mate:axb1B 2.axb8B Be4 3.Bxf4 Bxa8 4.Be3 mate:axb1N 2.axb8N Nxd2 3.Qc1 Ne4 4.Nc6 mate
Yarosh composed a completely different Babson task problem in 1983 and another in 1986. Several other Babsons have since been composed by other authors.
The cyclic Babson
Chess diagram|=
tright
Peter Hoffmann, "Die Schwalbe", 2003
= 8 |rl| | | |bl| |rd| |= 7 | | | | | | | |pl|= 6 |bl| | | |bd|kd|nl|pd|= 5 |pd| | | | | |pd|rl|= 4 |kl| | |nl| |pd|pl| |= 3 | | |pl| | |pl| | |= 2 |pl|bl|pd|pd|pd| | |ql|= 1 | | |bl| |nl| | | |= a b c d e f g h
Mate in 4.
In the August 2003 issue of the German problem magazine "Die Schwalbe", the problem to the right, a mate in four by Peter Hoffmann appeared. Hoffmann had previously published a number of conventional directmate Babsons, but this one is significant as it is the first cyclic Babson: rather than black promotions being matched by white, they are related in cyclic form: black promoting to a queen means white must promote to a bishop, black promoting to a bishop means white must promote to a rook, black promoting to a rook means white must promote to a knight, and black promoting to a knight means white must promote to a queen.
The key is 1.Nxe6, threatening 2.hxg8Q and 3.Qf7#. The thematic defences are:
:1...d1Q 2.hxg8B (2.hxg8Q? Qd7+ 3.Bxd7 is stalemate), threatening 3.c4+ Qmoves 4.BxQ#::2...Qd7+ 3.Bxd7 Kxg6 4.Rxh6#::2...Qxc1 3.Rxg5 (threat: 4.Rf5#) hxg5 4.Qh8#:1...d1B 2.hxg8R (2.hxg8Q? stalemate) Kxe6 3.Rd8 3.Kf6 Rd6#:1...d1R 2.hxg8N (2.hxg8Q? Rd4+ 3. c4 stalemate) Kxe6 3.Qxe2+ K-moves 4.Qe5#:1...d1N 2.hxg8Q Nxb2+ 3.moves and 4.Qf7#
There are also a number of sidelines.
As with Drumare's original Babson task, the problem uses promoted pieces and has a capturing key, but it is nonetheless remarkable for being the first published cyclic Babson.
In the September 2005 issue of "Schach", the first cyclic Babson without promoted pieces in the initial position was published. Again, the composer was Peter Hoffmann.
References
*Jeremy Morse, "Chess Problems Tasks and Records" (Faber and Faber, 1995, revised edition 2001) - contains a chapter on the Babson task
* [http://www.xs4all.nl/%7Etimkr/babson/babsonsons.htm A number of Babsons on Tim Krabbé's page]
* [http://www.xs4all.nl/~timkr/chess/babs.html A detailed analysis of Yarosh's second Babson by Tim Krabbé]
* [http://www.xs4all.nl/~timkr/admag/babson1.htm Some forerunners of the Babson task from Tim Krabbé] (in Dutch)
* [http://www.xs4all.nl/~timkr/admag/kazan.htm Tim Krabbé's trip to Kazan] (in Dutch)
* [http://www.chessbase.com/puzzle/puzz13a.htm A history of the Babson task from chessbase.com]
* [http://members.tripod.com/~JurajLorinc/chess/ti_b.htm#babso A few Babson problems at Chess Composition Microweb]
* [http://www.mi.ru/~yarosh/ Creative workshop of Leonid V. Jarosh] (currently in Russian only)
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Babson-Task — Der Babson Task ist eine Aufgabenstellung aus der Schachkomposition. Gefordert wird eine beiderseitige Allumwandlung in unmittelbar aufeinander folgenden Zügen jeweils durch denselben Bauern. Inhaltsverzeichnis 1 Entstehung der Aufgabenstellung 2 … Deutsch Wikipedia
• Drumare — Pierre Drumare (* 26. Oktober 1913; † 15. April 2001 in Paris) war ein französischer Schachkomponist. Komposition Als sich Drumare Anfang der 1950er Jahre für Schachkomposition zu interessieren begann, begann er im April 1952, den Cercle Caïssa… … Deutsch Wikipedia
• Chess problem terminology — This is a list of terms used in chess problems. For a list of unorthodox pieces used in chess problems, see fairy chess piece. For a list of terms used in chess is general, see chess terminology. NOTOC A *Actual play see post key play . *Albino a … Wikipedia
• Glossary of chess problems — This is a list of terms used in chess problems. For a list of unorthodox pieces used in chess problems, see fairy chess piece. For a list of terms used in chess is general, see chess terminology. Contents: Top · 0–9 · A B C D E F G H I… … Wikipedia
• Pierre Drumare — (* 26. Oktober 1913; † 15. April 2001 in Paris) war ein französischer Schachkomponist. Komposition Als sich Drumare Anfang der 1950er Jahre für Schachkomposition zu interessieren begann, begann er im April 1952, den Cercle Caïssa nahe dem Place… … Deutsch Wikipedia
• Promotion (chess) — … Wikipedia
• Promotion (jeu d'echecs) — Promotion (jeu d échecs) Pour les articles homonymes, voir Promotion. Au jeu d échecs, la promotion d un pion consiste à le remplacer par une autre pièce lorsqu il arrive sur la dernière rangée. Sommaire 1 Règles 2 … Wikipédia en Français
• Promotion (jeu d'échecs) — Pour les articles homonymes, voir Promotion. Au jeu d échecs, la promotion d un pion consiste à le remplacer par une autre pièce lorsqu il arrive sur la dernière rangée. Sommaire 1 Règles 2 … Wikipédia en Français
• Promouvoir — Promotion (jeu d échecs) Pour les articles homonymes, voir Promotion. Au jeu d échecs, la promotion d un pion consiste à le remplacer par une autre pièce lorsqu il arrive sur la dernière rangée. Sommaire 1 Règles 2 … Wikipédia en Français
• Бэбсон-таск — Бэбсон таск, таск Бабсона (англ. Babson task; буквально задача Бэбсона) рекордная задача на тему взаимного идентичного превращения 1 белой и 1 чёрной пешек во все фигуры. Этот замысел интересовал шахматных композиторов ещё в начале XX века,… … Википедия
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# Lect 22 Zaheer Abbas
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### Lect 22 Zaheer Abbas
1. 1. Created by Zaheer Abbas Aghani 2k9-152
2. 2. Non-Linear Data Structure Tree
3. 3. <ul><li>As list, Stack, & Queue, binary tree can also be implement in two ways. </li></ul><ul><li>Linked Representation </li></ul><ul><li>Array Representation. </li></ul><ul><li>LINKED REPRESENTATION OF BINARY TREE: if we implement tree data in linked list then every node of linked list has three member/parts. First member is for data, second and third member for left & right child . Second & third members are structure pointers which point to the same structure as for tree node. </li></ul>Prepared by: Shumaila Bashir Sheikh(Lecturer, ITC)
4. 4. <ul><li>Tree Structure : </li></ul><ul><li>Linked Representation of tree: </li></ul>A B C D E F Prepared by: Shumaila Bashir Sheikh(Lecturer, ITC) A B C D E F
5. 5. <ul><li>Tree Structure: </li></ul><ul><li>Array Representation of tree:- I f we implement tree data in array then we need a 2 dimension array to store that data in memory & a pointer variable that store the address of root node.. </li></ul>Data LN RN Root 3 1 2 3 4 5 6 A B D C F E 5 2 1 4 6 null null null null null null null Prepared by: Shumaila Bashir Sheikh(Lecturer, ITC) A B C D E F
6. 6. <ul><li>Four Basic Operations </li></ul><ul><ul><li>Traversing </li></ul></ul><ul><ul><li>Searching </li></ul></ul><ul><ul><li>Inserting </li></ul></ul><ul><ul><li>Deleting </li></ul></ul>
7. 7. <ul><li>In tree creation we take three parameters node, left child & right child, so traversing of binary tree means traversing of node, left subtree and right subtree. </li></ul><ul><li>There are three standard ways to traversing a binary tree. These three algorithms are </li></ul><ul><li>Preorder Traversal </li></ul><ul><li>Inorder Traversal </li></ul><ul><li>Postorder Traversal </li></ul>
8. 8. <ul><li>Visit the root. </li></ul><ul><li>Traverse the left subtree of root in preorder. </li></ul><ul><li>Traverse the right subtree of root in preorder. </li></ul><ul><li>If root is denoted as N, left subtree as L & right subtree as R then Preorder traversal is also called NLR Traversal. </li></ul>
9. 9. <ul><li>Preorder Traversal: ABDECFG </li></ul>A B C D E F G
10. 10. <ul><li>The nodes are visited in preorder as: ABDHECFIG </li></ul>I A B C D E F G H
11. 11. <ul><li>Traverse the left subtree of root in Inorder. </li></ul><ul><li>Visit the root. </li></ul><ul><li>Traverse the right subtree of root in Inorder. </li></ul><ul><li>Inorder traversal is also called LNR Traversal. </li></ul>
12. 12. <ul><li>Inorder Traversal: DBEAFCG </li></ul>A B C D E F G
13. 13. <ul><li>The nodes are visited in inorder as: DHBEAFCG </li></ul>A B C D E F G H
14. 14. <ul><li>Traverse the left subtree in postorder. </li></ul><ul><li>Traverse the right subtree in postorder. </li></ul><ul><li>Visit the Root. </li></ul><ul><li>Postorder traversal is also called LRN Traversal. </li></ul>
15. 15. <ul><li>Postorder Traversal: DEBFGCA </li></ul>A B C D E F G
16. 16. <ul><li>The nodes are visited in postorder as: HDEBFGCA </li></ul>A B C D E F G H
17. 17. <ul><li>In level order traversal, we traverse the nodes according to their levels. We start traversing with the level 0, then level traverse all the nodes of level 1, & then traverse all the nodes of level 2 & so on. </li></ul><ul><li>We traverse the nodes of a particular level from left to right. </li></ul>LEVEL 2 C K G <ul><li>The nodes are traversing in level-order as: ABECKG </li></ul>A E B LEVEL 0 LEVEL 1
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# Statistics: Five Probability multiple choice questions
1. A kindergarden class consists of 14 boys and 11 girls. If the teacher selects children from the class using random sampling:
a. what is the probability that the first child selected will be a girl?
b. if the teacher selects a random sample of n=3 children and the first two children are both boys, what is the probability that the third child selected will be a girl?
2. For each of the following z-scores, sketch a normal distribution and draw a vertical line at the location of the z-score. Then determine whether the body is to the right or left of the line and find the proportion in the body.
a. z= 0.75
b. z= -1.40
c. z= 0.85
d. z= -2.10
3. the distribution of IQ scores is normal with mean= 100 and standard deviation =15
What proportion of the population has IQ scores?
a. greater than 115?
b. greater than 130?
c. greater than 145?
4. A multiple choice test has 48 questions each with four response choices. If a student is simply guessing at the answers
a. what is the probability that a student would get more than 18 answers correct simply by guessing?
5. A trick coin has been weighted so that heads occurs with a probability of p=2/3, and p(tails) = 1/3. Iif you toss this coin 72 times
a. how many heads would you expect to get on average?
b. what is the probability of getting more than 50 heads?
c. what is the probability of getting exactly 50 heads?
#### Solution Preview
See the attached files.
1. Note that there are 25 students in the class
a) The probability of getting a girl on the first pick is 11/25
b) After two boys are selected, there are 12 boys left, 11 girls and 23 students in the class. Thus the probability of getting a girl on the third pick is 11/23.
2. To calculate the normal score porportion, you will need a normal calculator here http://stattrek.com/tables/normal.aspx
a) z = 0.75, body is on the left side. Porportion of the body is 0.7734
b) z = -1.40, body is on the right side, porportion of ...
#### Solution Summary
Statistics: Five Probability multiple choice questions
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# Specific Heat Formula
Specific Heat Formula
When heat energy is added to a substance, the temperature will change by a certain amount. The relationship between heat energy and temperature is different for every material, and the specific heat is a value that describes how they relate.
heat energy = (mass of substance)(specific heat)(change in temperature)
Q = mc∆T
Q = heat energy (Joules, J)
m = mass of a substance (kg)
c = specific heat (units J/kg∙K)
is a symbol meaning "the change in"
∆T = change in temperature (Kelvins, K)
Specific Heat Formula Questions:
1) The specific heat of gold is 129 J/kg∙K. What is the quantity of heat energy required to raise the temperature of 100 g of gold by 50.0 K?
Answer: The mass of gold is m = 100 g = 0.100 kg. The heat energy can be found using the formula:
Q = mc∆T
Q =(0.100 kg)(129 J/kg∙K)(50.0 K)
Q = 645 J
The energy required to raise the temperature of the piece of gold is 645 J.
