I recently rolled my own *Decimal to Roman Numeral converter* in Clojure. I was inspired by some tweets about a roman numeral code kata.

The Decimal to Roman Numeral converter takes a decimal number (like `2012`

) and returns a string of roman numerals (like `"MMXII"`

). The code that I came up with is around ten lines that look like this:

(def mapping (sorted-map-by > 1000 "M" 900 "CM" 500 "D" 400 "CD" 100 "C" 90 "XC" 50 "L" 40 "XL" 10 "X" 9 "IX" 5 "V" 4 "IV" 1 "I")) (defn roman-numeral [i] (loop [i i s ""] (if (zero? i) s (let [[[n c]] (filter #(>= i (first %)) mapping)] (recur (- i n) (str s c))))))

Let me try to explain it.

Overall, it’s pretty easy. There’s a `loop`

in which we check whether the number `i`

is zero or not. If it is, the current value of the string `s`

is returned. If the number `i`

is not zero, `let`

something something is being done, and we take another whirl in the loop using `recur`

, with the number `i`

decremented by some `n`

and the string `s`

appended with some `c`

.

There are a few things going on:

- mapping
- filter
- destructuring
- loop/recur

Let’s go through them in turn.

## mapping

I need the `mapping`

data structure to be a sorted map. Clojure has a literal syntax for hash maps, using curly braces (the `=>`

indicates what the code evaluates to):

{10 "X" 9 "IX" 5 "V" 4 "IV" 1 "I"} => {1 "I", 10 "X", 4 "IV", 5 "V", 9 "IX"}

Hash maps are not sorted, so obviously that won’t do. How about `sorted-map`

? Then it will be sorted by key, *ascending*:

(sorted-map 10 "X" 9 "IX" 5 "V" 4 "IV" 1 "I") => {1 "I", 4 "IV", 5 "V", 9 "IX", 10 "X"}

I want it to be sorted *descending*, so I use `sorted-map-by`

and pass in a `Comparator`

. The Clojure predicate `>`

(greater-than) implements `java.util.Comparator`

, so I can just pass that:

(sorted-map-by > 10 "X" 9 "IX" 5 "V" 4 "IV" 1 "I") => {10 "X", 9 "IX", 5 "V", 4 "IV", 1 "I"}

## filter

The `filter`

function returns a sequence containing those elements that fulfill the given predicate. Here is the code that performs the filtering:

(filter #(>= i (first %)) mapping)

There are a few funny things going on here. Firstly, the predicate function to `filter`

looks weird. Normally, anonymous functions in Clojure are written like this:

(fn [x] (>= i (first x)))

However, there is a shorter syntax that skips the `(fn ...)`

and the named argument, using `%`

(or `%1`

, `%2`

if more than one) as the argument placeholder. This is equivalent to the above, only shorter:

#(>= i (first %))

The other thing that might look weird is the call to `first`

. Wasn’t `mapping`

a map? What is passed in to the predicate, really? Well, it turns out that it’s quite convenient to be able to treat a map as a sequence. So, Clojure has a literal syntax for a map as a logical sequence. We can see that if we call `seq`

on it:

(seq (sorted-map-by > 10 "X" 9 "IX" 5 "V" 4 "IV" 1 "I")) => ([10 "X"] [9 "IX"] [5 "V"] [4 "IV"] [1 "I"])

Here we see that a map can logically be seen as a sequence of vectors containing key-value pairs. The filter predicate will be passed a vector containing a key-value pair. Therefore, `first`

in the predicate will return the key. So, the filter will return all key-value pairs that have a key that is less than or equal to the number `i`

. For example:

(filter #(>= 6 (first %)) mapping) => ([5 "V"] [4 "IV"] [1 "I"]) (filter #(>= 4 (first %)) mapping) => ([4 "IV"] [1 "I"])

## destructuring

Clojure supports a convenient form of picking apart a data structure and naming its parts, where you would normally bind the whole data structure to a single variable. Suppose we know that our call to filter returns a sequence of numbers:

(filter odd? [1 2 3 4 5]) => (1 3 5)

Say we’re only interested in the first and the second elements. Without destructuring, we would manually retrieve them like this:

(let [lst (filter odd? [1 2 3 4 5]) fst (first lst) snd (second lst)] (println fst snd)) => 1 3

