How can I reduce my if statements? - ruby

Question

Going through Chris Pine's Learn To Program and working on the number to roman numeral conversion project. The code below works, however it's pretty ugly w/ all those if (and end) statements. However, when I use elsif the program doesn't respond (appears to freeze up). Any thoughts would be helpful!
def calc input
roman_numeral = ''
while true
if input >= 1000
roman_numeral += 'M' * (input / 1000)
input = input - (1000 * (input / 1000))
if input <= 999 || input >= 500
roman_numeral += 'D' * (input / 500)
input = input - (500 * (input / 500))
if input <= 499 || input >= 100
roman_numeral += 'C' * (input / 100)
input = input - (100 * (input / 100))
if input <= 99 || input >= 50
roman_numeral += 'L' * (input / 50)
input = input - (50 * (input / 50))
if input <= 49 || input >= 10
roman_numeral += 'X' * (input / 10)
input = input - (10 * (input / 10))
if input <= 9 || input >= 5
roman_numeral += 'V' * (input / 5)
input = input - (5 * (input / 5))
if input <= 4 || input >= 1
roman_numeral += 'I' * (input / 1)
input = input - (1 * (input / 1))
puts roman_numeral
break
end
end
end
end
end
end
end
end
end
puts 'Give me a number, any number:'
input = gets.chomp.to_i
calc(input)

Answer 1

It's convenient to use the method Enumerable#find with an array:
ARR = [[1000,'M'], [ 500,'D'], [100,'C'], [50,'L'], [10,'X'], [5,'V'], [1,'I']]
def which(input)
ARR.find { |v,_| input >= v }
end
which(2) #=> [1, "I"]
which(7) #=> [5, "V"]
which(17) #=> [10, "X"]
which(77) #=> [50, "L"]
which(777) #=> [500, "D"]
which(7777) #=> [1000, "M"]
Assuming you are converting an integer to a roman numeral, consider making use of the method Fixnum#divmod. Suppose the integer were 2954 and you've already determined that there are two "M"'s and one "D" (so the beginning of your roman numeral string is "MMD"), and that 454 is left over. Then:
c, rem = 454.divmod(100)
#=>[4, 54]
c #=> 4
rem #=> 54
tells you there are four "C"'s with 54 left over.
Four "C"'s are written "CD" (not "CCCC"), however, so you may want to use a hash such as the following:
REP = {..., "C"=>["C", "CC", "CCC", "CD"], ...}
to convert the number of "C"'s to a roman numeral. Here you would append REP["C"][4-1] #=> "CD" to "MMD": "MMD" << "CD" #=> "MMDCD".

Answer 2

The answer from Cary Swoveland is an excellent way to decrease your if block nesting.
His answer tells you which numeral comes next, but not how many (as in your code). A natural way to tie it together is with a recursive function call:
class Romans
def self.calc(input, acc = "")
raise ArgumentError.new("Roman Numerals must be positve") if input < 0
raise ArgumentError.new("Roman Numerals must be integers") if ! input.is_a? Integer
return acc if input == 0
amount, numeral = which(input)
acc += numeral
input -= amount
calc(input, acc)
end
##ARR = [[1000,'M'], [ 500,'D'], [100,'C'], [50,'L'], [10,'X'], [5,'V'], [1,'I']]
def self.which(input)
##ARR.find { |v,_| input >= v }
end
end
In use:
pry(main)> (1..10).each{|i| puts "#{i}=> #{Romans.calc(i)}"}
1=> I
2=> II
3=> III
4=> IIII
5=> V
6=> VI
7=> VII
8=> VIII
9=> VIIII
10=> X
pry(main)> [Random.rand(1..100000)].each{|i| puts "#{i}=> #{Romans.calc(i)}"}
63124=> MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMCXXIIII
Be aware that ruby doesn't have TCO, and so will blow the stack with large enough numbers- but if you need the Roman numeral version of 8 million, you might want to make up some new letters.

Answer 3

Here's one that uses string multiplication. For example: ('M' * 3) = 'MMM'
def to_roman(number)
raise 'Must use positive numbers.' if number <= 0
roman = ''
roman << 'M' * (number / 1000)
roman << 'D' * (number % 1000 / 500)
roman << 'C' * (number % 500 / 100)
roman << 'L' * (number % 100 / 50)
roman << 'X' * (number % 50 / 10)
roman << 'V' * (number % 10 / 5)
roman << 'I' * (number % 5 / 1)
roman
end
puts to_roman(1234) # <-- test
Reference: Learn to Program 2nd by Chris Pine