2) A pot of water is heated by transferring 1676 kJ of heat energy to the water. If there is 5.00 kg of water in the pot, and the temperature is raised by 80.0 K, what is the specific heat of water?
Answer: The heat energy transferred to the water is 1676 kJ = 1 676 000 J. The specific heat can be found by rearranging the formula:
c = 4190 J/kg∙K
The specific heat of water is 4190 J/kg∙K.
Related Links: Forms of Energy (solar, heat, sound, and motion) Quiz Heat Quiz Heat Quiz Nature of Heat Quiz Thermal Energy Facts Thermal Energy Examples Convection Examples Conduction Examples Heat Energy Examples Heat Transfer Formula
Related Topics
Specific Heat Capacity Formula
Heat Transfer Formula
Heat Capacity Formula
Specific Heat Capacity Equation Calculator
Formulas: Physics Formulas and Math Formulas
Heat Quiz
Heat of Vaporization Formula
Enthalpy Formula
Physics Formulas
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# Coupling constant
(Redirected from Running coupling)
For the Murray–von Neumann coupling constant, see von Neumann algebra. For the coupling constant in NMR spectroscopy, see Nuclear magnetic resonance spectroscopy and Proton NMR. For coupling strength in bibliometrics, see Bibliographic coupling.
In physics, a coupling constant is a number that determines the strength of the force exerted in an interaction. Usually, the Lagrangian or the Hamiltonian of a system describing an interaction can be separated into a kinetic part and an interaction part. The coupling constant determines the strength of the interaction part with respect to the kinetic part, or between two sectors of the interaction part. For example, the electric charge of a particle is a coupling constant.
A coupling constant plays an important role in dynamics. For example, one often sets up hierarchies of approximation based on the importance of various coupling constants. In the motion of a large lump of magnetized iron, the magnetic forces are more important than the gravitational forces because of the relative magnitudes of the coupling constants. However, in classical mechanics one usually makes these decisions directly by comparing forces.
## Fine-structure constant
The coupling constant arises naturally in a quantum field theory. A special role is played in relativistic quantum theories by coupling constants which are dimensionless, i.e., are pure numbers. For example, the fine-structure constant,
$\alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c}$
(where $e$ is the charge of an electron, $\varepsilon_0$ is the permittivity of free space, $\hbar$ is the reduced Planck constant and $c$ is the speed of light) is such a dimensionless coupling constant that determines the strength of the electromagnetic force on an electron.
## Gauge coupling
In a non-Abelian Gauge theory, the gauge coupling parameter, $g$, appears in the Lagrangian as
$\frac1{4g^2}{\rm Tr}\,G_{\mu\nu}G^{\mu\nu}$
(where $G$ is the gauge field tensor) in some conventions. In another widely used convention, $G$ is rescaled so that the coefficient of the kinetic term is 1/4 and $g$ appears in the covariant derivative. This should be understood to be similar to a dimensionless version of the electric charge defined as
$\sqrt{4\pi\varepsilon_0\alpha}.$
## Weak and strong coupling
In a quantum field theory with a dimensionless coupling constant g, if g is much less than 1 then the theory is said to be weakly coupled. In this case it is well described by an expansion in powers of g, called perturbation theory. If the coupling constant is of order one or larger, the theory is said to be strongly coupled. An example of the latter is the hadronic theory of strong interactions (which is why it is called strong in the first place). In such a case non-perturbative methods have to be used to investigate the theory.
## Running coupling
One can probe a quantum field theory at short times or distances by changing the wavelength or momentum, k, of the probe one uses. With a high frequency (i.e., short time) probe, one sees virtual particles taking part in every process. This apparent violation of the conservation of energy can be understood heuristically by examining the uncertainty relation
$\Delta E\Delta t\ge\hbar,$
which allows such violations at short times. The previous remark only applies to some formulations of quantum field theory, in particular, canonical quantization in the interaction picture. In other formulations, the same event is described by "virtual" particles going off the mass shell. Such processes renormalize the coupling and make it dependent on the energy scale, $\mu$ at which one observes the coupling. The dependence of a coupling g(μ) on the energy-scale is known as running of the coupling. The theory of the running of couplings is known as the renormalization group.
### Beta functions
In quantum field theory, a beta function β(g) encodes the running of a coupling parameter, g. It is defined by the relation
$\beta(g) = \mu\frac{\partial g}{\partial \mu} = \frac{\partial g}{\partial \ln \mu},$
where μ is the energy scale of the given physical process. If the beta functions of a quantum field theory vanish, then the theory is scale-invariant.
The coupling parameters of a quantum field theory can flow even if the corresponding classical field theory is scale-invariant. In this case, the non-zero beta function tells us that the classical scale-invariance is anomalous.
### QED and the Landau pole
If a beta function is positive, the corresponding coupling increases with increasing energy. An example is quantum electrodynamics (QED), where one finds by using perturbation theory that the beta function is positive. In particular, at low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127.
Moreover, the perturbative beta function tells us that the coupling continues to increase, and QED becomes strongly coupled at high energy. In fact the coupling apparently becomes infinite at some finite energy. This phenomenon was first noted by Lev Landau, and is called the Landau pole. However, one cannot expect the perturbative beta function to give accurate results at strong coupling, and so it is likely that the Landau pole is an artifact of applying perturbation theory in a situation where it is no longer valid. The true scaling behaviour of $\alpha$ at large energies is not known.
### QCD and asymptotic freedom
In non-Abelian gauge theories, the beta function can be negative, as first found by Frank Wilczek, David Politzer and David Gross. An example of this is the beta function for Quantum Chromodynamics (QCD), and as a result the QCD coupling decreases at high energies.
Furthermore, the coupling decreases logarithmically, a phenomenon known as asymptotic freedom (the discovery of which was awarded with the Nobel Prize in Physics in 2004). The coupling decreases approximately as
$\alpha_s(k^2) \ \stackrel{\mathrm{def}}{=}\ \frac{g_s^2(k^2)}{4\pi} \approx \frac1{\beta_0\ln(k^2/\Lambda^2)},$
where β0 is a constant computed by Wilczek, Gross and Politzer.
Conversely, the coupling increases with decreasing energy. This means that the coupling becomes large at low energies, and one can no longer rely on perturbation theory.
### QCD scale
In quantum chromodynamics (QCD), the quantity Λ is called the QCD scale. The value is
$\Lambda_{MS} = 217^{+25}_{-23}{\rm\ MeV}.$
This value is to be used at a scale above the bottom quark mass of about 5 GeV. The meaning of ΛMS is given in the article on dimensional transmutation.
The proton-to-electron mass ratio is primarily determined by the QCD scale.
## String theory
A remarkably different situation exists in string theory[why?]. Each perturbative description of string theory depends on a string coupling constant. However, these coupling constants are not pre-determined, adjustable, or universal parameters; rather they are dynamical scalar fields that can depend on the position in space and time and whose values are determined dynamically.
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In the input of half wave rectifier was given 10 volt of voltage amplitude. Calculate the value of output of voltage amplitude?
• In the input of half wave rectifier was given 10 volt of voltage amplitude. Calculate the value of output of voltage amplitude?
In the input of half wave rectifier was given 10 volt of voltage amplitude. Calculate the value of output of voltage amplitude? Find answers now!
Positive: 20 %
In the input of half wave rectifier was given 10 volt of voltage amplitude. Calculate the value of output ... between the input voltage and the output ...
Positive: 17 %
More resources
Half Wave Diode Rectifier ... of half wave diode rectifier is given below, For positive half cycle For negative half cycle Input voltage and Output ...
Positive: 20 %
Lecture 4 ver2 diode_app ... occurs in the output voltage for a given change in input voltage ... to the input of a half-wave rectifier, ...
Positive: 15 %
... Positive Voltage Output Half Wave Rectifier. ... Ripple is the amplitude ... value approaching the average value of input voltage because of ...
| 229
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https://www.reddit.com/r/dailyprogrammer/comments/1epasu/052013_challenge_126_easy_realworld_merge_sort/
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This is an archived post. You won't be able to vote or comment.
[–] 39 points40 points (4 children)
Being a real life engineer, I'd copy paste into Excel, and sort there.. ;)
[–]1 2[S] 9 points10 points (3 children)
Clearly the best solution thus far!
[–]1 3 10 points11 points (2 children)
Summer time. Make the intern copy it into Excel and sort it. ;)
edit: paper potential for developing the intern_sort() algorithm hmmmm
[–] 1 point2 points (0 children)
Probably quicker to do it yourself than trying to explain to the intern what needs to be done though ~~
[–]1 2[S] 1 point2 points (0 children)
Mother of god... why haven't I thought about reaching out to a school and asking for interns? Ahhh! Alright, I'mma go to my local University and figure out if we can get help :-)
[–]0 0 11 points12 points (6 children)
C++11
``````#include <vector>
#include <algorithm>
template<class Cont>
void mergeAndSort(const Cont& a, Cont& b) {
// merge
auto begin = std::begin(b);
auto first = std::begin(a);
auto end = std::end(b);
for(; begin != end; ++begin) {
if(*begin == 0)
*begin = *first++;
else
break;
}
begin = std::begin(b);
// sort (assuming no built ins)
std::make_heap(begin,end);
while(begin != end)
std::pop_heap(begin, end--);
}
#include <iostream>
int main() {
std::vector<int> a = {692,1,32};
std::vector<int> b = {0,0,0,14,15,123,2431};
mergeAndSort(a,b);
for(auto&& i : b)
std::cout << i << ' ';
}
``````
Output
1 14 15 32 123 692 2431
[–]1 1 0 points1 point (4 children)
what is the Cont class?
[–]0 0 2 points3 points (3 children)
That's a template specification. In this case, it just means it can hold any data type we throw at it, i.e. it's similar to a Generic in C# or Java. If you've seen them before differently like `template<typename T>` then it's the equivalent of that.
[–]1 1 0 points1 point (2 children)
Ohh i've used templates before, but never the Cont classes. Is this a C++11 feature?
[–] 1 point2 points (0 children)
often you'll see code like:
``````template<typename T> ...
``````
typename and class are synonyms in this context. Cont is just an arbitrary identifier, evocative of "Container", as a hint to the person using the template that they're supposed to put some sort of container class in as a template arg. But it could have been T or anything else.
[–] 0 points1 point (0 children)
It can be called anything that's a legal identifier: T, Cont, FlyingMonkies, WhyDidYouSetTheCatOnFireThatWasReallyMeanAndIAmMakingThisIdentiferLongToMakeProgrammersSeppiku, etc...
[–] 0 points1 point (0 children)
Why merge by hand? Is there something I am missing that makes the entire merge section not just boil down to
``````std::copy(std::begin(a), std::end(a), std::begin(b));
``````
I also think it would be better to have seperate template types for a and b, as a can be any container, but b has to have random-access iterators.
[–] 6 points7 points (1 child)
python 3:
``````A = [int(x) for x in input("enter list A: ").split(" ")]
B = [int(x) for x in input("enter list B: ").split(" ")]
i = 0
while i < len(A):
j = 0
while A[i] > B[j]:
if B[j] != 0:
B[j-1] = B[j]
j += 1
B[j-1] = A[i]
i += 1
print(B)
``````
[–] 9 points10 points (0 children)
I like how u use "B[j]".
[–]-9 8 5 points6 points (0 children)
JavaScript, using insertion sort. Because insertion sort is adaptive it should perform well with B already sorted, especially if B is much larger than A, approaching O(n).
``````function insertionSort(array) {
for (var i = 1; i < array.length; i++) {
var hole, next = array[i];
for (hole = i; hole > 0 && next < array[hole - 1]; hole--) {
array[hole] = array[hole - 1];
}
array[hole] = next;
}
return array;
}
function merge(a, b) {
for (var i = 0; i < a.length; i++) {
b[i] = a[i];
}
return insertionSort(b);
}
``````
This isn't very JavaScripty due to the problem constraints, so I should have done it in C.
[–] 3 points4 points (0 children)
I don't know a thing about algorithms so here's a very naive sort (in Javascript):
``````var a = [692, 1, 32];
// given: b is sorted numerically
var b = [0, 0, 0, 14, 15, 123, 2431];
// Number of blanks at the start of b
var blanks = a.length;
for (var i = 0; i < a.length; i++) {
for (var j = blanks; j < b.length; j++) {
if (a[i] < b[j]) {
insert_into_b(a[i], j-1, blanks);
blanks--;
break;
}
}
}
function insert_into_b(val, index, num_blanks) {
for (var k = num_blanks - 1;k < index; k++){
b[k] = b[k+1];
}
b[index] = val;
}
console.log(b);
``````
[–] 1 point2 points (0 children)
My go using C++ (it's been a little while)
Critique appreciated.
