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Search found 107 results on 5 pages for 'primes'.

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• nth ugly number

  - by Anil Katti

  Numbers whose only prime factors are 2, 3 or 5 are called ugly numbers. Example: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... 1 can be considered as 2^0. I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large. I wrote a trivial program that computes if a given number is ugly or not. For n 500 - it became super slow. I tried using memoization - observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log - since that will reduce this problem from multiplication to addition - but, not much luck yet. Thought of sharing this with you all. Any interesting ideas? Using a concept similar to "Sieve of Eratosthenes" (thanks Anon) for (int i(2), uglyCount(0); ; i++) { if (i % 2 == 0) continue; if (i % 3 == 0) continue; if (i % 5 == 0) continue; uglyCount++; if (uglyCount == n - 1) break; } i is the nth ugly number. Even this is pretty slow. I am trying to find 1500th ugly number.

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• Simple Prime Generator in Python

  - by marc lincoln

  Hi, could someone please tell me what I'm doing wrong with this code. It is just printing 'count' anyway. I just want a very simple prime generator (nothing fancy). Thanks a lot. lincoln. import math def main(): count = 3 one = 1 while one == 1: for x in range(2, int(math.sqrt(count) + 1)): if count % x == 0: continue if count % x != 0: print count count += 1

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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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• Miller-rabin exception number?

  - by nightcracker

  Hey everyone. This question is about the number 169716931325235658326303. According to http://www.alpertron.com.ar/ECM.HTM it is prime. According to my miller-rabin implementation in python with 7 repetitions is is composite. With 50 repetitions it is still composite. With 5000 repetitions it is STILL composite. I thought, this might be a problem of my implementation. So I tried GNU MP bignum library, which has a miller-rabin primality test built-in. I tested with 1000000 repetitions. Still composite. This is my implementation of the miller-rabin primality test: def isprime(n, precision=7): if n == 1 or n % 2 == 0: return False elif n < 1: raise ValueError("Out of bounds, first argument must be > 0") d = n - 1 s = 0 while d % 2 == 0: d //= 2 s += 1 for repeat in range(precision): a = random.randrange(2, n - 2) x = pow(a, d, n) if x == 1 or x == n - 1: continue for r in range(s - 1): x = pow(x, 2, n) if x == 1: return False if x == n - 1: break else: return False return True And the code for the GMP test: #include <gmp.h> #include <stdio.h> int main(int argc, char* argv[]) { mpz_t test; mpz_init_set_str(test, "169716931325235658326303", 10); printf("%d\n", mpz_probab_prime_p(test, 1000000)); mpz_clear(test); return 0; } As far as I know there are no "exceptions" (which return false positives for any amount of repetitions) to the miller-rabin primality test. Have I stumpled upon one? Is my computer broken? Is the Elliptic Curve Method wrong? What is happening here? EDIT I found the issue, which is http://www.alpertron.com.ar/ECM.HTM. I trusted this applet, I'll contact the author his applet's implementation of the ECM is faulty for this number. Thanks. EDIT2 Hah, the shame! In the end it was something that went wrong with copy/pasting on my side. NOR the applet NOR the miller-rabin algorithm NOR my implementation NOR gmp's implementation of it is wrong, the only thing that's wrong is me. I'm sorry.

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• Fastest algorithm for primality test

  - by dada

  I need to test primality on intervals between numbers which are really big (in the range of long long), so i need some fast algorithm for checking if a number is prime or not. Please suggest your ideas.

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• Can you tell me why this generates time limit exceeded in spoj(Prime Number Generator)

  - by magiix

  #include<iostream> #include<string.h> #include<math.h> using namespace std; bool prime[1000000500]; void generate(long long end) { memset(prime,true,sizeof(prime)); prime[0]=false; prime[1]=false; for(long long i=0;i<=sqrt(end);i++) { if(prime[i]==true) { for(long long y=i*i;y<=end;y+=i) { prime[y]=false; } } } } int main() { int n; long long b,e; scanf("%d",&n); while(n--) { cin>>b>>e; generate(e); for(int i=b;i<e;i++) { if(prime[i]) printf("%d\n",i); } } return 0; } That's my code for spoj prime generator. Altought it generates the same output as another accepted code ..

