I'm learning Ruby and am working through this problem on Project Euler. When I try to run my Ruby file through the terminal for the answer, it just stalls (no error message, but just comes up blank like its thinking). I'm assuming it's an issue with calculating such a large number. I tried copying and pasting my code into repl.it, but came up with this error message:
(eval):12: (eval):12: negative allocation size (or too big)
(NoMemoryError)
Is there any way around this through the terminal (or any other suggestions)? Thanks.
Edit (added in my code):
#3. Largest prime factor
#The prime factors of 13195 are 5, 7, 13 and 29.
#What is the largest prime factor of the number 600851475143 ?
def largest_prime(num)
all_primes = []
(2...num).each do |p|
if (num % p == 0) && (is_prime?(p))
all_primes << p
end
end
all_primes.sort!
puts all_primes.last
end
def is_prime?(number)
(2...number).each do |x|
if number % x == 0
return false
else
next
end
end
true
end
puts largest_prime(600851475143)
The idea behind Project Euler is to build effective algorithms, otherwise the execution will just take too much time or resources - this is your case. Take a look at primality_test wiki article and you'll understand, why you shouldn't check (2...number) and why (2..Math.sqrt(num)) is enough. I didn't run your code, but suggest you to improve your is_prime? method first of all(by the way, you shouldn't prefix your method names with is_ when you have question mark in the end):
def prime?(num)
return false if num == 0 || num == 1
(2..Math.sqrt(num).to_i).each do |i|
return false if num % i == 0
end
true
end
It isn't MOST effective way of prime numbers detection, but its one of the simplest and effective enough for your needs.
Related
I am trying to solve a problem from http://www.beatmycode.com/challenge/5/take, and I wrote a script:
def is_prime?(num)
(2...num).each do |divisor|
return false if num % divisor == 0
end
true
end
def circular_prime(num)
circular_primes = []
if num < 2
return false
end
(2..num-1).each do |number|
is_prime?(number)
result = [number.to_s]
(0..number.to_s.size-2).each do |x|
var = result.last.split('')
result << var.unshift(var.pop).join if var.uniq.size != 1
end
circular_primes << result if result.all?{ |x| is_prime? x.to_i }
end
end
When I tested it with 10,100,1000, and 10_000, the script executed very fast, but when I tested with 100_000, the shell displayed an Interrupt error. Where is the weak point, and how can I fix it?
Your is_prime? method is too slow. Some minor improvement could be:
You don't have to test divisor all the way upto num, the square root of
num is enough.
You can skip all even numbers since 2 is the only even prime
number.
However, this is still not good enough because the algorithm is slow. Since you need to generate prime numbers among consecutive integers, consider using Sieve of Eratosthenes.
Of course there's prime standard library but I assume you want to do this manually.
This script's time complexity is O(n^2), so it's expected that it will take long time to finish for big input. Try using https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes to find primes and remembering which number is a prime in some array.
I am relatively new to Ruby but it seems simple enough as far as a language goes. I am working through the Euler Project with Ruby and I'm having a huge issue with speed on the following:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
My code:
beginning_time = Time.now
(1..10000).each { |i| i }
def isPrime(num)
factors = 0
primecount = 1
while primecount <= num
if (num%primecount == 0)
factors += 1
end
if (factors > 2)
return false
end
primecount += 1
end
return true
end
def make_sieve(num)
sieve = Array.new(num)
summation = 0
for i in 1..num
if(isPrime(i) == true)
summation += i
puts i
for x in i..num
if x%i == 0
# Go through entire array and make all multiples of i False
sieve[x] = false
else
sieve[i] = true
end
end
else
# If i is NOT prime, move to the next number. in the For Loop
next
end
end
puts summation
end
make_sieve(2000000)
end_time = Time.now
puts "Time elapsed #{(end_time - beginning_time)*1000} milliseconds"
I think I have the right idea with the sieve but I really have no clue what's going on that makes this program run so slow. I run it with 20,000 and it takes about 15 seconds, which seems slow even still, although the output comes out MUCH faster than when I put 2,000,000.
Am I going about this the wrong way logically or syntactically or both?
Your isPrime() test is very slow on primes; but you don't even need it. The key to sieve is, initially all the numbers are marked as prime; then for each prime we mark off all its multiples. So when we get to a certain entry in the sieve, we already know whether it is a prime or not - whether it is marked true for being prime, or it is marked false for being composite (a multiple of some smaller prime).
