This is a program that came from a textbook:
# Initialize our counter
i = 1
# i: [0, 100]
while (i <= 100)
# Initialize prime flag
prime_flag = true
j = 2
# Test divisibility of i from [0, i/2]
while (j <= i / 2)
# puts " i ==> " to i.to_s + " j ==> " + j.to_s
if (i % j == 0)
prime_flag = false
# break
end
j = j + 1
end
# We found a prime!
if prime_flag
puts "Prime ==> " + i.to_s
end
# Increment the counter
i += 1
end
The while (j <= i / 2) introduces a new loop. What if we are trying to find prime numbers. Why is this written? Prime numbers don't have square roots. What is the purpose of j being <= i / 2? I do not understand why j is introduced.
You are correct that you should be only checking numbers <= floor(sqrt(i)). The above code is unnecessarily checking numbers from ceil(sqrt(i)) through i/2. It would give the correct answer, however.
In addition, this is not very Ruby-like code. It's terrible and the author should feel terrible (unless they intended to show you something bad in order for you to be amazed when you see how you can write it better!).
Here's the same code done in a more Ruby-like manner. Note that prime? can definitely be a one-liner, but I split things on to more lines readability in the context of the question:
def prime?(i) # Define a function that takes 1 parameter `i`
MAX_NUM_TO_CHECK = Math.sqrt(i) # No need to check numbers greater than sqrt(i)
(2..MAX_NUM_TO_CHECK).all? do |j| # Return `true` if the following line is true for
# all numbers [2,MAX_NUM_TO_CHECK]
i % j != 0 # true if `i` is evenly not divisible by `j`. Any
# input that evaluates to false here is not prime.
end
end
# Test primality of numbers [1,100]
(1..100).each {|n| puts "Prime ==> #{n}" if prime? n}
I think the biggest differences between your book and this code are:
The algorithm is different in that we do not check all values, but rather limit the checks to <= sqrt(i). We also stop checking once we know a number is not prime.
We iterate over Ranges rather than keeping counters. This is slightly higher level and easier to read once you get the syntax.
I split the code into two parts, a function that calculates whether a parameter is prime or not, and then we iterate over a Range of inputs (1..100). This seems like a good division of functionality to me, helping readability.
Some language features used here not in your example:
If statements can go after expressions, and the expression is only evaluated if the predicate (the thing after the if) evaluates to true. This can make some statements more readable.
A range is written (x..y) and allows you to quickly describe a series of values that you can iterate over without keeping counters.
Code inside
do |param1, ..., paramN| <CODE>; end
or
{|param1, ..., paramN| <CODE>}
is called a block. It's an anonymous function (a function passed in as a parameter to another function/method). We use this with all? and each here.
each is used to run a block of code on each element of a collection, ignoring the return value
all? can be used to determine whether a block returns true for every item in a collection.
If you're not familiar with passing code into functions, this is probably a little confusing. It allows you to execute different code based on your needs. For example, each runs the yielded block for every item in the collection.You could do anything in that block without changing the definition of each... you just yield it a block and it runs that block with the proper parameters. Here is a good blog post to get you started on how this works and what else you can use it for.
Related
I need to DRY this code but I don't know how.
I tried to dry the if condition but I don't know how to put the while in this.
def sum_with_while(min, max)
# CONSTRAINT: you should use a while..end structure
array = (min..max).to_a
sum = 0
count = 0
if min > max
return -1
else
while count < array.length
sum += array[count]
count += 1
end
end
return sum
end
Welcome to stack overflow!
Firstly, I should point out that "DRY" stands for "Don't Repeat Yourself". Since there's no repetition here, that's not really the problem with this code.
The biggest issue here is it's unrubyish. The ruby community has certain things it approves of, and certain things it avoids. That said, while loops are themselves considered bad ruby, so if you've been told to write it with a while loop, I'm guessing you're trying to get us to do your homework for you.
So I'm going to give you a couple of things to do a web search for that will help start you off:
ruby guard clauses - this will reduce your if-else-end into a simple if
ruby array pop - you can do while item = array.pop - since pop returns nil once the array is empty, you don't need a count. Again, bad ruby to do this... but maybe consider while array.any?
ruby implicit method return - generally we avoid commands we don't need
It's worth noting that using the techniques above, you can get the content of the method down to 7 reasonably readable lines. If you're allowed to use .inject or .sum instead of while, this whole method becomes 2 lines.
(as HP_hovercraft points out, the ternary operator reduces this down to 1 line. On production code, I'd be tempted to leave it as 2 lines for readability - but that's just personal preference)
You can put the whole thing in one line with a ternary:
def sum_with_while(min, max)
min > max ? -1 : [*(min..max)].inject(0){|sum,x| sum + x }
end
This is one option, cleaning up your code, see comments:
def sum_with_while(range) # pass a range
array = range.to_a
sum, count = 0, 0 # parallel assignment
while count < array.length
sum += array[count]
count += 1
end
sum # no need to return
end
sum_with_while(10..20)
#=> 165
More Rubyish:
(min..max).sum
Rule 1: Choose the right algorithm.
