courseware | Courseware setup and information for instructors | Runtime Evironment library

 by   saasbook Ruby Version: Current License: No License

kandi X-RAY | courseware Summary

kandi X-RAY | courseware Summary

courseware is a Ruby library typically used in Server, Runtime Evironment, Deep Learning, Nodejs applications. courseware has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

a/k/a UC Berkeley CS (W)169(A,L) Software Engineering, a/k/a CS 169.(1,2,3)x on EdX.
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              courseware has a low active ecosystem.
              It has 235 star(s) with 205 fork(s). There are 62 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 0 open issues and 10 have been closed. On average issues are closed in 1533 days. There are 4 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of courseware is current.

            kandi-Quality Quality

              courseware has 0 bugs and 0 code smells.

            kandi-Security Security

              courseware has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              courseware code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              courseware does not have a standard license declared.
              Check the repository for any license declaration and review the terms closely.
              OutlinedDot
              Without a license, all rights are reserved, and you cannot use the library in your applications.

            kandi-Reuse Reuse

              courseware releases are not available. You will need to build from source code and install.
              courseware saves you 804 person hours of effort in developing the same functionality from scratch.
              It has 1847 lines of code, 36 functions and 157 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

            Top functions reviewed by kandi - BETA

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            Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of courseware
            Get all kandi verified functions for this library.

            courseware Key Features

            No Key Features are available at this moment for courseware.

            courseware Examples and Code Snippets

            No Code Snippets are available at this moment for courseware.

            Community Discussions

            QUESTION

            Finding the right amount to save away
            Asked 2021-May-25 at 17:24

            I am currently learning python and as a little test a friend of mine gave me a problem from the MIT open courseware python course. However, I am struggling on part C of the problem. It requires that you use a binary search to find the right amount you need to save if you want to buy a house in 3 years given a starting salary.

            Here is the problem pdf for further details (scroll to part C): https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/assignments/MIT6_0001F16_ps1.pdf

            I have technically "solved" it and am able to find the correct value according to the test cases given but the percents have many digits and the amount of bisection searches counted is much higher than that of the test cases.

            I was wondering if there is actually anything wrong with my code and if I am implementing the binary search correctly with regards to this problem.

            Example test cases: Test Case 1:

            Enter the starting salary:​ 150000

            Best savings rate:​ 0.4411

            Steps in bisection search:​ 12

            MY RESULTS:

            Enter the starting salary:​ 150000

            Best savings rate:​ 0.4411391177390328

            Steps in bisection search:​ 40

            Thanks for any help!

            Apologies in advance for this question I am still learning ;P

            My Code:

            ...

            ANSWER

            Answered 2021-May-25 at 17:24

            The line which causes too many iterations is

            Source https://stackoverflow.com/questions/67691361

            QUESTION

            Running Maven Tests
            Asked 2020-Dec-20 at 23:10

            I am learning Java, part of the courseware is Maven.

            I have created a simple Maven project. And in the test class I have added the following code:

            ...

            ANSWER

            Answered 2020-Dec-20 at 22:58

            Two reasons:

            1. Your test class name doesn't match the regex you have in your surefire configuration. You have explicitly said that the class name needs to match *Test*

            2. You are importing the @Test annotation from org.junit, not org.junit.jupiter.api.Test

            Source https://stackoverflow.com/questions/65384940

            QUESTION

            How can i run multiple queries on my code in php?
            Asked 2020-Dec-17 at 13:20

            I have tried this code, but i'm getting errors

            ...

            ANSWER

            Answered 2020-Dec-17 at 13:20

            You need to use the value of the $sql array inside your loop.

            Change the following...

            Source https://stackoverflow.com/questions/65341442

            QUESTION

            Why does the binary search algorithm works for this 1D" Peak finding" problem?
            Asked 2020-Jul-09 at 06:08

            I was looking into MIT's open courseware first lecture on an introduction to algorithms and there is something that is not terribly obvious to me. You cant start watching the lecture at 24:30 here and the lecture notes with all the details of the 1D peak problem definition and solution in here

            The problem goes:

            for an array of "n" integer elements find a peak

            and gives an example array of size 8: example = [6,7,4,3,2,1,4,5]

            Definition of a peak For the example array above example[1] and example[7] are "peaks" because those numbers are greater or equal than their adjacent element and a special condition for the last element of the array applies that it only needs to be greater than or equal to the element preceding it. that is:

            ...