``````#include <fstream>
#include <sstream>
#include <iostream>
#include <string>
using namespace std;
void
HoldQuitMessage()
{
cout << "Quit? (y/y)" << endl;
cin.get();
}
void
Merge(int* _listA, int _sizeA, int* _listB, int sizeB)
{
int aCurSort = 0;
int* pBCurPos = 0;
while (aCurSort < _sizeA)
{
pBCurPos = _listB; // Reset b pointer.
while (*pBCurPos == 0)
++pBCurPos; // Skip the zeros.
while (*pBCurPos < _listA[aCurSort])
{
*(pBCurPos - 1) = *pBCurPos; // Push smaller values backwards.
++pBCurPos;
}
// Pop in the current value just before here.
*(pBCurPos - 1) = _listA[aCurSort++];
}
}
bool
ParseLists(int*& _listA, int& _sizeA, int*& _listB, int& _sizeB, char* _filePath)
{
ifstream listFile;
listFile.open(_filePath);
if (listFile.is_open())
{
string curLine = "";
int lineCount = 0;
while (!listFile.eof() && lineCount < 3)
{
getline(listFile, curLine);
int numInts = 0, lastSpace = 0;
while (lastSpace != -1)
{
lastSpace = curLine.find(' ', lastSpace + 1);
++numInts;
}
int* newInts = new int[numInts];
istringstream tempstream(curLine);
for (int i = 0; i < numInts; ++i)
tempstream >> dec >> newInts[i];
if (lineCount++ % 2 == 0)
{
_listA = newInts;
_sizeA = numInts;
}
else
{
_listB = newInts;
_sizeB = numInts;
}
}
return (_sizeA > 0 && _sizeB > 0);
}
return (false);
}
void
Print(int* _intList, int _size)
{
for (int i = 0; i < _size; ++i)
cout << _intList[i] << ", ";
cout << endl;
}
int
main(int argc, char** argv)
{
int *listA = 0, *listB = 0;
int sizeA = 0, sizeB = 0;
if (argc < 2)
{
cout << "No file path." << endl;
HoldQuitMessage();
return (0);
}
if (!ParseLists(listA, sizeA, listB, sizeB, argv[1]))
{
cout << "File doesn't exist.";
}
Merge(listA, sizeA, listB, sizeB);
Print(listB, sizeB);
HoldQuitMessage();
return (0);
}
``````
Output:
``````1, 14, 15, 32, 123, 692, 2431,
Quit? (y/y)
``````
[–] 3 points4 points (0 children)
My solution in Javascript:
``````function mergeAndSort(unsortedArray, sortedArray) {
var uLength = unsortedArray.length,
sLength = sortedArray.length,
i = 0, j = 0,
currentElement = -1,
temp = 0;
for (i = 0; i < uLength; i++) {
sortedArray[0] = unsortedArray[i];
currentElement = 0;
for (j = 1;j < sLength; j++) {
if (sortedArray[currentElement] >= sortedArray[j]) {
temp = sortedArray[j];
sortedArray[j] = sortedArray[currentElement];
sortedArray[currentElement] = temp;
currentElement = j;
} else {
break;
}
}
}
return sortedArray;
}
``````
[–]0 1 3 points4 points (0 children)
D Language
``````import std.stdio, std.algorithm, std.range, std.random;
// Simple Heap Sort, O(1) space and O(n log n) time
// shamelessly ignores the fact that b is sorted
void heapSort(const(uint)[] a, uint[] b)
in {
assert(!a.canFind(0u));
assert(b.count!"a==0u" == a.length);
assert(b.isSorted);
}
out {
assert(!b.canFind(0u));
assert(b.isSorted);
}
body {
void percDown(uint[] arr, ulong root) {
while (root * 2 + 1 < arr.length) {
ulong child = root * 2 + 1;
ulong dest = root;
if (arr[dest] < arr[child]) {
dest = child;
}
if (child + 1 < arr.length && arr[dest] < arr[child+1]) {
dest = child+1;
}
if (dest != root) {
swap(arr[root], arr[dest]);
root = dest;
}
else {
break;
}
}
}
void heapSortImpl(uint[] arr) {
foreach_reverse(ulong start ; 0 .. arr.length/2) {
percDown(arr, start);
}
while (arr.length > 0) {
swap(arr.back, arr.front);
arr.popBack();
percDown(arr, 0);
}
}
b[0 .. a.length] = a[];
heapSortImpl(b);
}
``````
[–] 2 points3 points (8 children)
My attempt in C. Critique high appreciated. :)
``````#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(){
//Allocate arrays.
int list1[3];
memcpy(list1, (int[]) {692, 1, 32}, sizeof(list1));
int list2[7];
memcpy(list2, (int[]) {0, 0, 0, 14, 15, 123, 2431}, sizeof(list2));
int sizeA = sizeof(list1)/sizeof(int);
int sizeB = sizeof(list2)/sizeof(int);
//Display contents before.
printf("Before:\nList A: ");
for(int i = 0; i < sizeA; i++){
printf("%d, ", list1[i]);
}
printf("\nList B: ");
for(int i = 0; i < sizeB; i++){
printf("%d, ", list2[i]);
}
printf("\n");
//Merge one at a time, and sort into the list.
for(int i = sizeA-1; i >= 0; i--){
list2[i] = list1[i];
for(int j = i+1; j < sizeB; j++){
if(list2[j] >= list2[j-1]) break;
else{
list1[i] = list2[j];
list2[j] = list2[j-1];
list2[j-1] = list1[i];
}
}
}
//Display the results.
printf("After:\nList A: ");
for(int i = 0; i < sizeA; i++){
printf("%d, ", list1[i]);
}
printf("\nList B: ");
for(int i = 0; i < sizeB; i++){
printf("%d, ", list2[i]);
}
printf("\n");
//Peace out.
return(0);
}
``````
Output:
``````Before:
List A: 692, 1, 32,
List B: 0, 0, 0, 14, 15, 123, 2431,
After:
List A: 123, 1, 15,
List B: 1, 14, 15, 32, 123, 692, 2431,
``````
[–]1 2[S] 0 points1 point (2 children)
I don't have too much time to do a full review, but here are some comments I hope you find helpful! Overall the code is solid and you clearly have a good grasp of C as a language and approach:
1) Avoid initialization of variables with memcpy unless you really have to: you can make your code more readable and maintainable by changing this:
``````int list1[3];
memcpy(list1, (int[]) {692, 1, 32}, sizeof(list1));
``````
To this:
``````int list1[3] = {692, 1, 32};
``````
Hell, you can even omit the "[3]" and just let it be "[]" since what matters. You also get a slight performance boost by doing the single-line initialization because you don't have to do a function call or looped memory copy.
2) Avoid using the sizeof(...) function on lists to get the element count: again, it makes code hard to read and raises issues with maintainability. sizeof(<some primitive>) returns the byte-count, while sizeof(<some list>) is utterly different and returns the element count. See what I mean?
[–] 1 point2 points (1 child)
Thanks so much! I'm a little rusty on my C (I write too much Python), and didn't know there was a reasonably straight forward way to initialize arrays like that.
As far as sizeof, I think C's sizeof returns the number of bytes in the array rather than the number of elements, and should always do that. It's been a while since I took all my C programming classes, but my understanding (which could be entirely wrong) was that was kind of the canonical way to determine array lengths. Is there a better way to do this?
[–]1 2[S] 0 points1 point (0 children)
Actually, I've failed you! I've very poorly written what I wanted to say about it: basically you are 100% right. sizeof returns the number of bytes in an array, but there is a catch! A static (stack-allocated) array will return the total number of allocated bytes, while a dynamic (heap-allocated) array is generally just a single pointer, and thus a sizeof returns the pointer's size! To complicate things even more, some arrays allocated with the data type "char" can either be allocated as a single-byte per char or multiple-bytes per char (the C-language specification just says that a char has to be big enough to contain a character, so at minimum it is one byte but on many platforms it is 2 or even 4 bytes!).
[–][deleted] (4 children)
[deleted]
[–] 0 points1 point (3 children)
I didn't know C has library functions for fancy stuff like sorting.
By the way, what happens when you initialize an array like this?
``````int A[] = { 692, 1, 32, };
``````
I remember running into problems with undefined array lengths so now I always declare a fixed length or use malloc.
My solution is pretty similar to AstroCowboy's, it's over here.
[–][deleted] (2 children)
[deleted]
[–] 0 points1 point (1 child)
Thanks so much for your input, it's highly appreciated! I certainly have some learning to do... I am curious though, is using
``````int the_array [] = {1, 2, 3, 4};
``````
a C99 thing? (obviously some of the things I've done also require C99...)
[–] 3 points4 points (0 children)
Ruby solution
without input reading and cheating with shift and insert :)
``````for i in a do b.shift; b.insert(b.index{|j| j > i}, i) end
``````
whole without cheating:
``````a = readline().split(' ').map!(&:to_i)
for i in a do
for j in 0..b.length do
if b[j] > i
b[j-1] = i
break
else
b[j] = b[j+1]
end
end
end
puts b.join(" ")
``````
Result:
``````\$ ruby sort.rb < input
1 14 15 32 123 692 2431
``````
[–]1 3 2 points3 points (6 children)
If List B is pre-sorted wouldn't the input be:
692 1 32
0 0 0 14 15 123 2431
I think the 14 was out of place. But I get what you are saying.
[–]1 2[S] 4 points5 points (5 children)
Oh, errr, I'm slightly confused. Why is 14 out of place?
[–] 1 point2 points (4 children)
Cheeky. :)
[–]1 2[S] 1 point2 points (3 children)
Ha, no really, I want to know what the problem is. Is it a non-problem because I've changed the text?
[–] 1 point2 points (1 child)
It's a non-issue because you changed the text.
[–]1 2[S] 1 point2 points (0 children)
Sweet - sorry for the confusion then, I'm pretty thick sometimes.
[–]1 3 1 point2 points (0 children)
No problem now. Not the 14 I was looking for.
[–][deleted] (3 children)
[deleted]
[–] 1 point2 points (0 children)
Any code posted here is in spoilered automatically :)
[–]1 3 0 points1 point (0 children)
On reddit if you indent 4 spaces on each line it shows the text as is -- good for posting code. On this subreddit it detects this and makes it spoiler.
[–] 0 points1 point (0 children)
Hah, I wish half the code on here was as nice, clean and efficient as your solution.
[–] 2 points3 points (0 children)
Ruby
``````def merge(list1, list2)
list1.each do |number|
list2.each do |number2|
if number <= number2
list2.insert(list2.index(number2), number)
break;
end
end
end
list2
end
list1 = [692, 1, 32]
list2 = [14, 15, 123, 2431]
merge(list1, list2)
list2 is returned as [1, 14, 15, 32, 123, 692, 2431]
I don't have it finished for those leading 0s and the memory constraint, working on that
``````
[–] 2 points3 points (0 children)
python 2.7
``````listA = [691, 1, 32]
listB = [0, 0, 0, 14, 15, 123, 2431]
def mergeSort(x, b):
xLimit = 0
#find point of insertion for x
for i in range(len(b)):
if x < b[i]:
break
xLimit = i
#move all indexes before point of insertion backwards by one
for i in range(xLimit):
b[i] = b[i+1]
#insert number
b[xLimit] = x
def listMergeSort(a, b):
#insert each index from listA
for i in range(len(a)):
mergeSort(a[i], b)
listMergeSort(listA, listB)
print listB
``````
output:
[1, 14, 15, 32, 123, 691, 2431]
[–] 2 points3 points (2 children)
C99:
``````void sort(int *listA, int *listB, int lengthA)
{
int i = lengthA, j = 0;
for(int k = 0; k < lengthA; k++)
{
for(j = 0; listA[k] > listB[i + j]; j++) // find position to insert
listB[(i + j) - 1] = listB[i + j]; // shift B forward into one of the 0's
listB[(i-- + j) - 1] = listA[k]; // insert
}
}
``````
Usage: (simply made some arrays for testing because the format the lists are supplied in is not specified)
``````#include <stdio.h>
int main()
{
int listA[] = {692, 1, 32},
listB[] = {0, 0, 0, 14, 15, 123, 2431, 0};
int numElements = 7; // if lists have to be taken from a file or as arguments,
// put number of elements here
sort(listA, listB, 3);
printf("Result: ");
for(int i = 0; i < numElements - 1; i++)
printf("%d, ", listB[i]);
printf("%d\n", listB[numElements - 1]);
return 0;
}
``````
output: 1 14 15 32 123 692 2431
The program assumes the number of 0's at the start of list B equals the number of elements in list A
[–] 0 points1 point (1 child)
Interesting. Looks good (although, I'm still kind of trying to piece together the details of how your sort function works). I would be worried about using arrays of length 100 though...if you're using C99, TerribleCProgrammer mentioned the best way to do this is:
[–] 0 points1 point (0 children)
I made it kind of compact since most in this sub seem to like that, what I do is:
``````I go through listB, starting at the first nonzero value (listB[i]) copying each value
one place forward, so the last 0 gets overwritten and an open spot appears. While doing
this I compare to the current value of list A (the condition for the for loop), if it's
larger than the last listB value, I place it in the open spot and go to the next listA
value.