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• To find first N prime numbers in python

  - by Rahul Tripathi

  Hi All, I am new to the programming world. I was just writing this code in python to generate N prime numbers. User should input the value for N which is the total number of prime numbers to print out. I have written this code but it doesn't throw the desired output. Instead it prints the prime numbers till the Nth number. For eg.: User enters the value of N = 7. Desired output: 2, 3, 5, 7, 11, 13, 19 Actual output: 2, 3, 5, 7 Kindly advise. i=1 x = int(input("Enter the number:")) for k in range (1, (x+1), 1): c=0 for j in range (1, (i+1), 1): a = i%j if (a==0): c = c+1 if (c==2): print (i) else: k = k-1 i=i+1

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• assembly language programming (prime number)

  - by chris

  Prompt the user for a positive three digit number, then read it. Let's call it N. Divide into N all integer values from 2 to (N/2)+1 and test to see if the division was even, in which case N is instantly shown to be non-prime. Output a message printing N and saying that it is not prime. If none of those integer values divide evenly (remainder never is zero), then N is shown to be prime. Output a message printing N and saying that it is prime. Ask the user if he or she wants to test another number; if the user types "n" or "N", quit. If "y" or "Y", jump back and repeat. Comments in your code are essential. Hi. I am kinda in rush to do this.. please help me doing it. I'll be much appreciated. thank you

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• printing out prime numbers from array

  - by landscape

  I'd like to print out all prime numbers from an array with method. I can do it with one int but don't know how to return certain numbers from array. Thanks for help! public static boolean isPrime(int [] tab) { boolean prime = true; for (int i = 3; i <= Math.sqrt(tab[i]); i += 2) if (tab[i] % i == 0) { prime = false; break; } for(int i=0; i<tab.length; i++) if (( tab[i]%2 !=0 && prime && tab[i] > 2) || tab[i] == 2) { return true; } else { return false; } //return prime; } thanks both of you. Seems like its solved: public static void isPrime(int[] tab) { for (int i = 0; i < tab.length; i++) { if (isPrimeNum(tab[i])) { System.out.println(tab[i]); } } } public static boolean isPrimeNum(int n) { boolean prime = true; for (long i = 3; i <= Math.sqrt(n); i += 2) { if (n % i == 0) { prime = false; break; } } if ((n % 2 != 0 && prime && n > 2) || n == 2) { return true; } else { return false; } }

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• What is a convenient base for a bignum library & primality testing algorithm?

  - by nn

  Hi, I am to program the Solovay-Strassen primality test presented in the original paper on RSA. Additionally I will need to write a small bignum library, and so when searching for a convenient representation for bignum I came across this [specification][1]: struct { int sign; int size; int *tab; } bignum; I will also be writing a multiplication routine using the Karatsuba method. So, for my question: What base would be convenient to store integer data in the bignum struct? Note: I am not allowed to use third party or built-in implementations for bignum such as GMP. Thank you.

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• Fastest primality test

  - by Grigory Javadyan

  Hi. Could you suggest a fast, deterministic method that is usable in practice, for testing if a large number is prime or not? Also, I would like to know how to use non-deterministic primality tests correctly. For example, if I'm using such a method, I can be sure that a number is not prime if the output is "no", but what about the other case, when the output is "probably"? Do I have to test for primality manually in this case? Thanks in advance.

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• Recursion with Func

  - by David in Dakota

  Is it possible to do recursion with an Func delegate? I have the following, which doesn't compile because the name of the Func isn't in scope... Func<long, long, List<long>, IEnumerable<long>> GeneratePrimesRecursively = (number, upperBound, primeFactors) => { if (upperBound > number) { return new List<long>(); } else { if (!primeFactors.Any(factor => number % factor == 0)) primeFactors.Add(number); return GeneratePrimesRecursively(++number, upperBound, primeFactors); // breaks here. } };

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• Most Elegant Way to write isPrime in java

  - by Anantha Kumaran

  public class Prime { public static boolean isPrime1(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } for (int i = 2; i <= Math.sqrt(n) + 1; i++) { if (n % i == 0) { return false; } } return true; } public static boolean isPrime2(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } if (n % 2 == 0) { return false; } for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) { if (n % i == 0) { return false; } } return true; } } public class PrimeTest { public PrimeTest() { } @Test public void testIsPrime() throws IllegalArgumentException, IllegalAccessException, InvocationTargetException { Prime prime = new Prime(); TreeMap<Long, String> methodMap = new TreeMap<Long, String>(); for (Method method : Prime.class.getDeclaredMethods()) { long startTime = System.currentTimeMillis(); int primeCount = 0; for (int i = 0; i < 1000000; i++) { if ((Boolean) method.invoke(prime, i)) { primeCount++; } } long endTime = System.currentTimeMillis(); Assert.assertEquals(method.getName() + " failed ", 78498, primeCount); methodMap.put(endTime - startTime, method.getName()); } for (Entry<Long, String> entry : methodMap.entrySet()) { System.out.println(entry.getValue() + " " + entry.getKey() + " Milli seconds "); } } } I am trying to find the fastest way to check whether the given number is prime or not. This is what is finally came up with. Is there any better way than the second implementation(isPrime2).