            ANSWER

            Answered 2020-Jul-09 at 06:08

            Given the condition above in A why go to the left? instead of the right?

            If you would go to the right (without first checking condition B), there is a small probability that the values at the right will keep going down (from left to right), and you would not find a peak there.

            At the left side, however, you know that you cannot have that situation, as you have found at least one value that is higher (the neighbour) and could potentially even be a peak. Here is how you can prove that a peak exists at the left side (this is not a description of the algorithm; just a way to prove it):

            If the immediate neighbour is not a peak, then possibly the next one to its left is. If not, then possibly the next one to its left....etc. This series will end when finding a peak or arriving at the left most value. If none of the others were peaks, then this one must be it. This only happens when the values never decreased while looking further to the left.

            In short, whatever the situation at the left side, there is a peak somewhere there at that side, and that is all we need to know when choosing a side.

            Given the condition above in B why go to the right? instead of the left?

            This is of course the same reasoning, but mirrored.

            Note that you can decide to first check condition B and only then A. When both conditions are true at the same time, you can actually choose which side to go. This is where you got the feeling from that the choice looks "arbitrary". It is indeed arbitrary when both conditions A and B are true.

            But also think about the case where one of A and B is true and the other false. If you would go the wrong (downward) way, you have no guarantee that values would ever go up in that direction. And so there is a small probability that there is no peak on that side.

            Of course, there still could be a peak on that side, but since we are sure there is one on the other side, it is wise to go for certainty. We don't care about potentially discarding some peaks, as we only need to find one peak.

            The binary search algorithm assumes we start from a sorted array, so how come it makes sense to apply it to data that may be unsorted?

            Binary search for a particular value would only work in a sorted array, but here we are not looking for a particular value. The conditions of the value we are looking for are less strict. Instead of a particular value, we will be happy with any value that is a local peak.

            Source https://stackoverflow.com/questions/62793422

            QUESTION

            Adding Counter to recursive Fibonacci program
            Asked 2020-Jun-24 at 08:54

            I'm working with a basic Python MIT free courseware and I have run into a wall with a recursion exercise. The original program takes an integer and provides its Fibonacci using recursion. The book provides the script for the program, but the subsequent exercise asks to input a way for the program to recognize how many times fib(2) is executed on its way to calculating fib(n). ``

            Here is the code:

            ...

            ANSWER

            Answered 2020-Jun-24 at 08:54

            Explaining it by each line

            Source https://stackoverflow.com/questions/62544250

            QUESTION

            MergeSort problems parameters
            Asked 2020-May-03 at 05:25
            import java.lang.reflect.Array; 
            
            public class Sorting {
            public static void mergeSort(CompareInt[] arr) {
            
               for(int i=0;i<=arr.length-1;i++){
                    System.out.println("Initial Arr element:"+arr[i].val);
                    } 
                CompareInt[] arrAux;
                arrAux = new CompareInt[arr.length];
                mergeS(arr,0,arr.length-1,arrAux);       
            }
            
            public static void mergeS(CompareInt[] arr,int startI,int endI,CompareInt[] arrAux){
                //System.out.println("StartIndex:"+startI+" EndIndex:"+endI);
                if(endI-startI<=0)
                    return;
            
                int mid = (startI+endI)/2;
               // System.out.println("Midpoint value:"+mid);
            
                mergeS(arr,startI,mid);
                mergeS(arr,mid+1,endI);
            
              //  System.out.println("Inside mergeS");
              //  System.out.println("StartIndex:"+startI+" EndIndex:"+endI+" Midpoint value:"+mid);
             //   System.out.println("Arr length:"+arr.length);
            
                arrAux = merge(arr,mid);
            
                for(int i=0;i<=arr.length-1;i++){
                    arr[i]=arrAux[i];
                    System.out.println("Arr element:"+arr[i].val);
                    }
            }
            
            
            public static CompareInt[] merge(CompareInt[] arr,int midpoint){
            int i=0,j=0,k=0;
            int n1 = arr.length-midpoint;
            int n2 = arr.length-n1;
            
            //System.out.println("Midpoint value inside merge:"+midpoint);
            //System.out.println("N1:"+n1+" N2:"+n2);
            
            CompareInt[] L;
            CompareInt[] R;
            CompareInt[] resA;
            L = new CompareInt[n1];
            R = new CompareInt[n2];
            resA = new CompareInt[n1+n2];
            
            
            for(i=0;i
            ...