``````
You're right about the arrays, I just learned that and didn't implement it yet. Edited it in :)
[–] 2 points3 points (0 children)
This is my first time here. Not sure if I understood the problem correctly but here's my solution! C#
`````` using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Real_World_Merge_Sort
{
class Program
{
static void Main(string[] args)
{
Console.Write("Press any key to begin.");
Console.Clear();
mergeAndSort();
}
public static void mergeAndSort()
{
int[] listA = new int[] { 692, 1, 32 };
int[] listB = new int[] { 0, 0, 0, 14, 15, 123, 2431 };
int kount;
//Merge two lists
for (int k = 0; k < listA.Length; k++)
{
for (int i = 0; i < listB.Length; i++)
{
//If the int in listB equals 0 then
//add item from listA into listB
//decrement kount counter
//break to outer loop
if (listA[k] < listB[i])
{
for (kount = 0; kount < i; kount++)
{
listB[kount] = listB[kount + 1];
}
listB[i - 1] = listA[k];
break;
} if (i == listB.Length - 1)
{
for (kount = 0; kount < i; kount++)
{
listB[kount] = listB[kount + 1];
}
listB[i] = listA[k];
break;
}
}
}
for (int o = 0; o < listB.Length; o++)
{
Console.Write(listB[o] + " ");
}
}
}
}
``````
Output:
``````1 14 15 32 123 692 2431
``````
[–] 2 points3 points (9 children)
Again, I'm late to the party, but here's an original overly specialized and not too efficient sorter. To force myself avoid using a 'holder' variable, I had to write it in //comments first, so I just left those..:
``````public class ArraySorta {
//don't touch my gA[]
private final static int gA[] = { 692, 1, 32 };
//the following method requires as many preceding zeroes as there are elements in gA[]
private static int gB[] = { 0, 0, 0, 14, 15, 123, 2431 };
public static void main(String[] args){
// for each gA[r]
for (int r = 0; r < gA.length; r++) {
// until we find a spot for gA[r].......
for (int a = 0, b = 1; b < gB.length; a++, b++) {
// if we found a spot, use it
if (gB[a] < gA[r] & gB[b] > gA[r]) {
gB[a] = gA[r];
// now that gA[r] is placed, we need to get out of the loop
break;
}
// ........move gB[b] to the left in the array
else
gB[a] = gB[b];
}// end for a b loop
}// end for r loop
for (int k = 0; k < gB.length; k++) {
System.out.print(gB[k] + " ");
}
}// endmain
}// endclass
``````
Prints:
``````1 14 15 32 123 692 2431
``````
[–]1 2[S] 2 points3 points (1 child)
You are not late, nor should you ever feel late to post! :-) We're a community of programmers, and we all know that code isn't late, simply delivered when ready.
[–] 1 point2 points (0 children)
Yeah, I keep meaning to set up a real subscription, rather than just do these when they show up on my front page. Thanks for the submission!
[–] 0 points1 point (6 children)
(I'm self-teaching and progressing extremely slowly, so I don't actually have a functioning solution myself yet)
I was reading through your code to see the kind of logic applied, and tried running it with my own numbers and realised it doesn't work if any of the values in gA[] are greater than every value in gB[]. The problem is here:
``````& gB[b] > gA[r])
``````
but I don't know right now how to solve that without breaking the algo. I'll keep thinking it over.
Input:
``````gA[] = {465687, 654567, 4124, 654, 1, 3546}
gB[] = {0, 0, 0, 0, 0, 0, 48, 59, 348, 2687, 16548}
``````
Output:
``````1 48 59 348 654 2687 3546 4124 16548 16548 16548
``````
[–] 0 points1 point (5 children)
Good catch! You won't be able to fix it in that if statement, though!
It might make things easier to replace the if/else with a switch statement, because gB[a] never equals the final element of the array gB[]
``````gB[a] = gA[r]; //can never replace gB[b]!
``````
I'm guessing that you're trying to add an or operator, since you copied that line of code into your reply, something like
``````( (gB[a] < gA[r] & gB[b] > gA[r] ) | [we've reached the end] )
``````
But that won't work. If you need another clue, let me know.
[–] 0 points1 point (0 children)
I was actually just pointing out where the mistake was in case you'd moved no to other code and didn't feel like trawling through your own.
That said, I did indeed try to add an or statement there, but realised reasonably quickly that that wouldn't work.
[–] 0 points1 point (3 children)
I'm off to bed now and I'll have a further look (including reading up on switch statements) tomorrow. However, could you tell me why this:
``````if (gA[r] > gB[b] && b == gB.length) {
gB[b] = gA[r];
}
``````
wouldn't work to catch any of gA[] that make it all the way to the end?
[–] 0 points1 point (0 children)
It would work in the final instance, but would not have moved the previous final element to the left. =)
You'd need to move my original else statement up, and you should have it. =)
Edit: Using switch statements inside a for-loop is dicy.. it's easier to just write a bunch of if statements and then convert them for performance, in most cases, since the "break" line only gets out you out of the switch, but in this example it turned out you didn't need it. Sorry about the red herring.
[–] 0 points1 point (1 child)
I'm really glad you pointed out my error, so I've gone back and rewritten it. It seems like you were trying to understand something specific about the logic of my original post, though, so feel free to keep asking questions about that. Otherwise, if you want a spoiler of my improved version, check here
If you just want a clue, we only needed a single if/else statement in the nested for-loop.
I think you're a little confused about some basic java syntax, but I've already given you one red-herring and don't want to mislead you any more. When you get this, can you post a few of your earlier versions/attempts to pastebin(If you've never used pastebin, you don't need an account. You can just make it expire in a week, and leave it unlisted, I'll still be able to read it as long as you send a link.) and I'll see if I'm right? If so, I'll link you to the relevant page on the tutorial.
[–] 0 points1 point (0 children)
Your break really was the thing that threw me, to be honest.
I was taking it really inefficiently at first. I split the two into arrays, then tried to automate some way to compare each value in one array to every value in the other, then realised I'd be unable to order them using that anyway.
After that, I considered a mathematical approach to approximate largest values (might have worked with a probabilistic approach), checking them once I had a general order, but that seemed very inefficient to me.
Looking at your method, I realised that I had thrown away a very valuable part of the challenge - I ignored the fact that B was already ordered, choosing instead to try to sort A without any point of reference.
Thanks for the offer, of course, and I may take you up on it. Right now, however, it's half one and I have to be up in just under six hours, so enjoy your evening/night.
[–] 2 points3 points (0 children)
My solution in Java
``````public static int[] merge(int[] A, int[] B) {
for (int j = 0; j < A.length; j++) {
int i = B.length - 1;
while (A[j] < B[i] && i >= 0) {
i--;
}
for (int k = 0; k < i; k++) {
B[k] = B[k+1];
}
B[i] = A[j];
}
return B;
}
``````
[–] 2 points3 points (0 children)
PHP, assuming the leading zeros are always together.
``````<?php
return implode(' ',returnSortedArray(explode(' ',str_replace('0 0 0',\$toAdd,\$preSorted))));
}
function returnSortedArray(\$array){
sort(\$array);
return \$array;
}
echo RW_MergeSort('692 1 32','0 0 0 14 15 123 2431');
?>
``````
Not assuming a location/order for the zeros:
``````<?php
}
function returnSortedArray(\$array){
sort(\$array);
return \$array;
}
function RW_Merge(\$listA,\$listB){
\$numbersinA = substr_count(\$listA,' ') + 1;
for(\$i = 0; \$i < \$numbersinA; \$i++){
\$listB = preg_replace('/0/', explode(' ',\$listA)[\$i], \$listB, 1);
}
return(\$listB);
}
echo RW_MergeSort('692 1 32','0 0 0 14 15 123 2431');
?>
``````
Would love any critique on code practices/etc, perhaps whether or not I completed the challenge correctly. Thanks a ton, /r/DailyProgrammer!
[–] 2 points3 points (8 children)
Sample list B is unsorted when input description specifies it will be sorted?
Is this an intentional misdirection meant to simulate the miscommunication and mixed messages that are so common in the professional environments that work with legacy code like this?
[–]1 2[S] 3 points4 points (0 children)
Haha, my mistake, though you are right: in the real world of programming, working with a nasty premise is only half the trouble. The other half is just figuring out what in the world your boss or client really wants...
[–] 4 points5 points (6 children)
You will sort list B while list A is left unsorted.
[–] 1 point2 points (5 children)
Sample input used to be:
692 1 32
0 0 0 15 123 14 2431
The error has since been fixed. :)
[–] 2 points3 points (4 children)
I still don't see the need for the change or that it was an error, but okay. :)
[–] 0 points1 point (3 children)
If list B is already sorted, the merge sort can be made much faster than if neither list is sorted.
[–] 1 point2 points (2 children)
Well, yes. That's true. But it doesn't need to be sorted. Isn't the whole purpose of a merge sort being given one list, separating it into sublists and then merging the each smaller sublist in sorted order?
But even ignoring that, since that would make it a bit more complicated than this challenge intends, I think, it's just adding one extra step of sorting list B using whatever method you want and then merge sorting list A into list B.
[–] 0 points1 point (1 child)
Can you in-place split such a list into lists of size 1 without allocations and minimal local variables?
That sounds hard.
[–] 1 point2 points (0 children)
Yeah, which is why I said "ignoring that, since that would make it a bit more complicated than this challenge intends." Having list B unsorted to start just requires you to sort list B using whatever method you want before merging list A into list B in sorted order.
[–] 1 point2 points (1 child)
Forgive me if this is a stupid question, but is the idea to merge then sort B however, or to merge then merge-sort B?
[–] 0 points1 point (0 children)
Judging by the way the question was written, I suspect you can sort how you want, but I agree the title is a little misleading.
[–] 1 point2 points (0 children)
Crappy SBCL, I'd be happy for any comments or criticism
``````(defun list-merger (a b)
(loop for x in a do
(insert x b))
b)
(defun insert (a-element b)
(loop for i from 1 to (1- (length b)) do
(cond
((<= a-element (elt b i)) (setf (elt b (1- i)) a-element) (return-from insert t))
(t
(setf (elt b (1- i)) (elt b i)) (setf (elt b i) 0))))
(setf (elt b (1- (length b))) a-element))
``````
[–] 1 point2 points (1 child)
Is the goal here to write a sort and merge method without any built in sorting methods? If you can use system sort methods, you can just copy A into the buffer space and call sort() which does in-place sorting that doesn't allocate extra memory.
[–]1 2[S] 0 points1 point (0 children)
Correct; the real challenge here is to write your own merge-sort. If you look around at some of the general comments, you'll discover that there is a decently fast version ("fast" in the sense that it is still in O(N2 ) run-time, but the constant is much lower).
[–] 1 point2 points (0 children)
using javascript
``````function mergeSort(list1, list2){
var len2 = list2.length,
len1 = list1.length;
for(var i = 0, j, n, m; i < len1; i++){ // loop list 1
n = list1[i];
j = len2;
while(true){
m = list2[--j]; // going down list 2
if(m < n) { // start swapping
list2[j] = n;
n = m;
while(m !== 0){ // keep swapping until zero is swapped
m = list2[--j];
list2[j] = n;
n = m;
}
break;
}
}
}
return list2;
}
mergeSort([692,1,32],[0,0,0,14,15,123,2431]);
// [1,14,15,32,123,692,2431]
``````
[–] 1 point2 points (0 children)
Python solution:
``````def insertionSort(arr):
for i in range(1, len(arr)):
for j in range(i, 0, -1):
if arr[j] < arr[j-1]:
arr[j], arr[j-1] = arr[j-1], arr[j]
arrA = map(int, raw_input().split())
arrB = map(int, raw_input().split())
for i, a in enumerate(arrA):
arrB[i] = a
insertionSort(arrB)
print ' '.join(map(str, arrB))
``````
[–] 1 point2 points (0 children)
Python Solution:
``````def sortRange (myList, range1, range2):
while (((range1[1]-range1[0]) >= 0) and ((range2[1]-range2[0]) >= 0)):
if myList[range2[0]] < myList[range1[0]]:
temp = myList[range2[0]]
k = 0
while k <= range1[1]-range1[0]:
myList[range1[1]+1-k] = myList[range1[1]-k]
k += 1
myList[range1[0]] = temp
range2[0] += 1
range1[0] += 1
range1[1] += 1
else:
range1[0] += 1
return
def sorter(myList):
i = 1
while i < len(myList):
j = 0
while (j+i) < len(myList):
if (j+(2*i)-1) >= len(myList):
sortRange(myList, [j, j+i-1], [j+i, len(myList)-1])
else:
sortRange(myList, [j, j+i-1], [j+i, j+(2*i)-1])
j += (2*i)
i += i
return
listA = [692, 1, 32]
listB = [0, 0, 0, 14, 15, 123, 2431]
i = 0
for e in listA:
listB[i] = e
i += 1
sorter(listB)
print listB
``````
[–] 1 point2 points (0 children)
Here's my solution in Python:
``````def inline_merge(list1, list2):
for index,item in enumerate(list1):
for current in range(len(list2)-1):
next = current + 1
if list2[current] == 0:
list2[current] = item
if list2[current] <= list2[next]:
break
else:
list2[current] = list2[next]
list2[next]= item
``````
[–] 1 point2 points (0 children)
First time posting, Perl
``````my @first = (692, 1, 32);
my @second = (0, 0, 0, 14, 15, 123, 2341);
my \$i = 0;
foreach (@first)
{
\$second[\$i] = \$_;
\$i++;
}
\$i = 0;
while (\$i < \$#second)
{
if (\$second[\$i+1] < \$second[\$i])
{
my \$n = \$second[\$i];
\$second[\$i] = \$second[\$i+1];
\$second[\$i+1] = \$n;
\$i = 0;
}
\$i++;
}
print join (',', @second);
``````
[–] 1 point2 points (0 children)
First attempt at any of these challenges.
