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  • sum of logarithams of prime numbers

    - by nadi
    Write a program that computes the sum of the logarithms of all the primes from 2 to some number n, and print out the sum of the logs of the primes, the number n, and the ratio of these two quantities. Test this for different values of n.

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  • My Package Version Number Appears Greater Yet apt-get Doesn't Select It

    - by nutznboltz
    Backstory: It was determined that when using lxc container VMs the Nagios nrpe shutdown script when run on the host of the containers would kill the nrpe processes inside the containers. This was remediated by changing the script to use pidfiles instead of searching the process table for the nrpe process. Regrettably start-stop-daemon is a C program that resulted from translating a Perl script and it shows. There are far too many global varibles in start-stop-daemon.c and although there are some nice blocks of comments there are far to few comments that explain the intent behind variable names such as "schedule" (the string "schedule" appears in many contexts.) The manual page for start-stop-daemon strongly suggests that unless you use the "--retry" option the start-stop-daemon program may return before the process it sent a signal to actually calls exit() and terminates, however it doesn't actually state this in plain English. The obtuseness of start-stop-daemon is most likely the reason that the "fixed" version of the script includes a dubious comment indicating that sometimes the pid file has not been removed. I can easily see why someone would not understand that he left the --retry option missing. This bug also causes failures when the script is given the "restart" option; the nrpe daemon will shutdown but not start up again. Did I mention that since applying the update our nrpe servers started crashing over and over? Repairing this is why I am doing this work. I have been working on remediating the fix. You can see my current work in this PPA. Actual Question: The upstream version number of nagios-nrpe-server in lucid-updates is 2.12-4ubuntu1.10.04.1 My PPA uses this version number 2.12-4ubuntu1.10.04.1.1~ppa1~lucid1 I check the rules here and use this test program and I am lead to believe that the version number I use in my PPA is greater than the one in lucid-updates yet when I ran: sudo add-apt-repository ppa:nutznboltz/nrpe-unbreak-lp-600941 sudo apt-get update sudo aptitiude dist-upgrade The replacement package was not installed. I was able to install it using sudo aptitude install nagios-nrpe-server=2.12-4ubuntu1.10.04.1.1~ppa1~lucid1 Can anyone explain this behavior? Why didn't my version number appear greater to "aptitude dist-upgrade"? Thanks $ cat /etc/apt/preferences Package: * Pin: release a=lucid-backports Pin-Priority: 400 Package: * Pin: release a=lucid-security Pin-Priority: 990 Package: * Pin: release a=lucid-updates Pin-Priority: 900 Package: * Pin: release a=lucid-proposed Pin-Priority: 400 $ ls /etc/apt/preferences.d/ $ Should not make any difference as a PPA cannot be in any of those pockets. I went ahead and bumped the version number in the PPA to 2.12-4ubuntu1.10.04.2~ppa1~lucid1. I'll see if that makes a difference. I do notice that lintian complains: W: nagios-nrpe-server: debian-revision-not-well-formed 2.12-4ubuntu1.10.04.2~ppa1~lucid1

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  • Is bundler ready for prime time?

    - by schof
    I'm considering using bundler for deploying a Spree app on Heroku. My question is, is bundler ready for prime time? I know there are some rough edges but I guess I'd like to know more about what the current limitations are and figure out if this is an option for us. Specifically, I'd like to do the git repository stuff git "git://github.com/indirect/rails3-generators.git" gem "rails3-generator Does anyone want to encourage/discourage me from this course of action? Anybody have experience with this on Heroku in particular?

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  • Phone number in meta description bad or good for local rankings NAP