There is no need to test it being prime, at all.
And to find the multiples, we just count: for 5, it's each 5th entry after it; for 7 - each 7th. No need to test them with % operator, just set to false right away. No need to set any of them to true, because all numbers were set to true at the start.
You seem to be writing JavaScript code in Ruby, and are missing the subtleties that makes Ruby so elegant. You should take a look at something like Ruby Best Practices, which is quite a light read but deals with using Ruby idioms instead of imposing the concepts of another language.
As has been said, the whole point of an Eratosthenes sieve is that you just remove all compound numbers from a list, leaving just the primes. There is no need to check each element for primeness.
This is a Rubyish solution. It runs in about 1.5 seconds. It is a little complicated by the representing number N by array element N-1, so (i+i+1 .. num).step(i+1) is equivalent to (n * 2 .. num).step(n)
def make_sieve(num)
sieve = Array.new(num, true)
sieve.each_with_index do |is_prime, i|
next if i == 0 or not is_prime
(i+i+1 .. num).step(i+1) { |i| sieve[i] = false }
end
puts sieve.each_index.select { |i| sieve[i] }.map { |i| i+1 }.inject(:+)
end
make_sieve(2_000_000)
output
142913828923
This question references Project Euler Problem 5, so beware of spoilers!
Problem 5:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
I wrote the following code in Ruby as a solution to Problem 5.
num = 2520
until (1..20).all?{|x| num % x == 0}
num += 1
end
puts "#{num}"
However, whenever I run the script, it just hangs. Note that I tested the same method on the base case 2520 for the range 1 to 10 and it worked just fine.
Why does it work for the simpler case but not for the more advanced case? What can I do to fix what I have?
It's slow because the answer is more than 200 million, and you're counting up to it in steps of 1. That's going to take a while. You need a better algorithm.
You won't be able to brute force this problem as you have with others. You're going to need to find a more efficient solution for it.
SPOILER BELOW
This is a significantly more efficient way to do it (apologies if this isn't very Ruby-like):
def find_multiple
lcm = 1
(2..20).each do |i|
lcm *= i / gcd(lcm, i)
end
lcm
end
def gcd(a, b)
while b > 0
a %= b
return b if a == 0
b %= a
end
a
end
puts find_multiple
If you're looking for a more Ruby-like way to solve it, you can use the following (as suggested by steenslag in the comments):
(1..20).inject(:lcm)
A little late to the game but here's my input. If you like your code - which is succinct and pretty clear - you can make a couple of small adjustments to make it run faster. This worked for me without timing out:
num = 20
until (11..20).all?{ |i| num % i == 0 }
num +=20
end
puts num
Essentially you just increment by 20s since you know it needs to be divisible by 20, and you can skip iterating through anything in the lower 1/2 of the set.
Simplest and clean solution in ruby language.
(1..20).inject(:lcm) # output will be 232792560
Try this.
Code:
i=2520
max_product = (4..19).inject(1){|j,k| j*k}
puts max_product
while i < max_product
if( [3, 7].inject(true) do |memo, p|
memo && i%p==0
end)
if(
(4..20).inject(true) do |memo, p|
memo && i%p==0
end
)
puts i
end
end
i+=10
end
I'm NOT asking for a solution to this riddle.
So, I'm working through the Project Euler problems using Ruby.
I made a palindrome checker with the following method:
def is_palindrome(n)
if n.to_s == n.to_s.reverse
true
else
false
end
end
Using this, I use the next method to try and find the largest palindrome made with two 3 digit numbers.
x = 999
y = 998
while y > 100
if is_palindrome(x * y) == true
puts (x * y)
else
x-=1
y-=1
end
end
The result is the terminal throws a complete fit and gets stuck on 289982 and has to be Ctrl+C'd for it to stop.
As I said, I'm not looking for a solution to this riddle, but I want to know what it is my code is doing for this to happen. I'm still learning so my code is likely to be rather ugly, so extra karma points to anyone that can help me out with this.
In case a palindrome is found, you don't decrement your variables. Since the variables remain the same, the same palindrome is found again. And so it goes...
Amadan's solved your problem, so I'll go for the extra karma points :)
First thing, you can just do if something instead of if something == true.