C#
Code critique, shortcuts, general advice and pointers/tips/discussions would be greatly appreciated.
``````using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace c126e_realworldmergesort
{
class Program
{
static void MergeSort(List<int> a, List<int> b)
{
for (int i = a.Count() - 1; i >= 0; i--)
{
b[i] = a[i];
for (int k = i; k < b.Count() - 1; k++)
{
if (b[k] > b[k + 1])
{
int temp = b[k];
b[k] = b[k + 1];
b[k + 1] = temp;
}
else
break;
}
}
}
static void Main(string[] args)
{
List<int> listA = new List<int>() { 692, 1, 32 };
List<int> listB = new List<int>() { 0, 0, 0, 14, 15, 123, 692, 2431 };
MergeSort(listA, listB);
Console.Write("List A: ");
foreach (int integer in listA)
Console.Write(integer.ToString() + " ");
Console.WriteLine();
Console.Write("List B: ");
foreach (int integer in listB)
Console.Write(integer.ToString() + " ");
}
}
}
``````
OUTPUT:
List A: 692 1 32
List B: 1 14 15 32 123 692 692 2431
[–] 1 point2 points (0 children)
My code in JAVA. Merged then sorted.
``````public class Merge {
public Merge (int[] a, int[] b){
int k = 0;
for(int i = 0 ; i<b.length;i++){
if(b[i]==0){
k = i;
b[i] = a[k];
}
}
for(int i = 0; i<b.length;i++){
for(int j = i ; j<b.length;j++){
if(b[i] > b[j]){
int temp = b[i];
b[i] = b[j];
b[j] = temp;
}
}
}
}
public static void main(String args[]){
int[] a = {692,1,32};
int[] b = {0,0,0,14,15,123,2431};
new Merge(a,b);
for(int i = 0 ; i < b.length ; i++){
System.out.print(b[i] + " ");
}
}
}
``````
[–] 1 point2 points (2 children)
Will update this my answer later, but are we allowed to use any built-in function of a language like min()/max()?
UPDATE: Python 3.3 Implementation
Critique is welcome, thank you.
``````sampleA = [692,1,32]
sampleB = [0,0,0,14,15,123,2431]
def realWorldMergeSort(listA,listB):
"""\n
[05/20/13] Challenge #126 [Easy] Real-World Merge Sort\n
Return a merged and sorted list of elements of list A into list B
where you cannot allocate any extra space other than simple/trivial
local variables.
"""
for num in listA: # Loop through listA to locate a spot in listB
i = 0 # Initialize counter to loop through listB
while(num > listB[i]): # If the current number pulled from listA is bigger
if(listB[i] != 0): # than the element in listB and the element isn't 0,
listB[i - 1] = listB[i] # copy the current element into an earlier element in
i += 1 # the list. Once num is no longer greater than the
listB[i - 1] = num # current element place num in the previous element
# list and grab the next number in listA.
return listB # Return the list we've merged and sorted.
print(realWorldMergeSort(sampleA,sampleB))
``````
Code can also be located on my GitHub repository dedicated to r/dailyprogrammer.
[–]1 2[S] 1 point2 points (1 child)
Yes, but it's up to you to balance between using practical functions vs. "cheating" on the challenge. In such a case, min/max is really just fine.
[–] 0 points1 point (0 children)
I was trying to think of way to do this without helper functions in Python 3.3, but decided to do without min()/max() to pull this off.
[–] 1 point2 points (0 children)
I just started learning Perl, here's my Perl solution. Critique much appreciated.
``````sub insertion_sort
{
#list passed into subroutine
my @list = @_;
my \$index = 0;
my \$index_copy = 0;
foreach my \$item (@list)
{
if (\$list[\$index] > \$item)
{
while (\$list[\$index + 1] < \$list[\$index] and \$index >= 0)
{
my \$temp = \$list[\$index];
\$list[\$index] = \$list[\$index + 1];
\$list[\$index + 1] = \$temp;
\$index -= 1;
}
\$index_copy += 1;
\$index = \$index_copy;
}
}
#print solution
foreach my \$item (@list)
{
print(\$item . " ");
}
print("\n");
}
#read in both lists from files
open(FILE1, '<', \$ARGV[0]);
open(FILE2, '<', \$ARGV[1]);
my @listA = <FILE1>;
chomp(@listA);
my @listB = <FILE2>;
chomp(@listB);
my \$counter = 0;
#move numbers over from list A to list B
#empty spaces in B are all filled with zeros at start of list
foreach my \$item (@listB)
{
if (\$item == 0)
{
\$listB[\$counter] = \$listA[\$counter];
}
\$counter += 1;
}
#sort numbers in place
insertion_sort(@listB);
``````
Output: 1 14 15 32 123 692 2431
[–] 1 point2 points (0 children)
My solution in ruby, only uses 2 index Variables for listA and listB, i don't believe i'm creating any new arrays.
``````def mergeSort(listA, listB)
index = 0
until listA == [0, 0, 0] do
(listB.length-1).downto(0) do |i|
if listA[index] == 0
index+= 1
next
elsif listA[index] > listB[i]
listA[index], listB[i] = listB[i], listA[index]
end
end
end
return listB
end
listA = [ 692, 1, 32 ]
listB = [0, 0, 0, 14, 15, 123, 2431]
puts mergeSort(listA,listB)
``````
[–] 1 point2 points (0 children)
[–] 1 point2 points (0 children)
First time submitting, not sure how to format code. Also not sure if i'm following the premise correctly, but here's my try in Java:
``````public static void main(String[] args) {
int[] a = {691, 1, 32};
int[] b = {0, 0, 0, 14, 15, 123, 2431};
merge(a, b);
}
public static void merge(int[] a, int[] b) {
for (int elem : a) {
b[0] = elem;
for (int i = 0; i < b.length-1; i++) {
if (b[i] > b[i + 1]) {
int temp = b[i + 1];
b[i + 1] = b[i];
b[i] = temp;
} else {
break;
}
}
}
for (int elem2 : b) {
System.out.print(elem2 + " ");
}
}
``````
Output: 1 14 15 32 123 691 2431
[–] 1 point2 points (0 children)
code golfed c version @ 97 characters, feedback is appreciated:
``````typedef unsigned int k;c(k*a,k*b){return *a-*b;}s(k*a,k x,k*b,k y){qsort(memcpy(b,a,x*4),y,4,c);}
``````
you can run it with:
``````int main(int argc, char **argv) {
unsigned int a[]={692,1,32}, b[]={0,0,0,14,15,123,2431}, i;
s(a, 3, b, 7);
for (i = 0; i < 7; i++)
printf("%d ", b[i]);
}
``````
which outputs:
``````1 14 15 32 123 692 2431
``````
[–] 1 point2 points (0 children)
JavaScript solution
``````function merge ( arg1, arg2 ){
listA = arg1.split(" ");
listB = arg2.split(" ");
listB = listB.concat(listA);
var i = 0;
while( i < listB.length ){
listB[i] = parseInt(listB[i]);
if( listB[i] === 0 ) {
listB.splice(i, 1);
} else {
i++;
}
}
var x = 0;
while( x < listB.length ){
var track = false;
for( var j = x; j < listB.length; j++ ){
if( listB[x] > listB[j] ) {
track = j;
}
}
if( track ){
var val = listB[x];
listB.splice(x, 1);
listB.splice(track, 0, val);
} else {
x++;
}
track = false;
}
return listB;
}
console.log( merge( "692 1 32", "0 0 0 14 15 123 2431" ) );
``````
[–] 1 point2 points (0 children)
Done in Java... hope i read this question right. Its basically an insert and bubble sort.
``````public class Sort{
static int[] a;
static int[] b;
public static void main(String []args){
a = new int[] { 692,1,32 };
b = new int[] { 0,0,0,14,15,123,2431 };
sort();
}
public static void sort() {
for (int i = 0; i < b.length; i++) {
if (b[i] == 0) {
b[i] = a[i];
}
}
for (int i = 0; i < b.length - 1; i++) {
if (b[i] > b[i+1]) {
int temp = b[i];
b[i] = b[i+1];
b[i+1] = temp;
i = 0;
}
}
for (int i = 0; i < b.length; i++) {
System.out.print(b[i] + " ");
}
}
}
``````
[–] 1 point2 points (0 children)
public class easy126 {
``````/**
* Accepts an unSorted array and a Sorted array. Merges them,
* and arranges the product in numerical order.
* @param unSorted int array
* @param sorted int array
* @return merged int ArrayList
*/
public static ArrayList<Integer> mergeSort( int [] unSorted, int [] sorted){
ArrayList<Integer> merged = new ArrayList<Integer>();
boolean notDone = true;
/* first element is 0 so that if the unsorted element to add is less than
* the smallest sorted element, it is guaranteed to be added to the beginning of merged.
*/
//initialize merged with sorted list to add unsorted elements in their correct place.
for(int element:sorted){
}
for(int j=0 ; j<unSorted.length; j++){
notDone=true;
for(int i = merged.size()-1 ; i>=0 && notDone;i--){
if(unSorted[j]>merged.get(i)){
notDone = false;
}
}
//zeros are removed for clarity. (prompt removes zeros).
}for (int i=0; i<merged.size(); i++){
if (merged.get(i)==0){
merged.remove(i);
i--;
}
}
return merged;
}
public static void main(String[]args){
Scanner in = new Scanner(System.in);
//get unsorted list
System.out.println("Enter number of unsorted items.");
int numUnsorted=in.nextInt();
int [] unSorted = new int [numUnsorted];
System.out.println("Enter unsorted list: ");
for (int i=0; i<numUnsorted; i++){
unSorted[i]=in.nextInt();
}
//get sorted list
System.out.println("Enter number of Sorted Items.");
int numSorted = in.nextInt();
int [] sorted = new int [numSorted];
System.out.println("Enter sorted list: ");
for (int i=0; i<numSorted; i++){
sorted[i]=in.nextInt();
}
in.close();
//implement method and print to check.
ArrayList<Integer> result = new ArrayList<Integer>();
result = mergeSort(unSorted, sorted);
System.out.println(result);
}
``````
}
[–] 1 point2 points (0 children)
Edit: This is a Haskell implementation
``````intCMP :: Int -> Int -> Ordering
intCMP a b | a == b = EQ
| a < b = LT
| otherwise = GT
merge :: Ord a => (a -> a -> Ordering) -> [a] -> [a] -> [a]
merge _ [] ys = ys
merge _ xs [] = xs
merge cmp (x:xs) (y:ys)
| xx || xy = x : merge cmp xs (y:ys)
| yx = y : merge cmp (x:xs) ys
where yx = (cmp x y) == GT
xy = (cmp x y) == LT
xx = (cmp x y) == EQ
msort :: Ord a => (a -> a -> Ordering) -> [a] -> [a]
msort _ [] = []
msort _ [x] = [x]
msort cmp xs = merge cmp xs1 xs2
where xs1 = msort cmp (take (length xs `div` 2) xs)
xs2 = msort cmp (drop (length xs `div` 2) xs)
``````
[–] 1 point2 points (0 children)
Been racking my brain today over a Ruby quicksort. But after a few failed attempts (and learning several new things about the language), I eventually settled on bubblesort. I'm relatively new to Ruby and I've seen some sexy-looking minimalist solutions here already. If someone could explain how I should trim the fat from this, I'd really be grateful. Feedback highly encouraged and appreciated!