    - by bybe
    Once again I'm at it with increasing people's local rankings and I've learnt so much about local rankings in the past 2 weeks it feels like my brain is gonna pop anyway, question is fairly simple for someone who engages in local rankings and I appreciate the question may be a little guess work but isn't SEO mostly guessing anyway? From what I've read and learned that Google works of a system called nap for local rankings (With many other factors but this question is purely based on NAP). For people who care about local rankings NAP stands for Name of Business / Address of Business / Phone Number for Business. Now what what I've read you don't need the whole NAP to be on one website, a P or just a N can help towards your local rankings. It's believed that NAP rewards more than just P and N for example but knowing Google they might have a diversity checker which is my concern what your get to in a moment. Now of course sites weight differently where your business is posted, it's certainly going to be more credible if your NAP details are on your national phone book than say a blog site, so taking in this consideration too. Pure Guess (Not apart of the question but none the less makes a good read on my belief). Now my guess work would make me believe that the formula would look something like (N)+(A)+(P)x(T) So (N)name would be 1 or 0 to indicate present or not So (A)dress would be 1 or 0 to indicate present or not So (P)hone would be 1 or 0 to indicate present or not So (T)rust would be 1-100 to indicate level of trust So a phone number appearing on youtube might look something like 0+0+1x95= 95 and a NAP appearing on your national phone book might look something like 1+1+1x100= 300 Please note that I'm not saying this is the sole factor and I'm sure its way more complex that this with things like other factors on the page, off the page (Reviews, Links, Clicks) and so on but its still a contributor). The Question My question is fairly simple and I'd imagine hard to impossible to have an actual definite answer to this but maybe someone has seen official wording else where on this, is it bad to include address or phone number in the Meta Description? The reason I ask is that one of my competitors has these elements in the meta descriptions and their local rankings are absolutely superb, the problem I have with this is scrap bot sites like 'Similar Too' 'Seo Rankings' and 1,000's of the other scrap box networks that scrap site and then make urls with your site information are mostly limited to your meta description what this means that your phone number, address and sometimes even your company name if the domain is exact will appear as AP, and even NAP on thousands of websites. So, is it a bad strategy to include phone number and address in meta description, everything I read into would suggest its good of course with the downside of maybe lowering quality of description for click thoughts but top rankings would increase this 10 folds anyhow..

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  • Enumerating large (20-digit) [probable] prime numbers

    - by Paul Baker
    Given A, on the order of 10^20, I'd like to quickly obtain a list of the first few prime numbers greater than A. OK, my needs aren't quite that exact - it's alright if occasionally a composite number ends up on the list. What's the fastest way to enumerate the (probable) primes greater than A? Is there a quicker way than stepping through all of the integers greater than A (other than obvious multiples of say, 2 and 3) and performing a primality test for each of them? If not, and the only method is to test each integer, what primality test should I be using?

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  • Simple prime number program - Weird issue with threads C#

    - by Para
    Hi! This is my code: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading; namespace FirePrime { class Program { static bool[] ThreadsFinished; static bool[] nums; static bool AllThreadsFinished() { bool allThreadsFinished = false; foreach (var threadFinished in ThreadsFinished) { allThreadsFinished &= threadFinished; } return allThreadsFinished; } static bool isPrime(int n) { if (n < 2) { return false; } if (n == 2) { return true; } if (n % 2 == 0) { return false; } int d = 3; while (d * d <= n) { if (n % d == 0) { return false; } d += 2; } return true; } static void MarkPrimes(int startNumber,int stopNumber,int ThreadNr) { for (int j = startNumber; j < stopNumber; j++) nums[j] = isPrime(j); lock (typeof(Program)) { ThreadsFinished[ThreadNr] = true; } } static void Main(string[] args) { int nrNums = 100; int nrThreads = 10; //var threadStartNums = new List<int>(); ThreadsFinished = new bool[nrThreads]; nums = new bool[nrNums]; //var nums = new List<bool>(); nums[0] = false; nums[1] = false; for(int i=2;i<nrNums;i++) nums[i] = true; int interval = (int)(nrNums / nrThreads); //threadStartNums.Add(2); //int aux = firstStartNum; //int i = 2; //while (aux < interval) //{ // aux = interval*i; // i=i+1; // threadStartNums.Add(aux); //} int startNum = 0; for (int i = 0; i < nrThreads; i++) { var _thread = new System.Threading.Thread(() => MarkPrimes(startNum, Math.Min(startNum + interval, nrNums), i)); startNum = startNum + interval; //set the thread to run in the background _thread.IsBackground = true; //start our thread _thread.Start(); } while (!AllThreadsFinished()) { Thread.Sleep(1); } for (int i = 0; i < nrNums; i++) if(nums[i]) Console.WriteLine(i); } } } This should be a pretty simple program that is supposed to find and output the first nrNums prime numbers using nrThreads threads working in parallel. So, I just split nrNums into nrThreads equal chunks (well, the last one won't be equal; if nrThreads doesn't divide by nrNums, it will also contain the remainder, of course). I start nrThreads threads. They all test each number in their respective chunk and see if it is prime or not; they mark everything out in a bool array that keeps a tab on all the primes. The threads all turn a specific element in another boolean array ThreadsFinished to true when they finish. Now the weird part begins: The threads never all end. If I debug, I find that ThreadNr is not what I assign to it in the loop but another value. I guess this is normal since the threads execute afterwards and the counter (the variable i) is already increased by then but I cannot understand how to make the code be right. Can anyone help? Thank you in advance. P.S.: I know the algorithm is not very efficient; I am aiming at a solution using the sieve of Eratosthenes also with x given threads. But for now I can't even get this one to work and I haven't found any examples of any implementations of that algorithm anywhere in a language that I can understand.