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• Using the mpz_powm functions from the GMP/MPIR libraries with negative exponents

  - by Mihai Todor

  Please consider the following code: mpz_t x, n, out; mpz_init_set_ui(x, 2UL); mpz_init_set_ui(n, 7UL); mpz_init(out); mpz_invert(out, x, n); gmp_printf ("%Zd\n", out);//prints 4. 2 * 4 (mod 7) = 1. OK mpz_powm_ui(out, x, -1UL, n);//prints 1. 2 * 1 (mod 7) = 2. How come? gmp_printf ("%Zd\n", out); mpz_clear(x); mpz_clear(n); mpz_clear(out); I am unable to understand how the mpz_powm functions handle negative exponents, although, according to the documentation, it is supposed to support them. I would expect that raising a number to -1 modulo n is equivalent to inverting it modulo n. What am I missing here?

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• Explanation needed for sum of prime below n numbers

  - by Bala Krishnan

  Today I solved a problem given in Project Euler its problem no 10 and it took 7 hrs for my python program to show the result. But in that forum itself a person named lassevk posted solution for this and it took only 4 sec. And its not possible for me to post this question in that forum because its not discussion forum. So, think about this if you want to mark this question as non-constructive. marked = [0] * 2000000 value = 3 s = 2 while value < 2000000: if marked[value] == 0: s += value i = value while i < 2000000: marked[i] = 1 i += value value += 2 print s If any one understand this code please explain it simple as possible. Link to the Problem 10 question.

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• Prime Numbers in C?

  - by Ali Azam Rana

  FIRST PROGRAM #include<stdio.h> void main() { int n,c; printf("enter a numb"); scanf("%i",n); for(c=2;c<=n;c++) { if(n%c==0) break; } if(c==n) printf("\nprime\n"); else printf("\nnot prime\n"); getchar(); } SECOND PROGRAM #include <stdio.h> int main() { printf("Enter a Number\n"); int in,loop,rem,chk; scanf("%d",&in); for (loop = 1; loop <=in; loop++) { rem = in % loop; if(rem == 0) chk = chk +1; } if (chk == 2) printf("\nPRIME NUM ENTERED\n"); else printf("\nNUM ENTERED NOT PRIME\n"); getchar(); } the 2nd program works other was the one my friend wrote the program looks fine but on checking it by stepping into we found that the if condition in first program is coming true under every input so whats the logical error here please help me found out......

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• List of divisors of an integer n (Haskell)

  - by Code-Guru

  I currently have the following function to get the divisors of an integer: -- All divisors of a number divisors :: Integer -> [Integer] divisors 1 = [1] divisors n = firstHalf ++ secondHalf where firstHalf = filter (divides n) (candidates n) secondHalf = filter (\d -> n `div` d /= d) (map (n `div`) (reverse firstHalf)) candidates n = takeWhile (\d -> d * d <= n) [1..n] I ended up adding the filter to secondHalf because a divisor was repeating when n is a square of a prime number. This seems like a very inefficient way to solve this problem. So I have two questions: How do I measure if this really is a bottle neck in my algorithm? And if it is, how do I go about finding a better way to avoid repetitions when n is a square of a prime?

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• String encryption only with numbers?

  - by HH

  Suppose your bank clerk gives you an arbitrary password such as hel34/hjal0@# and you cannot remember it without writing it to a paper. Dilemma: you never write passwords to paper. So you try to invent an encryption, one-to-one map, where you write only a key to a paper, only numbers, and leave the rest junk to your server. Of course, the password can consist of arbitrary things. Implemention should work like hel34/hjal0#@ ---- magic ----> 3442 and to other way: 3442 ---- server magic ---> hel34/hjal0#@ [Update] mvds has the correct idea, to change the base, how would you implement it?