``````def bubble_sort(list)
loop do
swapped = false
0.upto(list.size - 2) do |i|
if list[i] > list[i + 1]
list[i], list[i + 1] = list[i + 1], list[i]
swapped = true
end
end
break unless swapped
end
end
# arrays
array_a = [692, 1, 32]
array_b = [0, 0, 0, 14, 15, 123, 2431]
# merge
array_a.each_with_index do |n, i|
if array_b[i] == 0
array_b[i] = n
end
end
# sort
bubble_sort(array_b)
# print
array_b.each do |n|
print "#{n} "
end
``````
Sample Input:
692 1 32
0 0 0 14 15 123 2431
Sample Output:
1 14 15 32 123 692 2431
[–] 1 point2 points (0 children)
I may have this wrong but here are two attempts in python 2.7. I think #2 may be correct, but I'm not sure I understand the requirement about not allocating extra space. Any comments would be appreciated.
solution 1:
``````def mergesort(lista, listb):
listb = listb + lista
listb = set(listb)
listb = list(listb)
listb.sort()
return listb
``````
Output:
``````[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 45]
``````
This one doesn't bring the zeros from list A:
``````a = [0, 0, 1, 3, 4, 8, 3, 6, 3, 7, 2, 8]
b = [7, 8, 5, 9, 3, 45, 5, 12, 6]
def mergesort(a, b):
#merge
for i in a:
if i == 0:
pass
else:
b.append(i)
#remove duplicates and sort
b = set(b)
b = list(b)
b.sort()
print(b)
mergesort(a, b)
``````
Output:
``````[1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 45]
``````
[–] 1 point2 points (0 children)
``````l1 = map(lambda x: int(x), raw_input().strip().split())
l1 = tuple(l1)
l2 = map(lambda y: int(y), raw_input().strip().split())
l2 = sorted(l2)
for n in xrange(len(l1)):
if l2[n] == 0:
l2[n] = l1[n]
l2 = sorted(l2)
for i in l2: print i,
``````
[–] 1 point2 points (0 children)
I THINK I might have taken this a bit too literally. For sake of the exercise, I left the 0's in the B list, since that's the list that is safe to modify.
In Python:
ListA = [692 1 32] ListB = [0 0 0 14 15 123 2431]
for x in ListA: ListB.append(x) for y in ListB: if y <= 0: ListB.remove(y) ListB.sort() print ListB
If you have suggestion on how to make this more simple/ letting me know I missed some instruction, that would be great!
[–] 1 point2 points (0 children)
C
``````#include <stdio.h>
#include <stdlib.h>
int compare (void const *, void const *);
int merge(int *, int const *, size_t);
void print(int const *, size_t);
int main(int argc, char **argv) {
int A[] = { 692, 1, 32 };
int B[] = { 0, 0, 0, 14, 15, 123, 2431 };
merge(B, A, 3);
qsort(B, 7, sizeof(int), compare);
print(B, 7);
return 0;
}
int compare (void const *a, void const *b) {
return (*(int*)a - *(int*)b);
}
int merge(int *dest, int const *src, size_t size) {
for (size_t i = 0; i < size; i++)
dest[i] = src[i];
return 0;
}
void print(int const *list, size_t size) {
for (size_t i = 0; i < size; i++)
printf("%d ", list[i]);
printf("\n");
}
``````
[–] 1 point2 points (0 children)
``````main = do
la <- fmap ((map readInt) . words) getLine
lb <- fmap ((map readInt) . words) getLine
putStrLn (show (merge la lb))
merge [] ys = ys
merge (x:xs) ys = merge xs (insert x ys)
where insert x (y:ys)
| x > y = y : insert x ys
| otherwise = x : y : ys
``````
[–] 1 point2 points (0 children)
Java:
`````` public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String[] Astr = scanner.nextLine().split(" ");
int[] A = new int[Astr.length];
for(int i = 0; i < Astr.length; i++)
A[i] = Integer.parseInt(Astr[i]);
String[] Bstr = scanner.nextLine().split(" ");
int[] B = new int[Bstr.length];
for(int i = 0; i < Bstr.length; i++)
B[i] = Integer.parseInt(Bstr[i]);
for(int i = A.length-1; i>=0;i--){
for(int j = i; j< B.length-1; j++){
if(A[i] <= B[j+1]){
B[j] = A[i];
break;
}
B[j] = B[j+1];
B[j+1] = 0;
}
}
System.out.println(Arrays.toString(B));
}
``````
[–] 2 points3 points (5 children)
``````def mergesort(a,b):
#Naively put A's elements into B
for i in range(0,len(a)):
b[i] = a[i]
#Loop over B's elements
for j in range(0,len(b)-1):
iMin = j
#Look for the minimum in the remainder of B
for i in range(j+1,len(b)):
if b[i] < b[iMin]:
iMin = i
#Swap places if needed
if iMin != j:
temp = b[j]
b[j] = b[iMin]
b[iMin] = temp
return b
k = [692,1,32]
l = [0,0,0,14,15,123,2431]
print(mergesort(k,l))
``````
My not-so-well-thought-out solution in Python, using an in-place selection sort (O(n2 ))
[–]1 2[S] 0 points1 point (4 children)
Big hint: you can do it in linear time :D
Edit: I take that back.. I can't get the run-time below O(N2 )
[–]0 1 5 points6 points (3 children)
I think the linear time solution would only apply if both lists were already sorted. So given that A is unsorted, the optimal run time would be O(nlogn) if we copy all elements from A into B then do an inplace merge or quicksort from there.
[–]1 2[S] 1 point2 points (0 children)
My thoughts were that we could do a faster (relatively speaking) merge-sort by just sorting A, and then doing a trivial linear-time merge. The issue though is the fastest we could sort A would be in O(n log(n)), thus our limiting factor.
[–] 0 points1 point (1 child)
Or use the extra space in B as the space needed for the mergesort. No need for an inplace merge, or use of a quicksort.
[–]0 1 0 points1 point (0 children)
But can you do that and keep it in O(nlogn)? If you just use the extra space in B for mergesort you have to do an O(n) search through A every time for the smallest element, which comes out to O(n2).
[–] 1 point2 points (0 children)
c++
``````#include <iostream>
#include <vector>
#include <algorithm>
using std::vector;
using std::cout;
using std::endl;
using std::sort;
vector <int> merge(vector <int> a, vector<int> b, int length)
{
for(int iii=0; iii<length;iii++){
b[iii]=a[iii];
}
return b;
}
int main()
{
vector <int> a={692,1,32};
vector <int> b={0,0,0,14,15,123,2431};
b=merge(a,b,3);
sort(b.begin(),b.end());
for(int iii=0;iii<7;iii++){
cout<<b[iii]<<endl;
}
}
``````
[–] 1 point2 points (0 children)
c#. May have been able to save a few iterations by checking if I'd hit the zero's & breaking but this works (I think) ;)
``````class Program {
static void Main(string[] args) {
int[] listA = {691, 1, 32};
int[] listB = {0, 0, 0, 14, 15, 123, 692, 2431 };
int lift, drop;
for (int a = 0; a < listA.Count(); a++) {
for (int b = listB.Count() - 1; b >= 0; b--) {
if (listA[a] >= listB[b]) {
drop = listA[a];
for (int y = b; y >= 0; y--) {
lift = listB[y];
listB[y] = drop;
drop = lift;
}
break;
}
else
continue;
}
}
foreach(int i in listB)
Console.Write(@"{0} ", i);
}
}
``````
Output
`````` 1 14 15 32 123 691 692 2431
``````
[–] 1 point2 points (0 children)
Here's some Groovy:
``````def a = [692, 1, 32]
def b = [0, 0, 0, 14, 15, 123, 2431]
b[0..<a.size] = a
println mergeSort(b)
def mergeSort(def list) {
if (list.size() <= 1) return list
def middle = list.size() / 2
merge(mergeSort(list[0..<middle]), mergeSort(list[middle..<list.size]))
}
def merge(def left, def right) {
def result = []
def lIndex = 0, rIndex = 0
def lSize = left.size()
def rSize = right.size()
while (lIndex < lSize && rIndex < rSize) {
result << ((left[lIndex] <= right[rIndex]) ? left[lIndex++] : right[rIndex++])
}
if (lIndex < lSize) {
left[lIndex..<lSize].each { result << it }
}
if (rIndex < rSize) {
right[rIndex..<rSize].each { result << it }
}
return result
}
``````
Output:
``````[1, 14, 15, 32, 123, 692, 2431]
``````
[–] 1 point2 points (1 child)
Here's my Python solution:
``````unsorted = (692, 1, 32)
sorted = [0, 0, 0, 14, 15, 123, 2431]
shift = False
for element in unsorted:
index = len(sorted)-1
temp = element
while index >= 0:
if not shift and temp > sorted[index]:
shift = True
if shift:
temp2 = sorted[index]
sorted[index] = temp
temp = temp2
index-=1
shift = False
print sorted
``````
I feel decently proud of this solution. Takes the elements in the unsorted list and inserts them into the sorted list by going over it backwards. Once it finds where it needs to insert, it changes from inserting to shifting the elements backwards. I could have saved a variable by doing the xor trick, but I didn't consider the loss in readability worth it.
[–] 1 point2 points (0 children)
And here it is updated for optimality using iterative (seemed like using the stack went against the spirit of the challenge) binary search to find our insertion point:
``````unsorted = (692, 1, 32)
sorted = [0, 0, 0, 14, 15, 123, 2431]
def binarySearchNoRecurse(element, sortedList, begin, end):
while end > begin:
if element < sortedList[(begin+end)/2]:
end = (begin+end)/2
else:
begin = (begin+end)/2
if end - begin == 1:
return begin
return end
for element in unsorted:
index = binarySearchNoRecurse(element, sorted, 0, len(sorted)-1)
temp = element
while index >= 0:
temp2 = sorted[index]
sorted[index] = temp
temp = temp2
index-=1
print sorted
``````
Running time of O(nlogn).
[–] 1 point2 points (0 children)
nodejs var net = require('net');
``````var server = net.createServer(function(socket){
socket.setEncoding('utf-8');
socket.on('data', function(d){
var inputs = JSON.parse(d);
while(inputs.B[0] === 0) { inputs.B.shift(); }
//Take from A
while(inputs.A.length !== 0) {
var element = inputs.A.shift();
var index = 0;
//Put in B
while(inputs.B[index] <= element){ index++; }
inputs.B.splice(index, 0, element);
}
socket.write(JSON.stringify(inputs));
});
socket.on('end', function(){
//Take this out if you wish to leave it open.
server.close();
});
console.log('127.0.0.1:1337 - net(utf-8)\n
takes JSON string "{ A: [unsorted to merge array], B [zero filled sorted array] }"'
}).listen(1337, "127.0.0.1");
var client = net.createConnection({port : 1337, host : "127.0.0.1"}, function(){
client.setEncoding('utf-8');
//Example of input from a client
var data = { A: [692, 1, 32], B: [0, 0, 0, 14, 15, 123, 2431] };
client.write(JSON.stringify(data));
});
client.on('data', function(d){
console.log(d);
client.destroy();
});
``````
Edited for line length
[–] 0 points1 point (1 child)
So my theory is that you
``````take the first non-zero value of B (target), then check each element of A.
If you find an element which is less than the target and place it before the target,
and make it the new target. Continue searching A from the same index. IF you reach the end of list A,
place the target at the lowest unfilled index. Repeat until list A is empty.
``````
Am I on the right track?
[–]1 2[S] 0 points1 point (0 children)
Yep! That's a very solid approach, if not the most "correct" approach. Now, we can take this algorithm and try to clasify its complexity. My understanding is that this approach is still O(N2 ) because of the two loops required (one over list A when attempting to insert into B, and then over B while doing the overall iteration).
[–] 0 points1 point (0 children)
Scala:
``````import scala.collection.mutable.LinkedList
object MergeSort {
for (ae <- a) {
for ((be, bi) <- b.view.zipWithIndex if be < ae && bi > 0) {
b(bi - 1) = be
b(bi) = ae
}
}
b
}
def main(args: Array[String]) {
val a = List(692, 1, 32)
val b = LinkedList(0, 0, 0, 14, 15, 123, 2431)
val result = mergeSort(a, b)
println(result)
}
}
``````
[–] 0 points1 point (0 children)
Java. I'm a first time submitter and a software engineer of a few years experience. However, I've become so used to just working in established applications that I need the practise of challenges with the fundamentals.
``````public static void main( String[] args ) {
int[] firstArray = new int[] { 692, 1, 32 };
int[] finalArray = new int[] { 0, 0, 0, 14, 15, 123, 2431 };
// prettyPrint( firstArray );
int idx = -1;
// Find first non-zero element of the merge array
for ( int i = 0; i < finalArray.length; i++ ) {
if ( finalArray[i] > 0 ) {
idx = i;
break;
}
}
if ( idx == -1 ) {
System.err.println( "No zero elements" );
System.exit( 1 );
}
for ( int e : firstArray ) {
finalArray = mergeSort( finalArray, e, idx-- );
}
prettyPrint( finalArray );
}
/**
* Merge sorts the insert element into the given array, starting from the
* given index.