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  • Where does the version number come from?

    - by Robert Schneider
    I have a version control system (e.g. Subversion) and now I'd like to set up a build process. Now I have to create a version number and insert it into the system. But where does the version number come from and get into? Assume I want to use this common <major.<minor.<bugfix/revision scheme. Should I pass a number to the build script? Or should I pass arguments like increaseMajor, increaseMinor, increaseRevision? Or would you recommend to create a branch with the number which will be detected by the build script? I could imagine that the major and minor version number have to be put in manually somewhere. The revision number could be increased automaically. But still I don't know where I would place the major and minor number. In my case I have some php files that I would like to zip, but before I have to insert some version numbers into php file. I have edited this post to try to make my request clearer: I do not use Subversion, that was just an example. And I don't want to discuss the version number scheme. Imagine I want to create version 3.5.0 or 3.5.1. Would I pass this version number to a build script? Would the script create the branch in the repository with this number or would it expect that someone has already created this branch? Manually? Or would the build script look for name of the branch (e.g. '3.5.1) and use it for further things? And does the version number come from my brain or is it automatically created (I guess the major/minor number it comes from my little brain and revision number is created)? Or would you place the number into a file that may gets inserted into the repository? I guess if would use a release management tool I would insert the version number there. But I don't use one yet.

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  • Trying to calculate the 10001st prime number in Java.

    - by user247679
    Greetings. I am doing Problem 7 from Project Euler. What I am supposed to do is calculate the 10001st prime number. (A prime number being a number that is only divisible by itself and one.) Here is my current program: public class Problem7 { public static void main(String args []){ long numberOfPrimes = 0; long number = 2; while (numberOfPrimes < 10001){ if(isPrime(number)){ numberOfPrimes++; } number++; } System.out.println("10001st prime: "+ number); } public static boolean isPrime(long N) { if (N <= 1) return false; else return Prime(N,N-1); } public static boolean Prime(long X,long Y) { if (Y == 1) return true; else if (X % Y == 0) return false; else return Prime(X, Y-1); } } It works okay with finding, say the 100th prime number, but when I enter very large numbers such as 10001 it causes a stack overflow. Does anyone know of a way to fix this? Thanks for reading my problem!

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  • Is it possible to get a google voice number without already having a phone number?

    - by boost
    I'm sorry if this is the wrong place to post this, but I couldn't find a more suitable website in the stackexchange network. I currently have no phone number; I can make outgoing calls via the widget in gmail, but I cannot receive calls (as far as I know). I know that you can set up a google voice number to forward to google chat, and this is exactly what I want. The problem is, I can't get a google voice number in the first place, because it first requires that I have an existing phone number, even if I wouldn't use it. So, it is possible to skip providing an existing phone number, and just get a google voice number that forwards to google chat? Alternatively, is there any free phone service that can be used without a phone and lets you receive calls from any number? I realize I'm kind of asking for free candy, but, if it's out there, I'd be a fool

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  • is there a such thing as a randomly accessible pseudo-random number generator? (preferably open-sour

    - by lucid
    first off, is there a such thing as a random access random number generator, where you could not only sequentially generate random numbers as we're all used to, assuming rand100() always generates a value from 0-100: for (int i=0;i<5;i++) print rand100() output: 14 75 36 22 67 but also randomly access any random value like: rand100(0) would output 14 as long as you didn't change the seed rand100(3) would always output 22 rand100(4) would always output 67 and so on... I've actually found an open-source generator algorithm that does this, but you cannot change the seed. I know that pseudorandomness is a complex field; I wouldn't know how to alter it to add that functionality. Is there a seedable random access random number generator, preferably open source? or is there a better term for this I can google for more information? if not, part 2 of my question would be, is there any reliably random open source conventional seedable pseudorandom number generator so I could port it to multiple platforms/languages while retaining a consistent sequence of values for each platform for any given seed?

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  • Unix: millionth number in the serie 2 3 4 6 9 13 19 28 42 63 ... ?

