I would like to pass an array of numbers to my is_prime? method and return if the numbers are valid or not. I do not want to use:
require 'prime'
a = [1,2,3,4,5]
Hash[a.zip(a.map(&Prime.method(:prime?)))]
This is learning experience. My current code is only outputing the first number in the array. Can someone help me understand what I am doing wrong? Thanks!
def is_prime?(*nums)
i = 2
nums.each do |num|
while i < num
is_divisible = ((num % i) == 0)
if is_divisible == false
x = "#{num}: is NOT a prime number." #false
else
x = "#{num}: is a prime number." #true
end
i +=1
end
return x
end
end
puts is_prime?(27,13,42)
You are returning in the loop.
A few bugs in your method:
def is_prime?(*nums)
nums.each do |num|
return false if num == 1
next if num == 2 # 2 is the only even prime
i = 2 # needs to be reset for each num
while i < num
return false if num % i == 0 # num is not prime
i += 1
end
end
true # We'll reach here only if all the numbers are prime
end
This will return your results in the same format as your usage of the prime library with the same logic as your custom function:
def is_prime?(*nums)
nums.each_with_object({}) do |num, hsh|
hsh[num] = num > 1 && 2.upto(num - 1).none? { |i| num % i == 0 }
end
end
puts is_prime?(27,13,42)
# => {27=>false, 13=>true, 42=>false}
Since you mention this is just for learning, I'm assuming you know that a sieve is a better way to go for this than brute force iteration.
If you want an explanation of how the above code works or further help understanding why your current code doesn't, let me know in the comments.
Related
I'm just learning Ruby :) and Im trying to create a simple prime-number program where all the primes of a number are printed.
I'm getting errors where the prime and nonprime numbers are mixed up
(ie: input of 9 will get you all nonprimes).
I'm sorry for such a beginner question - I'm struggling alot and need some help :)
puts "Enter a number please "
num = gets.chomp.to_i
i = 2
number = 2
while i < num
if number % i == 0
prime = false
end
i += 1
end
if prime == true
puts "#{number} is prime"
else
puts "#{number} is not prime"
end
number += 1
end
while i < num
if number % i == 0
prime = false
end
i += 1
end
# ...
It looks like that first end is meant to be an else.
It's easier to catch these things when you simplify your code, e.g. this method reduces to this (although there are other issues with it):
i = 2
number = 2
while i < num
(number % i).zero? ? prime = false : i += 1
puts "#{number} is #{'not ' unless prime}prime"
number += 1
end
End error is because of else
while i < num
if number % i == 0
prime = false
else
i += 1
end
if you have a short If - neater is writing it like:
if-condition ? 1 : 0
and in your case while is.. not the nicest choice - you should use range
(1...3).map{|x| puts(x) }
{} - allows multiline(with do end end)
this prints [1,2]
(1..3).map{|x| x*2 }
would be [2,4,9]
This should be enough hints of how to play around with your code without ruining the process.
All right. I think I have the right idea to find the solution to Euler #23 (The one about finding the sum of all numbers that can't be expressed as the sum of two abundant numbers).
However, it is clear that one of my methods is too damn brutal.
How do you un-brute force this and make it work?
sum_of_two_abunds?(num, array) is the problematic method. I've tried pre-excluding certain numbers and it's still taking forever and I'm not even sure that it's giving the right answer.
def divsum(number)
divsum = 1
(2..Math.sqrt(number)).each {|i| divsum += i + number/i if number % i == 0}
divsum -= Math.sqrt(number) if Math.sqrt(number).integer?
divsum
end
def is_abundant?(num)
return true if divsum(num) > num
return false
end
def get_abundants(uptonum)
abundants = (12..uptonum).select {|int| is_abundant?(int)}
end
def sum_of_two_abunds?(num, array)
#abundant, and can be made from adding two abundant numbers.
array.each do |abun1|
array.each do |abun2|
current = abun1+abun2
break if current > num
return true if current == num
end
end
return false
end
def non_abundant_sum
ceiling = 28123
sum = (1..23).inject(:+) + (24..ceiling).select{|i| i < 945 && i % 2 != 0}.inject(:+)
numeri = (24..ceiling).to_a
numeri.delete_if {|i| i < 945 && i % 2 != 0}
numeri.delete_if {|i| i % 100 == 0}
abundants = get_abundants(ceiling)
numeri.each {|numerus| sum += numerus if sum_of_two_abunds?(numerus, abundants) == false}
return sum
end
start_time = Time.now
puts non_abundant_sum
#Not enough numbers getting excluded from the total.
duration = Time.now - start_time
puts "Took #{duration} s "
Solution 1
A simple way to make it a lot faster is to speed up your sum_of_two_abunds? method:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
array.each do |abun2|
current = abun1+abun2
break if current > num
return true if current == num
end
end
return false
end
Instead of that inner loop, just ask the array whether it contains num - abun1:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
return true if array.include?(num - abun1)
end
false
end
That's already faster than your Ruby code, since it's simpler and running faster C code. Also, now that that idea is clear, you can take advantage of the fact that the array is sorted and search num - abun1 with binary search:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
return true if array.bsearch { |x| num - abun1 <=> x }
end
false
end
And making that Rubyish:
def sum_of_two_abunds?(num, array)
array.any? do |abun1|
array.bsearch { |x| num - abun1 <=> x }
end
end
Now you can get rid of your own special case optimizations and fix your incorrect divsum (which for example claims that divsum(4) is 5 ... you should really compare against a naive implementation that doesn't try any square root optimizations).