*
* @param finalArray
* @param insert
* @param startIndex
* @return
*/
private static int[] mergeSort( int[] finalArray, int insert, int startIndex ) {
// From the start index, iterate through until we find the first element
// that's bigger than the insert.
int insertIndex = -1;
// prettyPrint( finalArray );
for ( int i = startIndex; i < finalArray.length; i++ ) {
if ( finalArray[i] > insert ) {
insertIndex = i - 1;
break;
}
}
// Shunt everything from startIndex to insertIndex back a position
for ( int i = startIndex - 1; i < insertIndex; i++ ) {
finalArray[i] = finalArray[i + 1];
}
finalArray[insertIndex] = insert;
// prettyPrint( finalArray );
return finalArray;
}
// Useful for debugging, though it does kinda ruin the idea of using primitives only!
public static void prettyPrint( int[] toPrint ) {
System.out.println( ArrayUtils.toString( toPrint ) );
}
``````
[–] 0 points1 point (0 children)
Nothing fancy or elegant, but here it is in Python:
``````def merge(a, b):
i = 0;
while i < len(a):
b[i] = a[i]
i+=1
sort(b)
def sort(b):
lowest = c1 = 0
c2 = c1 + 1
while c1 < len(b):
while c2 < len(b):
if b[lowest] > b[c2]:
lowest = c2
c2+=1
temp = b[c1]
b[c1] = b[lowest]
b[lowest] = temp
c1+=1
c2 = c1+1
lowest = c1
print b
merge([3,4,8,10,14], [0,0,0,0,0,1,2,6,9,20])
``````
Output:
``````[1, 2, 3, 4, 6, 8, 9, 10, 14, 20]
``````
[–] 0 points1 point (0 children)
I've been without my laptop for a week now, it stopped working and is being fixed by Apple, and I have been chaffing at the bit to program.
Pythonista + iPad + Apple Wireless Keyboard = Semi-acceptable solution... :D
Without further ado, heres my solution in python 2.7 (Pythonista):
``````def arr_isort(A, B):
for a in A:
for i in range(len(B)):
if B[i] == 0:
continue
B[i-1] = B[i]
if a <= B[i]:
B[i-1] = a
break
if __name__ == '__main__':
A = map(int, raw_input().strip().split())
B = map(int, raw_input().strip().split())
arr_isort(A, B)
print ' '.join(map(str, B))
``````
Output is as expected:
``````1 14 15 32 123 692 2431
``````
Not sure how I could simplify it any further. Any comments will be gladly appreciated.
[–] 0 points1 point (0 children)
in Go first time posting, any kind of critique is well met
``````package main
import "fmt"
func merge_sort(a, b []int) []int {
for _, no := range a {
i := 0
for no > b[i] {
if b[i] != 0 {
b[i-1] = b[i]
}
i += 1
}
b[i-1] = no
}
return b
}
func main() {
listA := []int{123, 5, 28}
listB := []int{0, 0, 0, 43, 64, 72, 490}
fmt.Println(merge_sort(listA, listB))
}
``````
[–] 0 points1 point (0 children)
``````def mergeSort(A, B):
for i in range(0,len(A)): #Iterator for A
e = 0 #iterator for B
while(B[e] < A[i] and e < len(B)-1): #While element of A is greater than the e element of B, keep moving through B (or until reach end of B)
e+=1
if(e != len(B) - 1): #Correct if exited previous while loop due to larger value found at that spot
e-=1 #(It won't correct if exited previous while loop because e reached end of B)
holdValue = A[i] #Will remain between iterations of the for loop
while(e > 0): #Iterate backwards on B (shifting values 1 to the left)
tempHold = B[e]
B[e] = holdValue
holdValue = tempHold
e-=1
B[e] = holdValue #final shift (for B's 0th element)
return B
``````
[–] 0 points1 point (6 children)
Pretty decent Python solution, perhaps?
``````def mergeSort(a, b):
for i in range(len(a)):
b[i] = a[i]
b.sort()
print b
``````
Sorry, is that cheating?
[–]-9 8 3 points4 points (3 children)
Well, the goal is to perform the operation with a fixed amount of memory. The `range` function allocates a new array. Even ignoring that, Python uses Timsort, which allocates a stack.
[–] 0 points1 point (0 children)
Ahhh! I get the goal of the problem now... but, in understanding it, I've just realized that I don't know how to do it...
[–] 0 points1 point (0 children)
In Python 3, `range` acts like Python 2.x's `xrange` and only returns an iterator.
I started down a similar road, basically Rapptz's approach, but trying to use as much of `heapq` as possible. But I couldn't figure out a way to pop and re-heapify in place, so I had to do it myself:
``````import heapq
def merge_sort(a, b):
b[0:len(a)] = a
n = len(b)
# Stupid hack to use heapify() with < ordering.
for i in xrange(n):
b[i] = -b[i]
heapq.heapify(b)
# Repeatedly swap "largest" element to end of list, and re-heapify.
for size in xrange(n - 1, -1, -1):
b[0], b[size] = b[size], -b[0]
j = 0
while True:
left = 2 * j + 1
right = left + 1
swap = j
if left < size and b[left] < b[swap]:
swap = left
if right < size and b[right] < b[swap]:
swap = right
if swap != j:
b[j], b[swap] = b[swap], b[j]
j = swap
else:
break
a = [692, 1, 32]
b = [0, 0, 0, 14, 15, 123, 2431]
merge_sort(a, b)
print(b)
``````
[–] 0 points1 point (0 children)
Depends on the Python version. In 3 range does the function of xrange from Python 2, which is using a generator.
[–] 0 points1 point (1 child)
Well if you wanted to cheat and use a built in sort:
``````def merge_sort2(a, b):
b = sorted(a + b[len(a):])
return b
``````
[–] 0 points1 point (0 children)
That'd be better... I couldn't figure out how to sort it without either making another list or a really complicated while loop...
[–]0 1 0 points1 point (0 children)
i told myself i'd use dailyprogrammer to learn new languages but here i am going back to Java import java.util.*;
``````public class Easy {
public static void easy126(List<Integer> a, List<Integer> b){
Collections.sort(a); // O(nlogn), n=|a|
int bBegin = 0, aBegin = 0;
while(b.get(++bBegin)==0);
for(int j = 0; j < b.size(); j++){ // O(m) m=|b|
if(aBegin < a.size() && bBegin < b.size())
if(a.get(aBegin) < b.get(bBegin))
b.set(j, a.get(aBegin++));
else
b.set(j, b.get(bBegin++));
else if(aBegin < a.size())
b.set(j, a.get(aBegin++));
else
b.set(j, b.get(bBegin++));
}
prettyPrint(b);
}
public static void swap(List<Integer> list, int i, int j){
int t = list.get(i);
list.set(i, list.get(j));
list.set(j, t);
}
public static void prettyPrint(List<Integer> list){
for(Integer i : list)
System.out.print(i + " ");
System.out.println();
}
public static void main(String[] args){
List<Integer> a = new ArrayList<Integer>();
List<Integer> b = new ArrayList<Integer>();
easy126(a,b);
}
}
``````
output:
``````1 14 15 32 123 692 2431
``````
[–]1 3 0 points1 point (1 child)
Objective-C using Apple's Cocoa framework -- Data not hardcoded and can read in lists of any size (makes solution longer to handle flexible i/o)
Approach: made an object to handle the work. It holds 2 mutable arrays (A and B) which are collections of NSNumber objects (they are just wrapper objects that can hold basic C variable types - like ints).
It would be wierd to actually read in the 0 and make an NSNumber object for it and place the pointer in the B list (which is nothing more than an mutable array of NSNumber pointers). So I ignore the 0s.
I add the pointers in A to B. I insert the pointers in the beginning of the B list in sorted order. So this is a custom insertion sort of the A values into the first half of my B list. This is at worse case O( size of A2 ) and a best case of O( size of A) if A was already in sorted order. Then I walk this new B list 1 time and do a merge with the two sorted halves which is O (size of B) worse case and if all values in list A are < values in list B then it will be O (size of A)
``````// Numbers.h
#import <Foundation/Foundation.h>
@interface Numbers : NSObject
{
NSMutableArray *A;
NSMutableArray *B;
}
- (void) mergeAndSort;
- (void) displayA;
- (void) displayB;
@end
// Numbers.m
#import "Numbers.h"
@implementation Numbers
- (id) init {
self = [super self];
if (self) {
A = [[NSMutableArray alloc] initWithCapacity: 0];
B = [[NSMutableArray alloc] initWithCapacity: 0];
}
return self;
}
- (void) addToA: (int) n {
NSNumber *number = [NSNumber numberWithInt: n];
}
- (void) addToB: (int) n {
if (n > 0) {
NSNumber *number = [NSNumber numberWithInt: n];
}
}
void showNumbersIn(NSMutableArray *list) {
for (NSNumber *n in list)
printf("%d ", [n intValue]);
printf("\n");
}
- (void) displayA {
showNumbersIn(A);
}
- (void) displayB {
showNumbersIn(B);
}
- (void) mergeAndSort {
int bIndex;
int aIndex;
int moved = 0;
NSNumber *number;
// copy A values to buffer in B and do an insertion sort
// O( size(a)^2 worse case -- O( size (a)) for best case)
for (number in A) {
if (moved == 0) {
[B insertObject: number atIndex: 0];
moved++;
}
else {
bIndex = 0;
while ( (bIndex < moved) && ([number intValue] > [[B objectAtIndex: bIndex] intValue]) ) {
bIndex++;
}
[B insertObject: number atIndex: bIndex];
moved++;
}
}
// essentially we walk B once and merge the 2 sorted halves of B.
// O( size(B) )
bIndex = moved;
aIndex = 0;
while (moved > 0) {
if ([[B objectAtIndex: aIndex] intValue] < [[B objectAtIndex: bIndex] intValue]) {
aIndex++;
moved--;
} else {
while ([[B objectAtIndex: bIndex] intValue] < [[B objectAtIndex: aIndex] intValue]) {
bIndex++;
if (bIndex == [B count]) {
break;
}
}
NSNumber *popNumber = [B objectAtIndex: aIndex];
[B removeObjectAtIndex: aIndex];
[B insertObject: popNumber atIndex: (bIndex - 1)];
moved--;
}
}
}
@end
// main.m
#import <Foundation/Foundation.h>
#import "Numbers.h"
int main(int argc, const char * argv[])
{
@autoreleasepool {
char buffer[11];
char dummy;
Numbers *n = [[Numbers alloc] init];
scanf("%s%c", &buffer[0], &dummy);
while (dummy != '\n') {
scanf("%s%c", &buffer[0], &dummy);
}
scanf("%s%c", &buffer[0], &dummy);
while (dummy != '\n') {
scanf("%s%c", &buffer[0], &dummy);
}
printf("===========\n");
[n mergeAndSort];
[n displayB];
} // autoreleasepool
return 0;
}
``````
My test Input/Output:
``````692 1 32 5000
0 0 0 0 14 15 32 33 691 693 2500
===========
1 14 15 32 32 33 691 692 693 2500 5000
``````
Challenge input/Output:
``````692 1 32
0 0 0 14 15 123 2431
===========
1 14 15 32 123 692 2431
``````
[–]1 3 0 points1 point (0 children)
Javascript with jquery -- just started learning so it is very basic stuff. I like to be able to enter data and see it so I made a index.html and load in the javascript. Below are both the index.html and 126.js I gotta figure out a better way to test javascript out or find some cool IDE.
``````<!--index.html 126 challenge -->
<!DOCTYPE html>
<html>
<title></title>
<script src="./126.js"></script>
<script>
</script>
<body>
</body>
</html>
// 126.js
var A;
var B;
var mergeAndSort = function () {
var temp;
var i;
var j;
var moved;
for (i = 0; i < A.length; i++)
B[i] = A[i];
for (i = 1; i < A.length; i++) {
temp = B[i];
j = i - 1;
while (j >= 0 && B[j] > temp) {
B[j+1] = B[j];
j--;
}
B[j+1] = temp;
}
i = 0;
j = A.length;
moved = j;
while (moved > 0) {
if (B[i] < B[j]) {
i++;
moved--;
} else {
while (B[i] > B[j]) {
j++;
if (j === B.length)
break;
}
temp = B.splice(i,1);
B.splice(j-1, 0, temp);
moved--;
}
}
};
var main_loop = function() {
var outputString = "";
A = prompt("Enter A values").split(" ");
B = prompt("Enter B values").split(" ");
for (var i = 0; i < A.length; i++)
A[i] = parseInt(A[i]);
for (var i = 0; i < A.length; i++)
B[i] = parseInt(B[i]);
mergeAndSort();
for (var i = 0; i < B.length; i++)
outputString = outputString + B[i] + " ";
};
main_loop();
});
``````
[–] 0 points1 point (0 children)
First time doing these. Not sure I fully understand the problem, tbh, but here's my python 2.7 fudge:
``````def merge_sort(a, b):
# Fill the slots in b with a
for i, j in enumerate(a):
b[i] = j
pos = 0
while pos < len(b):
j = pos - 1
while j >= 0:
if b[j + 1] < b[j]:
temp = b[j + 1]
b[j + 1] = b[j]
b[j] = temp
j = j - 1
pos += 1
return b
print merge_sort([692, 1, 32], [0, 0, 0, 14, 15, 123, 2431])
``````
output:
``````[1, 14, 15, 32, 123, 692, 2431]
``````
Probably both sub-optimal and unpythonic :)
[–] 0 points1 point (5 children)
My Python solution without using built in sort.