    - by HH
    It takes about minute to achieve 3000 in my comp but I need to know the millionth number in the serie. The definition is recursive so I cannot see any shortcuts except to calculate everything before the millionth number. How can you fast calculate millionth number in the serie? Serie Def n_{i+1} = \floor{ 3/2 * n_{i} } and n_{0}=2. Interestingly, only one site list the serie according to Goolge: this one. Too slow Bash code #!/bin/bash function serie { n=$( echo "3/2*$n" | bc -l | tr '\n' ' ' | sed -e 's@\\@@g' -e 's@ @@g' ); # bc gives \ at very large numbers, sed-tr for it n=$( echo $n/1 | bc ) #DUMMY FLOOR func } n=2 nth=1 while [ true ]; #$nth -lt 500 ]; do serie $n # n gets new value in the function throught global value echo $nth $n nth=$( echo $nth + 1 | bc ) #n++ done

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  • How to make a random number generator in matlab that is based on percentages?

    - by Ben Fossen
    I am currently using the built in random number generator. for example nAsp = randi([512, 768],[1,1]); 512 is the lower bound and 768 is the upper bound, the random number generator chooses a number from between these two values. What I want is to have two ranges for nAsp but I want one of them to get called 25% of the time and the other 75% of the time. Then gets plugged into he equation. Does anyone have any ideas how to do this or if there is a built in function in matlab already? for example nAsp = randi([512, 768],[1,1]); gets called 25% of the time nAsp = randi([690, 720],[1,1]); gets called 75% of the time

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  • Millionth number in the serie 2 3 4 6 9 13 19 28 42 63 ... ?

    - by HH
    It takes about minute to achieve 3000 in my comp but I need to know the millionth number in the serie. The definition is recursive so I cannot see any shortcuts except to calculate everything before the millionth number. How can you fast calculate millionth number in the serie? Serie Def n_{i+1} = \floor{ 3/2 * n_{i} } and n_{0}=2. Interestingly, only one site list the serie according to Goolge: this one. Too slow Bash code #!/bin/bash function serie { n=$( echo "3/2*$n" | bc -l | tr '\n' ' ' | sed -e 's@\\@@g' -e 's@ @@g' ); # bc gives \ at very large numbers, sed-tr for it n=$( echo $n/1 | bc ) #DUMMY FLOOR func } n=2 nth=1 while [ true ]; #$nth -lt 500 ]; do serie $n # n gets new value in the function throught global value echo $nth $n nth=$( echo $nth + 1 | bc ) #n++ done

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  • Possibility Program for number of pieces

    - by Brad
    I would like to put a program together to calculate the number of 60' pieces would be needed from a list of shorter pieces. For example, I sell rebar cut to length from our standard length of 60'-0". Now the length the customer requires are as follows: 343 pc @ 12.5' 35 pc @ 13' 10 pc @ 15' 63 pc @ 15.5'....... There are 56 total lengths ranging from 12.5' to 30.58' The idea is to limit the amount of waste from the 60' piece. The input from the user would be: number of differnt lengths Length of piece to cut from count of different lengths The result would be the number of prime pieces needed to fulfill the order. What well-known algorithms exist that could help me solve this problem?

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  • C#, finding the largest prime factor of a number

    - by Juan
    Hello! I am new at programming and I am practicing my C# programming skills. My application is meant to find the largest prime factor of a number entered by the user. But my application is not returning the right answer and I dont really know where the problem is. Can you please help me? using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { class Program { static void Main(string[] args) { Console.WriteLine("Calcular máximo factor primo de n. De 60 es 5."); Console.Write("Escriba un numero: "); long num = Convert.ToInt64(Console.ReadLine()); long mfp = maxfactor(num); Console.WriteLine("El maximo factor primo es: " + num); Console.Read(); } static private long maxfactor (long n) { long m=1 ; bool en= false; for (long k = n / 2; !en && k > 1; k--) { if (n % k == 0 && primo(k)) { m = k; en = true; } } return m; } static private bool primo(long x) { bool sp = true; for (long i = 2; i <= x / 2; i++) { if (x % i == 0) sp = false; } return sp; } } }

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  • Go and a bad prime number algorithm