You wish to compute an arithmetic series.1
def sum_with_while(min, max)
max >= min ? (max-min+1)*(min+max)/2 : -1
end
sum_with_while(4, 4)
#=> 4
sum_with_while(4, 6)
#=> 15
sum_with_while(101, 9999999999999)
#=> 49999999999994999999994950
1. An arithmetic series is the sum of the elements of an arithmetic sequence. Each term of the latter is computed from the previous one by adding a fixed constant n (possibly negative). Heremax-min+1 is the number of terms in the sequence and (min+max)/2, if (min+max) is even, is the average of the values in the sequence. As (max-min+1)*(min+max) is even, this works when (min+max) is odd as well.
I am doing a ruby problem that wants a method to find all divisors of a number except itself with the output being a sorted array. If the number is prime, list that it is prime.
I am currently trying to teach myself recursion. Simple recursive problems like finding the factorial of a number is pretty basic to understand but I wanted to know if this particular problem could be done recursively. It seems it fits the criteria of one that could but I could not figure it out.
Example n = 15, divisors besides itself are [3,5].
My code that solved the problem.
require 'prime'
def divisors(n)
return "#{n} is prime" if Prime.prime?(n)
x = n/2
arr = []
until x == 1
arr << x if n % x == 0
x -= 1
end
arr.sort
end
Any help doing this recursively would be great or just letting me know it's not a problem that can be done this way would be helpful too.
def divisors(n, x=nil)
return "#{n} is prime" if Prime.prime?(n)
x ||= n/2
arr = []
return arr if x == 1
if n % x == 0
arr << x
end
(arr.concat divisors(n, x - 1)).sort
end
The function is refactored to handle three things:
the initial call (x ||= /2)
base cases (early returns)
iteration logic done through recursion.
An important thing is that the variable which changes during the iteration (x) is placed as a parameter for the method (with a default value, so it can essentially be used as a private parameter)
By the way, I personally found learning Elixir very helpful in understanding recursion. With pattern matching and multiple functional clauses, the initial call, base case, and iteration can be split into their own methods.
I don't quite understand how to "initialize a multidimensional array to equal 1" as the initial for loops seem to suggest here. I haven't learned to properly read pseudocode, and I don't fully understand how this program works.
function countRoutes(m,n)
grid ← array[m + 1][n + 1]
for i = 0 to m do
grid[i][0] ← 1
end for
for j = 0 to n do
grid[0][j] ← 1
end for
for i = 1 to m do
for j = 1 to n do
grid[i][j] ← grid[i − 1][j] + grid[i][j − 1]
end for
end for
return grid[m][n]
end function
Thanks for your help!
This isn't hard to translate.. Ruby uses = instead of left arrow for assignment, and def instead of function to define a subroutine (which it calls a method instead of a function), but that's about it. Let's go through it.
function countRoutes(m,n)
That's beginning a function definition. In Ruby we use a method instead, and the keyword to define such a thing is def. It's also convention in Ruby to use snake_case for multiword names instead of camelCase:
def count_routes(m, n)
Now to create the grid:
grid ← array[m + 1][n + 1]
Ruby arrays are dynamic, so you don't normally need to specify the size at creation time. But that also means you don't get initialization or two-dimensionality for free. So what we have to do here is create an array of m+1 arrays, each of which can be empty (we don't need to specify that the sub-arrays need to hold n+1 items). Ruby's Array constructor has a way to do just that:
grid = Array.new(m+1) do [] end
Now the initialization. Ruby technically has for loops, but nobody uses them. Instead, we use iterator methods. For counting loops, there's a method on integers called times. But the pseudocode counts from 0 through m inclusive; times also starts at 0, but only counts up to one less than the invocant (so that way when you call 3.times, the loop really does execute "three times", not four). In this case, that means to get the behavior of the pseudocode, we need to call times on m+1 instead of m:
(m+1).times do |i|
grid[i][0] = 1
end
As an aside, we could also have done that part of the initialization inside the original array creation:
grid = Array.new(m+1) do [1] end
Anyway, the second loop, which would be more awkward to incorporate into the original creation, works the same as the first. Ruby will happily extend an array to assign to not-yet-existent elements, so the fact that we didn't initialize the subarrays is not a problem:
(n+1).times do |j|
grid[0][j] = 1
end
For the nested loops, the pseudocode is no longer counting from 0, but from 1. Counting from 1 through m is the same number of loop iterations as counting from 0 through m-1, so the simplest approach is to let times use its natural values, but adjust the indexes in the assignment statement inside the loop. That is, where the pseudocode starts counting i from 1 and references i-1 and i, the Ruby code starts counting i from 0 and references i and i+1 instead.
m.times do |i|
n.times do |j|
grid[i+1][j+1] = grid[i][j+1] + grid[i+1][j]
end
end
And the return statement works the same, although in Ruby you can leave it off:
return grid[m][n]
end
Putting it all together, you get this:
def count_routes(m, n)
grid = Array.new(m+1) do [1] end
(n+1).times do |j|
grid[0][j] = 1
end
m.times do |i|
n.times do |j|
grid[i+1][j+1] = grid[i][j+1] + grid[i+1][j]
end
end
return grid[m][n]
end
The notation grid[i][j] ← something means assigning something to the element of grid taking place on i-th line in j-th position. So the first two loops here suggest setting all values of the first column and the first row of the grid (correspondingly, the first and the second loops) to 1.
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
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