问题
I am learning Ruby and doing some math stuff. One of the things I want to do is generate prime numbers.
I want to generate the first ten prime numbers and the first ten only. I have no problem testing a number to see if it is a prime number or not, but was wondering what the best way is to do generate these numbers?
I am using the following method to determine if the number is prime:
class Integer < Numeric
def is_prime?
return false if self <= 1
2.upto(Math.sqrt(self).to_i) do |x|
return false if self%x == 0
end
true
end
end
回答1:
In Ruby 1.9 there is a Prime class you can use to generate prime numbers, or to test if a number is prime:
require 'prime'
Prime.take(10) #=> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
Prime.take_while {|p| p < 10 } #=> [2, 3, 5, 7]
Prime.prime?(19) #=> true
Prime implements the each method and includes the Enumerable module, so you can do all sorts of fun stuff like filtering, mapping, and so on.
回答2:
If you'd like to do it yourself, then something like this could work:
class Integer < Numeric
def is_prime?
return false if self <= 1
2.upto(Math.sqrt(self).to_i) do |x|
return false if self%x == 0
end
true
end
def next_prime
n = self+1
n = n + 1 until n.is_prime?
n
end
end
Now to get the first 10 primes:
e = Enumerator.new do |y|
n = 2
loop do
y << n
n = n.next_prime
end
end
primes = e.take 10
回答3:
require 'prime'
Prime.first(10) # => [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
回答4:
Check out Sieve of Eratosthenes. This is not Ruby specific but it is an algorithm to generate prime numbers. The idea behind this algorithm is that you have a list/array of numbers say
2..1000
You grab the first number, 2. Go through the list and eliminate everything that is divisible by 2. You will be left with everything that is not divisible by 2 other than 2 itself (e.g. [2,3,5,7,9,11...999]
Go to the next number, 3. And again, eliminate everything that you can divide by 3. Keep going until you reach the last number and you will get an array of prime numbers. Hope that helps.
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
回答5:
People already mentioned the Prime class, which definitely would be the way to go. Someone also showed you how to use an Enumerator and I wanted to contribute a version using a Fiber (it uses your Integer#is_prime? method):
primes = Fiber.new do
Fiber.yield 2
value = 3
loop do
Fiber.yield value if value.is_prime?
value += 2
end
end
10.times { p primes.resume }
回答6:
# First 10 Prime Numbers
number = 2
count = 1
while count < 10
j = 2
while j <= number
break if number%j == 0
j += 1
end
if j == number
puts number
count += 1
end
number += 1
end
回答7:
Implemented the Sieve of Eratosthene (more or less)
def primes(size)
arr=(0..size).to_a
arr[0]=nil
arr[1]=nil
max=size
(size/2+1).times do |n|
if(arr[n]!=nil) then
cnt=2*n
while cnt <= max do
arr[cnt]=nil
cnt+=n
end
end
end
arr.compact!
end
Moreover here is a one-liner I like a lot
def primes_c a
p=[];(2..a).each{|n| p.any?{|l|n%l==0}?nil:p.push(n)};p
end
Of course those will find the primes in the first n numbers, not the first n primes, but I think an adaptation won't require much effort.
回答8:
Here is a way to generate the prime numbers up to a "max" argument from scratch, without using Prime or Math. Let me know what you think.
def prime_test max
primes = []
(1..max).each {|num|
if
(2..num-1).all? {|denom| num%denom >0}
then
primes.push(num)
end
}
puts primes
end
prime_test #enter max
回答9:
I think this may be an expensive solution for very large max numbers but seems to work well otherwise:
def multiples array
target = array.shift
array.map{|item| item if target % item == 0}.compact
end
def prime? number
reversed_range_array = *(2..number).reverse_each
multiples_of_number = multiples(reversed_range_array)
multiples_of_number.size == 0 ? true : false
end
def primes_in_range max_number
range_array = *(2..max_number)
range_array.map{|number| number if prime?(number)}.compact
end
回答10:
class Numeric
def prime?
return self == 2 if self % 2 == 0
(3..Math.sqrt(self)).step(2) do |x|
return false if self % x == 0
end
true
end
end
With this, now 3.prime? returns true, and 6.prime? returns false.
Without going to the efforts to implement the sieve algorithm, time can still be saved quickly by only verifying divisibility until the square root, and skipping the odd numbers. Then, iterate through the numbers, checking for primeness.
Remember: human time > machine time.
回答11:
I did this for a coding kata and used the Sieve of Eratosthenes.
puts "Up to which number should I look for prime numbers?"
number = $stdin.gets.chomp
n = number.to_i
array = (1..n).to_a
i = 0
while array[i]**2 < n
i = i + 1
array = array.select do |element|
element % array[i] != 0 || element / array[i] == 1
end
end
puts array.drop(1)
回答12:
Ruby: Print N prime Numbers http://mishra-vishal.blogspot.in/2013/07/include-math-def-printnprimenumbernoofp.html
include Math
def print_n_prime_number(no_of_primes=nil)
no_of_primes = 100 if no_of_primes.nil?
puts "1 \n2"
count = 1
number = 3
while count < no_of_primes
sq_rt_of_num = Math.sqrt(number)
number_divisible_by = 2
while number_divisible_by <= sq_rt_of_num
break if(number % number_divisible_by == 0)
number_divisible_by = number_divisible_by + 1
end
if number_divisible_by > sq_rt_of_num
puts number
count = count+1
end
number = number + 2
end
end
print_n_prime_number
回答13:
Not related at all with the question itself, but FYI:
- if someone doesn't want to keep generating prime numbers again and again (a.k.a. greedy resource saver)
- or maybe you already know that you must to work with subsequent prime numbers in some way
- other unknown and wonderful cases
Try with this snippet:
require 'prime'
for p in Prime::Generator23.new
# `p` brings subsequent prime numbers until the end of the days (or until your computer explodes)
# so here put your fabulous code
break if #.. I don't know, I suppose in some moment it should stop the loop
end
fp
If you need it, you also could use another more complex generators as Prime::TrialDivisionGenerator or Prime::EratosthenesGenerator. More info
回答14:
def get_prime(number)
(2..number).each do |no|
if (2..no-1).all? {|num| no % num > 0}
puts no
end
end
end
get_prime(100)
来源：https://stackoverflow.com/questions/11673968/how-do-i-generate-the-first-n-prime-numbers