    - by anonymous
    I wrote this prime number sieving algorithm and it doesn't run properly. I can't find the error in the algorithm itself. Could someone help me? This is what it's supposed to print: [2 3 5 7 11 13 17 19 23 29] Versus what it actually prints: [3 5 7 11 13 17 19 23 25 29] . package main import "fmt" func main() { var primes = sieve(makeNumbers(29)) fmt.Printf("%d\n", primes); } func makeNumbers(n int) []int { var numbers = make([]int, n - 1) for i := 0; i < len(numbers); i++ { numbers[i] = i + 2 } return numbers } func sieve(numbers []int) []int { var numCopy = numbers var max = numbers[len(numbers)-1] var sievedNumbers = make([]int, 0) for i := 0; numCopy[i]*numCopy[i] <= max; i++ { for j := i; j < len(numCopy); j++ { if numCopy[j] % numCopy[i] != 0 || j == i { sievedNumbers = append(sievedNumbers, numCopy[j]) } } numCopy = sievedNumbers sievedNumbers = make([]int, 0) } return numCopy }

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  • Getting the number of fragments which passed the depth test

    - by Etan
    In "modern" environments, the "NV Occlusion Query" extension provides a method to get the number of fragments which passed the depth test. However, on the iPad / iPhone using OpenGL ES, the extension is not available. What is the most performant approach to implement a similar behaviour in the fragment shader? Some of my ideas: Render the object completely in white, then count all the colors together using a two-pass shader where first a vertical line is rendered and for each fragment the shader computes the sum over the whole row. Then, a single vertex is rendered whose fragment sums all the partial sums of the first pass. Doesn't seem to be very efficient. Render the object completely in white over a black background. Downsample recursively, abusing the hardware linear interpolation between textures until being at a reasonably small resolution. This leads to fragments which have a greyscale level depending on the number of white pixels where in their corresponding region. Is this even accurate enough? Use mipmaps and simply read the pixel on the 1x1 level. Again the question of accuracy and if it is even possible using non-power-of-two textures. The problem wit these approaches is, that the pipeline gets stalled which results in major performance issues. Therefore, I'm looking for a more performant way to accomplish my goal. Using the EXT_OCCLUSION_QUERY_BOOLEAN extension Apple introduced EXT_OCCLUSION_QUERY_BOOLEAN in iOS 5.0 for iPad 2. "4.1.6 Occlusion Queries Occlusion queries use query objects to track the number of fragments or samples that pass the depth test. An occlusion query can be started and finished by calling BeginQueryEXT and EndQueryEXT, respectively, with a target of ANY_SAMPLES_PASSED_EXT or ANY_SAMPLES_PASSED_CONSERVATIVE_EXT. When an occlusion query is started with the target ANY_SAMPLES_PASSED_EXT, the samples-boolean state maintained by the GL is set to FALSE. While that occlusion query is active, the samples-boolean state is set to TRUE if any fragment or sample passes the depth test. When the occlusion query finishes, the samples-boolean state of FALSE or TRUE is written to the corresponding query object as the query result value, and the query result for that object is marked as available. If the target of the query is ANY_SAMPLES_PASSED_CONSERVATIVE_EXT, an implementation may choose to use a less precise version of the test which can additionally set the samples-boolean state to TRUE in some other implementation dependent cases." The first sentence hints on a behavior which is exactly what I'm looking for: getting the number of pixels which passed the depth test in an asynchronous manner without much performance loss. However, the rest of the document describes only how to get boolean results. Is it possible to exploit this extension to get the pixel count? Does the hardware support it so that there may be hidden API to get access to the pixel count? Other extensions which could be exploitable would be debugging features like the number of times the fragment shader was invoked (PSInvocations in DirectX - not sure if something simila is available in OpenGL ES). However, this would also result in a pipeline stall.

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  • How does a cryptographically secure random number generator work?

    - by Byron Whitlock
    I understand how standard random number generators work. But when working with crytpography, the random numbers really have to be random. I know there are instruments that read cosmic white noise to help generate secure hashes, but your standard PC doesn't have this. How does a cryptographically secure random number generator get its values with no repeatable patterns?

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  • Optimizing list comprehension to find pairs of co-prime numbers

    - by user3685422
    Given A,B print the number of pairs (a,b) such that GCD(a,b)=1 and 1<=a<=A and 1<=b<=B. Here is my answer: return len([(x,y) for x in range(1,A+1) for y in range(1,B+1) if gcd(x,y) == 1]) My answer works fine for small ranges but takes enough time if the range is increased. such as 1 <= A <= 10^5 1 <= B <= 10^5 is there a better way to write this or can this be optimized?

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