## Roman Numerals

### March 6, 2009

Roman numerals are a number-notation system developed in classical Rome, chiefly used today to indicate the year in which a motion picture was made, or the sequence number of a Super Bowl.

Roman numerals use letters of the alphabet to indicate numerical value, according to the following code:

I 1    V 5    X 10    L 50    C 100    D 500    M 1000

For example, the number 1732 is represented by Roman numerals as MDCCXXXII, and the number 1956 is represented by Roman numerals as MDCCCCLVI. Letter symbols are normally written from the largest symbol to the smallest, left to right, so the numeric values are additive. However, in order to conserve space, it is permissible to replace four of the same symbol written all in a row in a subtractive manner to the left of a higher-value symbol, so that 1956 may also be represented as MCMLVI, where the CM symbol, with C before M, indicates that C is subtracted from M, and thus indicates the numeric value 900. Wikipedia and MathWorld explain the common usage of Roman numerals.

Your task is to write a function that takes two roman numerals (character strings) as input and returns their sum as a roman numeral as output. Be sure that input can be given in either the additive or subtractive forms of Roman numerals; give output using the subtractive form. What is `add-roman("CCCLXIX", "CDXLVIII")`?

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### 11 Responses to “Roman Numerals”

1. Roger said

I started to program thinking that constructions like “IIX” for 8 are allowed. Wikipedia says that this kind of subtractive notation exists, but it seems very rare. Whatever, here is my roman->decimal.

```
(define roman->decimal
(lambda (x)
(decode-roman
(string->list x))))

(define decode-roman
(lambda (chars)
(letrec
((decode-roman-helper
(lambda (fir res akk cur)
(cond
((null? res) (+ akk cur))
((< (single-roman->decimal fir) (single-roman->decimal (car res)))
(decode-roman-helper (car res)
(cdr res)
(- akk cur)
(single-roman->decimal (car res))))
((> (single-roman->decimal fir) (single-roman->decimal (car res)))
(decode-roman-helper (car res)
(cdr res)
(+ cur akk)
(single-roman->decimal (car res))))
(else (decode-roman-helper
(car res)
(cdr res)
akk
(+ (single-roman->decimal (car res)) cur)))))))
(decode-roman-helper (car chars) (cdr chars) 0 (single-roman->decimal (car chars))))))

(define single-roman->decimal
(lambda (str)
(cond
((char=? str #\M) 1000)
((char=? str #\D) 500)
((char=? str #\C) 100)
((char=? str #\L) 50)
((char=? str #\X) 10)
((char=? str #\V) 5)
((char=? str #\I) 1)
(else 0))))
```
2. FalconNL said

Haskell (someone more experienced than me can probably turn these into one-liners):

import Data.Map (fromList, (!))
import Data.Char
import Data.List

data Roman = I | V | X | L | C | D | M deriving (Enum, Eq, Ord, Read, Show)

main = print \$ addRoman “CCCLXIX” “CDXLVIII”

values :: [(Roman, Int)]
values = [(M, 1000), (D, 500), (C, 100), (L, 50), (X, 10), (V, 5), (I, 1)]

fromRoman :: String -> Int
fromRoman = fromRoman’ . map (read . return) where
fromRoman’ (x:y:xs) = (if x String
toRoman = map toLower . concatMap show . subtractiveStyle . toRoman’ values where
toRoman’ [] _ = []
toRoman’ ((r, v):xs) n = replicate (div n v) r ++ toRoman’ xs (mod n v)

subtractiveStyle :: [Roman] -> [Roman]
subtractiveStyle (x:y:ys) | y == pred x && isPrefixOf [y,y,y] ys
= y : succ x : subtractiveStyle (drop 3 ys)
subtractiveStyle xs = xs

addRoman :: String -> String -> String
addRoman a b = toRoman \$ fromRoman a + fromRoman b

3. FalconNL said

Evidently there’s only a limited selection of languages that will trigger the right formatting. My apologies.

```import Data.Map (fromList, (!))
import Data.Char
import Data.List

data Roman = I | V | X | L | C | D | M deriving (Enum, Eq, Ord, Read, Show)

main = print \$ addRoman "CCCLXIX" "CDXLVIII"

values :: [(Roman, Int)]
values = [(M, 1000), (D, 500), (C, 100), (L, 50), (X, 10), (V, 5), (I, 1)]

fromRoman :: String -> Int
fromRoman = fromRoman' . map (read . return) where
fromRoman' (x:y:xs) = (if x < y then -1 else 1) * val x + fromRoman' (y:xs)
fromRoman' xs = sum \$ map val xs
val c = fromList values ! c

toRoman :: Int -> String
toRoman = map toLower . concatMap show . subtractiveStyle . toRoman' values where
toRoman' []          _ = []
toRoman' ((r, v):xs) n = replicate (div n v) r ++ toRoman' xs (mod n v)

subtractiveStyle :: [Roman] -> [Roman]
subtractiveStyle (x:y:ys) | y == pred x && isPrefixOf [y,y,y] ys
= y : succ x : subtractiveStyle (drop 3 ys)
subtractiveStyle xs = xs

addRoman :: String -> String -> String
addRoman a b = toRoman \$ fromRoman a + fromRoman b
```
4. programmingpraxis said

FalconNL: Haskell is not one of the supported languages for the WordPress sourcecode tag. See my HOWTO page for more about posting source code in comments.

