Integer to Roman while loop explanation-Python

Question

I am trying to understand the while loop and sorted call in the program to convert a number to Roman numerals below.

numerals = { 1 : "I", 4 : "IV", 5 : "V", 9 : "IX", 10 : "X", 40 : "XL",
             50 : "L", 90 : "XC", 100 : "C", 400 : "CD", 500 : "D", 900 : "CM", 1000 : "M" }

num = 58  # LVIII

roman = ''

for k, v in sorted(numerals.items(), reverse=True):
    while num >= k:
        roman += v
        num -= k

print(roman)

Questions: 1) Why doesn't the code work if numerals.items() is used instead of sorted(numerals.items(), reverse=True)? (For example, 58 would result in IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII instead of LVIII.) When using a breakpoint on that line it looks like the order remains same both with and without sorted.

2) The first Roman numeral is L. Why? When debugging I noticed it starts counting down from 1000. When it reaches 50, I see that roman == 'L'. The code tests if num >= k. 1000 (M) is also greater than 58. Why does the condition num >= k result in L being the first digit?

Answer 1

1. sorted(numerals.items(), reverse=True) not only means sorted, but also means sorted in reverse. The key point is reverse=True. Because we must compare to the largest Roman first. From M to I, not from I to M.
2. "The logic says num>=k. 1000,M is also greater than num 58." I think you have misunderstanding here. for while num>=k: it find less or equal than num, not large, so L will be the first one.

Hope that will help you.