Also, you don't need to do
if something == something_else
true
else
false
end
because the == operator returns either true or false already, so your is_palindrome method could be as simple as
def is_palindrome(n)
n.to_s == n.to_s.reverse
end
Also, in Ruby, you will usually see methods like this one (a boolean check) named with a question mark at the end, because it's really like you're asking it a yes/no question! So usually you'd see that called is_palindrome?(n). Similar ones you might see elsewhere are Array#empty? and Hash#has_key?(k).
Not answering your question, but your use of booleans is horrendous. Remember boolean expressions return a boolean itself, you don't need another comparison. i.e.
def is_palindrome(n)
if n.to_s == n.to_s.reverse
true
else
false
end
end
Should be:
def is_palindrome(n)
n.to_s == n.to_s.reverse
end
And
while y > 100
if is_palindrome(x * y) == true
Should be
while y > 100
if is_palindrome(x * y)
Here's my solution:
# Find largest palindrome made from the product of two 3-digit numbers
largest = 0
for i in 100..999
for j in i..999
product = j*i
# Check for palindrome and determine largest
largest = product.to_s == product.to_s.reverse && product > largest ? product : largest
end
end
p largest
I am trying to create a program that will test whether a value is prime, but I don't know how. This is my code:
class DetermineIfPrime
def initialize (nth_value)
#nth_value = nth_value
primetest
end
def primetest
if Prime.prime?(#nth_value)
puts ("#{#nth_value} is prime")
else
puts ("This is not a prime number.")
end
rescue Exception
puts ("#{$!.class}")
puts ("#{$!}")
end
end
And every time I run that it returns this.
NameError
uninitialized constant DetermineIfPrime::Prime
I tried other ways to do the job, but I think this is the closest I can get.
I also tried this:
class DetermineIfPrime
def initialize (nth_value)
#nth_value = nth_value
primetest
end
def primetest
for test_value in [2, 3, 5, 7, 9, 11, 13] do
if (#nth_value % test_value) == 0
puts ("#{#nth_value} is not divisible by #{test_value}")
else
puts ("This is not a prime number since this is divisible by #{test_value}")
break
end
end
end
end
Or am I just doing something wrong?
Ruby has built in method to check if number is prime or not.
require 'prime'
Prime.prime?(2) #=> true
Prime.prime?(4) #=> false
def is_prime?(num)
return false if num <= 1
Math.sqrt(num).to_i.downto(2).each {|i| return false if num % i == 0}
true
end
First, we check for 0 and 1, as they're not prime. Then we basically just check every number less than num to see if it divides. However, as explained here, for every factor greater than the square root of num, there's one that's less, so we only look between 2 and the square root.
Update
def is_prime?(num)
return if num <= 1
(2..Math.sqrt(num)).none? { |i| (num % i).zero? }
end
The error you are getting is because you haven't required Primein your code, You need to do require Prime in your file.
One cool way I found here, to check whether a number is prime or not is following:
class Fixnum
def prime?
('1' * self) !~ /^1?$|^(11+?)\1+$/
end
end
10.prime?
From an algorithmic standpoint, checking if a number is prime can be done by checking all numbers up to and including (rounding down to previous integer) said number's square root.
For example, checking if 100 is prime involves checking everything up to 10.
Checking 99 means only going to 9.
** Another way to think about it **
Each factor has a pair (3 is a factor of 36, and 3's pair is 12).
The pair is on the other side of the square root (square root of 6 is 36, 3 < 6, 12 > 6).
So by checking everything until the square root (and not going over) ensures you check all possible factors.
You can make it quicker by having a list of prime numbers to compare, as you are doing. If you have a maximum limit that's reasonably small, you could just have a list of primes and do a direct lookup to see if that number is prime.
def is_prime?(num)
Math.sqrt(num).floor.downto(2).each {|i| return false if num % i == 0}
true
end
lol sorry for resurrecting a super old questions, but it's the first one that came up in google.
Basically, it loops through possible divisors, using the square root as the max number to check to save time on very large numbers.
In response to your question, while you can approach the problem by using Ruby's Prime I am going to write code to answer it on its own.
Consider that all you need to do is determine a factor that is smaller than the integer's square root. Any number larger than the integer's square root as a factor requires a second factor to render the number as the product. (e.g. square root of 15 is approx 3.8 so if you find 5 as a factor it is only a factor with the factor pair 3 and 5!!)
def isPrime?(num)
(2..Math.sqrt(num)).each { |i| return false if num % i == 0}
true
end
Hope that helps!!