And then it should finish in well under a minute (about 11 seconds on my PC).
Solution 2
Or you could instead ditch sum_of_two_abunds? entirely and just create all sums of two abundants and nullify their contribution to the sum:
def non_abundant_sum
ceiling = 28123
abundants = get_abundants(ceiling)
numeri = (0..ceiling).to_a
abundants.each { |a| abundants.each { |b| numeri[a + b] = 0 } }
numeri.compact.sum
end
That runs on my PC in about 3 seconds.
I'm trying to create a recursive method sum_of_digits(i) that takes the sum of integers, i.e. '456' = 4+5+6 = 15
However, I receive a NoMethodError for chr.to_i in the following code:
def sum_of_digits(i)
input = i.to_s
if i == 0
return 0
elsif input.length == 1
return i
else
for n in 1..input.length
sum += input[i].chr.to_i % 10^(n-1)
end
end
return sum
end
Thank you!
String indexes are zero-based in ruby. The problem is here:
for n in 1..input.length
it should be written as
for n in 0..input.length-1
BTW, call to chr is superfluous as well, since you already have a string representation of a digit there. As well, sum must be declared in advance and set to zero.
Also, the whole code is not ruby idiomatic: one should avoid using unnecessary returns and for-loop. The modified version (just in case) would be:
def sum_of_digits(i)
input = i.to_s
case
when i == 0 then 0 # return zero
when input.length == 1 then i # return i
else
sum = 0
input.length.times do |index|
sum += input[index].to_i % 10^index
end
sum
end
end
or, even better, instead of
sum = 0
input.length.times do |index|
sum += input[index].to_i % 10^index
end
sum
one might use inject:
input.length.times.inject(0) do |sum, index|
sum += input[index].to_i % 10^index
end
I am trying to write a concise is_prime method, which determines if an integer is prime. (I've written the method below). I've used a 'while' loop to finish off the method, but what I really want to use is a 'for' loop – similar to how it would be coded in java. (Java for loop is included below).
My immediate question is how to make my ruby is_prime method more concise and use a for loop instead of a while. My secondary question is how to translate java for loops into ruby (could I use an each method)?
//java (replaces only the while loop below and not the entire is_prime method)
for(int d=3; d*d<=n; d+=2){
if(n%d==0) return false;
}
#ruby
def is_prime(n)
if (n < 2)
return false
end
if (n == 2)
return true
end
if (n%2 == 0)
return false
end
d = 3
while (d*d < n) do
if(n%d==0)
return false
end
d+=2
end
return true;
end
puts is_prime(5) # true
puts is_prime(13) # true
puts is_prime(10) # false
puts is_prime(28) # false
Maybe you're just writing this method for fun, but in case you need to get work done, Ruby has a built in prime library.
require 'prime'
Prime.prime?(2**31-1)
#=> true
def is_prime n
for d in 2..(n - 1)
if (n % d) == 0
return false
end
end
true
end
Hope this will help you.
Here is another way to write it:
def is_prime?(number)
(2..number-1).each {|n| return false if number <= 1 || number % n == 0}
return true
end
You can use ruby for loop like this:
def is_prime(n)
return false if n < 2 or n % 2 == 0
return true if n == 2
for d in 3..Math.sqrt(n).to_i
return false if n % d == 0
end
return true;
end
You will know how to use loop in ruby from this link.
I need some feedback to figure out why I cant puts or print anything from my methods on the screen. This is a simple script I wrote to solve the problem of finding the 1001st prime number. Thanks
def primes
# iterates through numbers until it has the 1001th prime number and returns it.
# I chose to create the num_primes variable instead of counting the number of
# elements in in_prime_array every iteration
# based upon a guess that it would be faster to check.
is_prime_array = []
num_primes = 0
i = 2
loop do
is_prime_array << i && num_primes += 1 if is_prime?(i) == true
i += 1
break if num_primes == 1001
end
is_prime_array[1001]
end
def is_prime? (num)
# Checks to see if the individual number given is a prime number or not.
i = 2
loop do
if i == num
return true
elsif num % i == 0
return false
else
i += 1
end
end
end
Thanks for any help!
EDIT
I took your advice and tried this pice of code:
def is_prime? (num)
# Checks to see if the individual number given is a prime number or not.
i = 2
loop do
if i == num
return true
elsif num % i == 0
return false
else
i += 1
end
end
end
i = 0
count = 0
loop do
count += 1 if is_prime?(x)
puts "#{i}" if count == 1001
break
end
It still returns nothing. Hummm
i = 0
count = 0
loop do
if is_prime(i)
count += 1
end
if count == 10001
puts "#{i}"
break
end
end
Simple method :)
It's an off-by-one error. If you have 1001 elements in an array, the last element will be at index 1000.
Where you have
is_prime_array[1001]
Change it to
is_prime_array[1000]
And you can do this:
puts primes
=> 7927
You could also have
is_prime_array.last
instead of a specific index number.
What are you trying to "puts"? The first thing I notice is that there is no call to primes in the file, so nothing will happen if you try to run this code by itself. Maybe that's why you don't see anything printed.
Here's an example of how to print a few variables inside your loop:
loop do
...
puts "At iteration #{i}, we have prime=#{is_prime?(i)}"
If you don't know, enclosing a statement with #{<statement goes here>} inside a string is the same as appending the return value of <statement goes here> to the string at that position. This is the same as "Str " + blah + " rest of str" in a language like Java.