``````def merge_sort(a, b):
for k, v in enumerate(a):
for s in range(len(a) - k, len(b)):
if v <= b[s]:
b = b[k:s] + [v] + b[s:]
break
else:
b = b[1:] + [v]
return b
``````
[–] 0 points1 point (4 children)
Don't slices return new lists?
Also in Python < 3, range() creates a new list. You can use xrange() to avoid having to put a new list in memory.
[–] 0 points1 point (3 children)
In Python 3, range is xrange.
[–] 0 points1 point (2 children)
I restricted my comment about range() to Python < 3 for that very reason.
[–] 1 point2 points (1 child)
Oh, I thought you were declaring love for Python!
[–] 0 points1 point (0 children)
Hahaha.
[–] -1 points0 points (1 child)
Ruby Solution:
``````def sort(a,b)
b.push(*a).sort!.delete(0)
puts b
end
a = 692, 1, 32
b = 0, 0, 0, 14, 15, 123, 2431
sort(a,b)
``````
Output:
1 14 15 32 123 692 2431
EDIT: If you downvote this solution please let me know where I could improve the code.
[–] 1 point2 points (0 children)
While I understand why you did it like you did, point is to create your own sort, not use sort!.
you must implement your sort / merge function
If you use sort, you dont have to do it new method, you can just join arrays and sort the result.
``````(a+b).sort!.drop(a.length)
``````
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http://www.ask.com/web?q=What+Is+45+Divided+By+4%3F&o=2603&l=dir&qsrc=3139&gc=1
| 1,472,144,011,000,000,000
|
text/html
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crawl-data/CC-MAIN-2016-36/segments/1471982293692.32/warc/CC-MAIN-20160823195813-00139-ip-10-153-172-175.ec2.internal.warc.gz
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| 15,383
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Web Results
Mar 4, 2009 ... Remainder When 45 is divided by 4 QID 599 ... Uploaded on Mar 4, 2009 ... Pre- Calculus - How to divide polynomials using long division ...
www.algebra.com/algebra/homework/equations/Equations.faq.question.572284.html
45 divided by 9=___ divided by 8 12 divided by ____=15 divided by 5 ... is 4. Therefore 12 divided by 4 =15 divided by 5. Answer by scott8148(6628) · About Me ...
The number 45 is divided into four unequal parts such that when 2 is ...
The first part is 8, the second part is 12, the third part is 5, and the fourth part is 20 . The number ... E - 2 + E + 2 + E / 2 + 2 * E = 4 * E + E / 2 = (9 * E) / 2 = 45
What is the remainder when ... - Quora
www.quora.com/What-is-the-remainder-when-123456789101112131415161718192021222324252627282930313233343536373839404142434481-is-divided-by-45
The remainder that you get when you divide a number by n is called that number modulo n. ... For example, 23 modulo 4 is 3, since when you divide 4 into 23, you ... ... Note that 45 is the product of 5 and 9, two relatively prime numbers.
Division ÷ | Basics of Arithmetic | SkillsYouNeed
www.skillsyouneed.com/num/division.html
5 does not divide by 4 to leave a whole number, as 5 is greater than 4. We are only ... 5 goes into 45 9 times, but not exactly as it leaves a remainder of 3.
How to Divide - Instructables
www.instructables.com/id/How-to-Divide-1/
I forgot how to divide and I have always sucked at long division. .... so much 4 passing 8th grad so hard i faild a "simple" divison test i am now failing cant do ...
Writing a Remainder as a Reduced Fraction - Math Forum - Ask Dr. Math
mathforum.org/library/drmath/view/58860.html
... have two problems to solve: 1,652 divided by 12 and 43,144 divided by 45. ... Let's divide 23 by 4: ___5_ 4 ) 23 20 -- 3 This means that if we ...
Dividing by Zero - Math is Fun
www.mathsisfun.com/numbers/dividing-by-zero.html
Don't divide by zero or this could happen! Just kidding. Divide by ... So they get 4 each: 12/3 = 4 ... We can't share among zero people, and we can't divide by 0.
Division - Math is Fun
www.mathsisfun.com/numbers/division.html
Answer: 12 divided by 3 is 4: they get 4 each. Symbols. ÷ /. We use the ÷ symbol, or sometimes the / symbol to mean divide: ...
Long Division With Remainders - Math is Fun
www.mathsisfun.com/long_division2.html
4 ÷ 25 = 0 remainder 4, The first number of the dividend is divided by the divisor. The whole number result is placed at the top. Any remainders are ignored at ...
45 / 4 = 11.25
What is 45 divided by 4 - Answers.com
What is the answer for 720 divided by 45? ... 45 divided by 9 is 4 no it is not four it is 5 5 times 8 is 40 so 5 times 9 must be 45, right? so 45 divided by 9 is 5 you ...
Divide Two Numbers- WebMath
www.webmath.com
Divide Two Numbers - powered by WebMath. ... solution for the division of two numbers. Fill in the division problem with your numbers, then click "Divide." ...
“fair sharing” because when you divide - Developmental Math Topic ...
www.montereyinstitute.org
12 ÷ 3 = 4 (with a division symbol; this equation is read “12 divided by 3 equals 4. ”) .... 45 ÷ 6. 6 • 7 = 42. How many sixes are there in 45? Try 7. 45 – 42 = 3.
| 985
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| 3.953125
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CC-MAIN-2016-36
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longest
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en
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https://antoncao.me/blog/voting
| 1,725,825,269,000,000,000
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crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00596.warc.gz
| 92,865,607
| 4,525
|
One Way to Think about Voting
Feb 20, 2020
It can be difficult to motivate oneself to vote in elections when the voter base is large because, in the grand scheme of things, one vote doesn't matter. However, if a lot of people operate under this mindset then it does matter, since the group of people who end up voting might not be representative of the population as a whole. I'd like to propose a different way of thinking about voting in order to counteract this mindset. Specifically, what I'm proposing is meant to be a helpful philosophy, not a rigorous (or even accurate) mathematical model.
The main idea is to believe that individual votes are not independent. If you do vote, then there is a higher chance that people like you will also vote. The opposite is true if you don't. Therefore, the effect of your vote is magnified by the group of people you belong to.
More formally, let $$G$$ be the population that is eligible to vote in an election. Partition $$G$$ into disjoint groups $$g_1, g_2, \cdots g_n$$ of "similar" people. By "similar", I mean that people in the same group have the same probability of voting. Now, a critical assumption to make is that "similar" people will also vote for the same candidate. I think this is a reasonable assumption because I believe that both a person's likelihood of voting and political preference are strongly determined by their background[1].
For each group $$g_i$$, let $$p_i$$ be the probability that someone in $$g_i$$ votes. We'll have an estimate of $$p_i$$ based on historical data, but conceptually we can think of $$p_i$$ as being described by an unknown probability distribution due to uncertainty.
Finally, let $$g_i$$ be the group that you are a part of, and $$v$$ be the indicator variable representing whether you vote ($$1$$) or not ($$0$$). By Bayes' theorem, $Pr(p_i = p | v = 1) = \frac{Pr(v = 1 | p_i = p)Pr(p_i = p)}{Pr(v = 1)}$ The right hand side simplifies to $\frac{p \cdot Pr(p_i = p)}{ \mathbb{E}[p_i] }$ because of the definition of $$v$$. This implies that $$Pr(p_i = p | v = 1)$$ grows with respect to $$Pr(p_i = p)$$ as $$p$$ increases, which means $$\mathbb{E}[p_i | v = 1] > \mathbb{E}[p_i]$$. Similarly, $$\mathbb{E}[p_i | v = 0] < \mathbb{E}[p_i]$$. Because the amount of people from group $$g_i$$ expected to vote is $$\mathbb{E}[p_i] \cdot | g_i |$$, if you vote then you are effectively increasing the number of votes for your desired candidate by $$(\mathbb{E}[p_i | v = 1] - \mathbb{E}[p_i | v = 0]) \cdot | g_i |$$, which is probably much bigger than $$1$$.
This approach scales because the size of your group $$|g_i|$$ will be roughly proportional to the population $$|G|$$ that is participating in the election, so it can be applied to elections of all sizes, local and national.
Footnotes
1. For example, I consider myself to belong to a group consisting of Asian-American college students who grew up in California. To a first approximation of political views, almost all of us would probably vote for the Democrat candidate. Using a very rough back-of-the-envelope calculation, there are at least 30,000 people in this group who are eligible to vote. If the fact that I vote increases expected probability (discussed in the second to last paragraph) by 1%, then I am effectively contributing 300 votes to my desired candidate. Back to text.
Acknowledgements
Thanks to Ganesh for helping me proofread this post!
| 866
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| 3.671875
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CC-MAIN-2024-38
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en
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https://twiki.cern.ch/twiki/bin/view/Sandbox/ContentSigmaQualityTest?sortcol=2;table=1;up=0
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crawl-data/CC-MAIN-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00112.warc.gz
| 616,713,813
| 11,028
|
-- ChadFreer - 2017-12-19
# Content Sigma Quality Test
This twiki is meant only to describe the Content Sigma quality test.
For documentation on all quality tests see Quality tests
## What is the content sigma quality test
The content sigma quality test is designed to test histogram bins/blocks of bins against their neighboring bins to compare the statistics of a run (visualization below).
The workflow for the test:
1. Get the content of bin or block to be tested
2. Get the average of neighbors with the associated deviation
3. compare content to neighbor average using poisson statistics and thresholds set in the xml file
1. Noisy/Dead Probability = poissonCDF[Content,Neighbor+unc]
4. calculate tolerance
1. Tolerance =normalCDF(tolerance) {set in xml file}
5. Calculate if block fails
Test uses poisson statistics to test thresholds and standard deviation to calculate uncertainty.
## Test Parameter and Flags
toleranceNoisy: The poisson threshold for noisy bins/blocks
toleranceDead: The poisson threshold for dead bins/blocks
Xblocks: the number of blocks for x axis that will be used. For example XBlocks=5 means that the x bins will be split into 5 blocks.
• FLAG: XBlocks=999 sets it to binwise
Yblocks: the number of blocks for y axis that will be used. For example YBlocks=5 means that the y bins will be split into 5 blocks.
• FLAG: YBlocks=999 sets it to binwise
neighboursX: the number of neighbor blocks in x that you wich to compare against. Neighbors do not wrap around histogram edges, they merely stop as shown above
• FLAG: neighboursX=999 means that neighbors in that dimension will be suppressed
neighboursY: the number of neighbor blocks in y that you wich to compare against. Neighbors do not wrap around histogram edges, they merely stop as shown above
• FLAG: neighboursY=999 means that neighbors in that dimension will be suppressed
noisy: 1 means this will use noisy test. 0 supresses noisy test
dead: 1 means this will use dead test. 0 supresses dead test
error: the percentage of failed bins/blocks required for the test to throw an error flag (between 0-1)
warning: the percentage of failed bins/blocks required for the test to throw a warning flag (between 0-1)
xMin: specify a specific lower x value for histogram. Anything lower will be ignored
• FLAG: xMin=0 uses lowest natural bin for test
xMax: specify a specific upper x value for histogram. Anything higher will be ignored
• FLAG: xMax=0 uses highest natural bin for test
yMin: specify a specific lower y value for histogram. Anything lower will be ignored
• FLAG: yMin=0 uses lowest natural bin for test
yMax: specify a specific upper y value for histogram. Anything higher will be ignored
• FLAG: yMax=0 uses highest natural bin for test
## XML declaration and Implentation
#### The following is the creation of a content sigma test
```<QTEST name="ContentSigma_Noisy">
<TYPE>ContentSigma</TYPE>
<PARAM name="toleranceNoisy">3</PARAM>
<PARAM name="Xblocks">999</PARAM>
<PARAM name="Yblocks">999</PARAM>
<PARAM name="neighboursX">10</PARAM>
<PARAM name="neighboursY">10</PARAM>
<PARAM name="noisy">1</PARAM>
<PARAM name="error">0.94</PARAM>
<PARAM name="warning">0.97</PARAM>
<PARAM name="xMin">0</PARAM> <!-- specify area of histogram to be analyzed -->
<PARAM name="xMax">0</PARAM>
<PARAM name="yMin">0</PARAM>
<PARAM name="yMax">0</PARAM>
</QTEST>```
#### The following is the application of the test on specific histograms
```<LINK name="*emtfTrackPhi">
<TestName activate="true">ContentSigma_Noisy</TestName>
| 873
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