            ANSWER

            Answered 2020-May-03 at 05:25

            This code will sort out your problem :

            Source https://stackoverflow.com/questions/61569883

            QUESTION

            for loop for the two-dimentional String array
            Asked 2020-Mar-28 at 18:59

            I want to have the following outcome:

            and i was writing the following code for this purpose:

            ...

            ANSWER

            Answered 2020-Mar-28 at 18:42

            The nulls at the first line are happening because you have 2 repeated conditional statements if (i == 0) and the other nulls on each line are happening because you are concatenating the char with the string using += which is giving nullA etc so instead do this:

            Source https://stackoverflow.com/questions/60905017

            QUESTION

            read_csv() adds "\r" to *.csv input
            Asked 2020-Mar-26 at 10:41

            I'm trying to read in a small (17kb), simple csv file from EdX.org (for an online course), and I've never had this trouble with readr::read_csv() before. Base-R read.csv() reads the file without generating the problem.

            A small (17kb) csv file from EdX.org

            ...

            ANSWER

            Answered 2020-Mar-26 at 10:41

            In a nutshell, the characters are inside the file (probably by accident) and read_csv is right to not remove them automatically: since they occur within quotes, this by convention means that a CSV parser should treat the field as-is, and not strip out whitespace characters. read.csv is wrong to do so, and this is arguably a bug.

            You can strip them out yourself once you’ve loaded the data:

            Source https://stackoverflow.com/questions/60864957

            QUESTION

            Initializing integer variables for comparisons
            Asked 2020-Feb-25 at 15:29

            I'm working my way through the MIT Open Courseware class Introduction to Computer Science and Programming in Python and I've spent an embarrassing amount of time trying to wrap my head around the "Finger Exercise" from the textbook:

            Write a program that examines three variables—x, y, and z—and prints the largest odd number among them. If none of them are odd, it should print a message to that effect.

            I wrote a couple of solutions that didn't quite work, missing odd numbers if there were larger even numbers. I finally threw in the towel and searched here for solutions that others people working on this class had asked. This solution from AFDev seemed to be the simplest to me (in the context of what the intent of the exercise was and how concise the solution was.) I combined that with my user input to get the following:

            ...

            ANSWER

            Answered 2018-Aug-05 at 19:01

            None seems fine if you're going to do it this way. (I'd say it's safer than using a large negative number.) You can get round the code duplication with None as e.g.

            Source https://stackoverflow.com/questions/51697030

            QUESTION

            Using weblate for translating content other than software
            Asked 2019-Oct-31 at 13:37

            I'm interested in translating the subtitles of a course in MIT Open Courseware to my native language. I want to do it in collaborative fashion so that it would be faster. I have found the weblate. However I'm not sure whether it's possible to use it for translations of material other than software.

            Do you have any experience using weblate to translate documents that are not necessarily software?

            ...

            ANSWER

            Answered 2019-Oct-31 at 13:37

            As you've probably already figured out, the many subtitle formats are directly supported by Weblate, so yes it's possible to translate non software projects.

            Weblate can be used for localizing any kind of content - I've seen localizing projects for printed media or books as well.

            Source https://stackoverflow.com/questions/58617754

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install courseware

            You can download it from GitHub.
            On a UNIX-like operating system, using your system’s package manager is easiest. However, the packaged Ruby version may not be the newest one. There is also an installer for Windows. Managers help you to switch between multiple Ruby versions on your system. Installers can be used to install a specific or multiple Ruby versions. Please refer ruby-lang.org for more information.

            Support

            [Open-ended response questions](https://github.com/saasbook/open-response-exam-questions) suitable for manual grading, along with reference solutions.
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