5. kawas said

At first I thought my solution was long and ugly and than I read the praxis’ Scheme solution :)
I read Roger’s Scheme but it is incomplete : no encode-roman
I read FalconNL’s Haskell and subtractiveStyle doesn’t work on numbers like 1904

So I guess I’ll roll my ugly clojure solution :

```(def *romans* {\I 1, \V 5, \X 10, \L 50, \C 100, \D 500, \M 1000})

(defn from-roman [rs]
(loop [rs (reverse rs) prev 0 dnum 0]
(let [v (*romans* (first rs))]
(cond
(nil? v)   dnum
(< v prev) (recur (next rs) v (- dnum v))
:else      (recur (next rs) v (+ dnum v))))))

(defn to-roman [d]
(let [dq (quot d 1000) dr (rem d 1000)
r (vec (repeat dq \M))
romans (reverse (sort-by second *romans*))]
(loop [d dr romans romans r r]
(if (zero? d) (apply str r)
(let [[[u10 v10] [u5 v5] [u1 v1]] romans dq (quot d v1) dr (rem d v1)]
(cond
(= dq 9) (recur dr (nnext romans) (conj r u1 u10))
(> dq 4) (recur dr (nnext romans) (apply conj r u5 (repeat (- dq 5) u1)))
(= dq 4) (recur dr (nnext romans) (conj r u1 u5))
(> dq 0) (recur dr (nnext romans) (apply conj r (repeat dq u1)))
:else    (recur d (nnext romans) r)))))))

(to-roman (apply + (map from-roman rs))))
```

Some use cases :

```user=> (add-roman "CCCLXIX" "CDXLVIII")
"DCCCXVII"

"MMMMCMIV"
```
6. Hello everybody,

I briefly walked through all posted examples and figured out that noone completely solved this task.
kawas was very close to solution however even in his second use case user=> (add-roman “MMCCCII” “MMDCII”) result seemed to be wrong because there cannot be 4 M in a row. Please see wikipedia as a proof-link http://en.wikipedia.org/wiki/Roman_numerals

```The symbols "I", "X", "C", and "M" can be repeated three times in succession, but no more. (They may appear four times if the third and fourth are separated by a smaller value, such as XXXIX.) "D", "L", and "V" can never be repeated.
```

So here is my solution in Python:

```rom_arab_dict = {'I' : 1, 'V' : 5, 'X' : 10, 'L' : 50, 'C' : 100, 'D' : 500, 'M' : 1000}
limit_sum = 3999

arab_sum = roman_to_arabic(a) + roman_to_arabic(b)
if arab_sum <= limit_sum:
arabic_to_roman(arab_sum)
print('Roman sum: %s + %s = %s' % (a, b, arabic_to_roman(arab_sum)))
print('Arab sum(verification): %s + %s = %s' % (roman_to_arabic(a), roman_to_arabic(b), arab_sum))
else:
print('Result sum out of range of Roman numbers.\n Should be less or equal to %d' % limit_sum)

def roman_to_arabic(roman_num):
sum = 0
arab_lst = [rom_arab_dict[x] for x in roman_num for y in rom_arab_dict.keys() if x == y]
for index, item in enumerate(arab_lst):
if index + 1 < len(arab_lst):
if item < arab_lst[index + 1]:
arab_lst[index] = arab_lst[index + 1] - arab_lst[index]
del arab_lst[index + 1]
sum += arab_lst[index]

return sum

def arabic_to_roman(arab_num):
# dict {'key':val} to list [(v, k)]
items = [(v, k) for k, v in rom_arab_dict.items()]
items.sort()
# reverse order from high to low
items.reverse()

rom_lst = []
for index, item in enumerate(items):
v, k = item
while arab_num - v >= 0:
arab_num -= v
tmp_v, tmp_k = items[index]
rom_lst.append(tmp_k)
if rom_lst.count(tmp_k) == 4:
rom_lst = rom_lst[:-3]
tmp_v, tmp_k = items[index - 1]
rom_lst.append(tmp_k)

return ''.join(rom_lst)

#Test cases
```

Results:

Roman sum: CCCLXIX + CDXLVIII = DCCCXVII
Arab sum(verification): 369 + 448 = 817
Roman sum: CDXXVIII + DLXXVIII = MVI
Arab sum(verification): 428 + 578 = 1006
Roman sum: MDCCL + MDCLXX = MMMCDXX
Arab sum(verification): 1750 + 1670 = 3420

Cheers,
Pavel

7. Dan Prager said

A nice additional constraint to the problem — at least for Roman numerals in additive form — is to forbid conversion back to decimal for the purposes of carrying out the addition. This seems more authentic, given that conversion to decimal was not an option available to the classical Romans!

I discuss this approach to a solution along with some other (arguably) interesting connections on my new blog. Straight to the Python code.

8. wilornel said

How to chose between VIV and IX? I can’t seem to think of a way to follow this convention.