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• How to find the largest square in the number (Java)

  - by Ypsilon IV

  Hello everyone, I want to find the largest square in the number, but I am stuck at some point. I would really appreciate some suggestions. This is what I've done so far: I take the number on the input, factorize into prime numbers, and put the sequence of prime numbers to ArrayList. Numbers are sorted, in a sense, thus the numbers in the sequence are increasing. For example, 996 is 2 2 3 83 1000 is 2 2 2 5 5 5 100000 is 2 2 2 2 2 5 5 5 5 5 My idea now is to count number of occurrences of each elements in the sequence, so if the number of occurrences is divisible by two, then this is the square. In this way, I can get another sequence, where the right most element divisible by two is the largest square. What is the most efficient way to count occurrences in the ArrayList? Or is there any better way to find the largest square? Many thanks in advance!

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• Interview question : What is the fastest way to generate prime number recursively ?

  - by hilal

  Generation of prime number is simple but what is the fastest way to find it and generate( prime numbers) it recursively ? Here is my solution. However, it is not the best way. I think it is O(N*sqrt(N)). Please correct me, if I am wrong. public static boolean isPrime(int n) { if (n < 2) { return false; } else if (n % 2 == 0 & n != 2) { return false; } else { return isPrime(n, (int) Math.sqrt(n)); } } private static boolean isPrime(int n, int i) { if (i < 2) { return true; } else if (n % i == 0) { return false; } else { return isPrime(n, --i); } } public static void generatePrimes(int n){ if(n < 2) { return ; } else if(isPrime(n)) { System.out.println(n); } generatePrimes(--n); } public static void main(String[] args) { generatePrimes(200); }

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• Please explain how Trial Division works for Primality Test

  - by mister_dani

  I came across this algorithm for testing primality through trial division I fully understand this algorithm static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i <= Math.sqrt(N); i++) if (N % i == 0) return false; return true; } It works just fine. But then I came across this other one which works just as good but I do not fully understand the logic behind it. static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i * i<N; i++) if (N % i == 0) return false; return true; } It seems like i *i < N behaves like i <= Math.sqrt(N). If so, why?

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• Project Euler 7: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Improving python code

  - by cobie

  I just answered the question on project euler about finding circular primes below 1 million using python. My solution is below. I was able to reduce the running time of the solution from 9 seconds to about 3 seconds. I would like to see what else can be done to the code to reduce its running time further. This is strictly for educational purposes and for fun. import math import time def getPrimes(n): """returns set of all primes below n""" non_primes = [j for j in range(4, n, 2)] # 2 covers all even numbers for i in range(3, n, 2): non_primes.extend([j for j in range(i*2, n, i)]) return set([i for i in range(2, n)]) - set(non_primes) def getCircularPrimes(n): primes = getPrimes(n) is_circ = [] for prime in primes: prime_str = str(prime) iter_count = len(prime_str) - 1 rotated_num = [] while iter_count > 0: prime_str = prime_str[1:] + prime_str[:1] rotated_num.append(int(prime_str)) iter_count -= 1 if primes >= set(rotated_num): is_circ.append(prime) return len(is_circ)

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• Fast, Vectorizable method of taking floating point number modulus of special primes?

  - by caffiend

  Is there a fast method for taking the modulus of a floating point number? With integers, there are tricks for Mersenne primes, so that its possible to calculate y = x MOD 2^31 without needing division. Can any similar tricks be applied for floating point numbers? Preferably, in a way that can be converted into vector/SIMD operations, or moved into GPGPU code. The primes I'm interested in would be 2^7 and 2^31, although if there are more efficient ones for floating point numbers, those would be welcome.

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• Python recursive function error: "maximum recursion depth exceeded"

  - by user283169

  I solved Problem 10 of Project Euler with the following code, which works through brute force: def isPrime(n): for x in range(2, int(n**0.5)+1): if n % x == 0: return False return True def primeList(n): primes = [] for i in range(2,n): if isPrime(i): primes.append(i) return primes def sumPrimes(primelist): prime_sum = sum(primelist) return prime_sum print (sumPrimes(primeList(2000000))) The three functions work as follows: isPrime checks whether a number is a prime; primeList returns a list containing a set of prime numbers for a certain range with limit 'n', and; sumPrimes sums up the values of all numbers in a list. (This last function isn't needed, but I liked the clarity of it, especially for a beginner like me.) I then wrote a new function, primeListRec, which does exactly the same thing as primeList, to help me better understand recursion: def primeListRec(i, n): primes = [] #print i if (i != n): primes.extend(primeListRec(i+1,n)) if (isPrime(i)): primes.append(i) return primes return primes The above recursive function worked, but only for very small values, like '500'. The function caused my program to crash when I put in '1000'. And when I put in a value like '2000', Python gave me this: RuntimeError: maximum recursion depth exceeded. What did I do wrong with my recursive function? Or is there some specific way to avoid a recursion limit?

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