Today I made myself proud...
Problem 5 - Smallest Multiple
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?
My Groovy solution:
class Test extends GroovyTestCase {
def solveFor(upperBound) {
def answer = 0
def temp = upperBound * (upperBound + 1)
while(true) {
answer = temp
(2..upperBound).each {
if(temp % it != 0) {
answer = 0
return
}
}
if(answer != 0) {
return answer
}
temp += upperBound
}
}
void test_base_case() {
assertEquals 2520, solveFor(10)
}
void test_solution() {
def start = System.currentTimeMillis()
println solveFor(20)
def end = System.currentTimeMillis()
println "Elapsed time ${end - start} milliseconds"
}
}
To this Ruby solution:
require "test/unit"
def solve_for(upper)
answer = upper
until divisible_by_numbers_up_to?(answer, upper) do
answer += upper
end
answer
end
def divisible_by_numbers_up_to?(number, upper)
(2..upper).all? { |i| number % i == 0 }
end
class SolveProblemXTests < Test::Unit::TestCase
def test_simple_case
solution = solve_for(2)
assert_equal(2, solution, "Solution was #{solution} not 2520") end
def test_base_case
solution = solve_for(10)
assert_equal(2520, solution, "Solution was #{solution} not 2520")
end
end
answer = solve_for(20)
puts "Answer is #{answer}"
After some thinking I felt it was still a little clunky and I actually wanted to try to make it more terse for fun. So I left the tests etc. but changed the implementation at top of file to this:
def solve_for(upper)
(upper..Float::INFINITY).step(upper).find do |num|
(2..upper).all? { |i| num % i == 0 }
end
end
Not sure it got more confusing to read. Earlier code was not necessarily easiest to read but I learned some stuff about Ruby and coding in general. Honestly it was my exposure to Clojure that made me think I could use a lazy loaded list of numbers going to infinity.
Mission Accomplished!