(To first answer the question: yes, you are doing something wrong. As BLUEPIXY mentions, you need to put require 'prime' somewhere above the line that calls Prime.prime?. Typically on line 1.)
Now, a lot of answers have been given that don't use Prime.prime?, and I thought it might be interesting to benchmark some of them, along with a possible improvement of my own that I had in mind.
###TL;DR
I benchmarked several solutions, including a couple of my own; using a while loop and skipping even numbers performs best.
Methods tested
Here are the methods I used from the answers:
require 'prime'
def prime1?(num)
return if num <= 1
(2..Math.sqrt(num)).none? { |i| (num % i).zero? }
end
def prime2?(num)
return false if num <= 1
Math.sqrt(num).to_i.downto(2) {|i| return false if num % i == 0}
true
end
def prime3?(num)
Prime.prime?(num)
end
def prime4?(num)
('1' * num) !~ /^1?$|^(11+?)\1+$/
end
prime1? is AndreiMotinga's updated version. prime2? is his original version (with the superfluous each method removed). prime3? is Reboot's, using prime library. prime4? is Saurabh's regex version (minus the Fixnum monkey-patch).
A couple more methods to test
The improvement I had in mind was to leverage the fact that even numbers can't be prime, and leave them out of the iteration loop. So, this method uses the #step method to iterate over only odd numbers, starting with 3:
def prime5?(num)
return true if num == 2
return false if num <= 1 || num.even?
3.step(Math.sqrt(num).floor, 2) { |i| return false if (num % i).zero? }
true
end
I thought as well that it might be interesting to see how a "primitive" implementation of the same algorithm, using a while loop, might perform. So, here's one:
def prime6?(num)
return true if num == 2
return false if num <= 1 || num.even?
i = 3
top = Math.sqrt(num).floor
loop do
return false if (num % i).zero?
i += 2
break if i > top
end
true
end
Benchmarks
I did a simple benchmark on each of these, timing a call to each method with the prime number 67,280,421,310,721. For example:
start = Time.now
prime1? 67280421310721
puts "prime1? #{Time.now - start}"
start = Time.now
prime2? 67280421310721
puts "prime2? #{Time.now - start}"
# etc.
As I suspected I would have to do, I canceled prime4? after about 60 seconds. Presumably, it takes quite a bit longer than 60 seconds to assign north of 6.7 trillion '1''s to memory, and then apply a regex filter to the result — assuming it's possible on a given machine to allocate the necessary memory in the first place. (On mine, it would seem that there isn't: I went into irb, put in '1' * 67280421310721, made and ate dinner, came back to the computer, and found Killed: 9 as the response. That looks like a SignalException raised when the process got killed.)
The other results are:
prime1? 3.085434
prime2? 1.149405
prime3? 1.236517
prime5? 0.748564
prime6? 0.377235
Some (tentative) conclusions
I suppose that isn't really surprising that the primitive solution with the while loop is fastest, since it's probably closer than the others to what's going on under the hood. It is a bit surprising that it's three times faster than Prime.prime?, though. (After looking at the source code in the doc it is less so. There are lots of bells and whistles in the Prime object.)
AndreiMotinga's updated version is nearly three times as slow as his original, which suggests that the #none? method isn't much of a performer, at least in this context.
Finally, the regex version might be cool, but it certainly doesn't appear to have much practical value, and using it in a monkey-patch of a core class looks like something to avoid entirely.
If you are going to use any Prime functions you must include the Prime library. This problem can be solved without the use of the prime library however.
def isPrime?(num)
(2..Math.sqrt(num)).each { |i|
if num % i == 0 && i < num
return false
end
}
true
end
Something like this would work.
Try this
def prime?(num)
2.upto(Math.sqrt(num).ceil) do |i|
break if num%i==0
return true if i==Math.sqrt(num).ceil
end
return false
end
So most of the answers here are doing the same thing in slightly different ways which is one of the cool things about Ruby, but I'm a pretty new student (which is why I was looking this up in the first place) and so here's my version with comment explanations in the code:
def isprime n # starting with 2 because testing for a prime means you don't want to test division by 1
2.upto(Math.sqrt(n)) do |x| # testing up to the square root of the number because going past there is excessive
if n % x == 0
# n is the number being called from the program
# x is the number we're dividing by, counting from 2 up to the square root of the number
return false # this means the number is not prime
else
return true # this means the number is prime
end
end
end