With destructuring, we can directly get and name those parts in the `let`

binding by placing the names in a vector, in which place we previously had just the variable name `lst`

:

(let [ [fst snd] (filter odd? [1 2 3 4 5])] (println fst snd)) => 1 3

Suppose now that the `filter`

function instead returns a sequence of key-value pairs, as in our `roman-numeral`

:

(filter #(>= 4 (first %)) mapping) => ([4 "IV"] [1 "I"])

Destructuring can be done with arbitrary depth, so if we want to: a) get the first element of a list of key-value pairs, and then: b) break that element apart and name the key `n`

and the value `c`

, we’ll write:

(let [ [ [n c] ] (filter ...)] (recur ...))

So, in just seven characters, `[[n c]]`

, we have taken the first element of the filter result, and also named that element’s parts `n`

and `c`

respectively. Thanks to the sorted map, the first element will be the highest matching number and its corresponding roman numeral.

## loop/recur

Without going into too much details, `loop`

and `recur`

is Clojure’s way of working around the lack of tail call optimization in the JVM. We’ll use `recur`

to perform a recursive call without any risk of emptying the stack. We’ll use `loop`

to explicitly provide a recursion point with the number of variables we need, and to initialize those variables.

(defn roman-numeral [i] (loop [i i s ""] ...)

Here we say that we want two loop variables, `i`

and `s`

, where `i`

will be initialized to the value of the parameter `i`

, and `s`

will be initialized to the empty string. Variables have lexical scope, so there’s no problem shadowing the *parameter* `i`

with the *looping variable* `i`

. Inside `loop`

, `i`

will always refer to the looping variable `i`

. I could have used another looping variable, like `j`

or `k`

, but I didn’t.

(recur (- i n) (str s c))

Here we say that we want to call the recursion point with new values of `i`

and `s`

, where `i`

will be the current value of `i`

minus the `n`

we got from destructuring the key-value pair from `filter`

, and `s`

will be the value of the current `s`

concatenated with the `c`

that we got from the destructuring.

### I still don’t get the `loop/recur`

thing…

Perhaps it’ll be more clear if I begin by defining a separate private helper function for the recursion, and just call it with some initial values from the `roman-numeral`

function:

(defn- roman-helper [i s] (if (zero? i) s (let [[[n c]] (filter #(>= i (first %)) mapping)] (roman-helper (- i n) (str s c))))) (defn roman-numeral [i] (roman-helper i ""))

The recursive call from `roman-helper`

to itself *will* require stack space. We could change the recursive call to a `recur`

instead, thus eliminating the risk for stack overflow. This works since `recur`

will use the top of the function if there is no corresponding `loop`

. Note that `roman-helper`

has been replaced with `recur`

:

(defn- roman-helper [i s] (if (zero? i) s (let [[[n c]] (filter #(>= i (first %)) mapping)] (recur (- i n) (str s c))))) (defn roman-numeral [i] (roman-helper i ""))

We can place the helper function inside `roman-numeral`

, as a local anonymous function that we’ll bind to a variable name in a `let`

:

(defn roman-numeral [i] (let [roman-helper (fn [i s] (if (zero? i) s (let [[[n c]] (filter #(>= i (first %)) mapping)] (recur (- i n) (str s c)))))] (roman-helper i "")))

Remember that `loop`

not only defines an explicit recursion point, but also initializes any loop variables. From here, it’s a small step to replace the helper function and the call to it with just a `loop`

that initializes `i`

and `s`

:

(defn roman-numeral [i] (loop [i i s ""] (if (zero? i) s (let [[[n c]] (filter #(>= i (first %)) mapping)] (recur (- i n) (str s c))))))

## Example: (roman-numeral 6)

Let’s run through an example where we call `roman-numeral`

with 6.

The first round, `i`

will be initialized to `6`

, and `s`

will be initialized to `""`

. Of course, `i`

is not zero, so `filter`

will be called and return ([5 “V”] [4 “IV”] [1 “I”]) . The `let`

will destructure the first element into `n=5`

and `c="V"`

. Finally, `recur`

will be called with (- 6 5) and (str “” “V”) .

The next round in the loop will have `i=1`

and `s="V"`

. The call to `filter`

will return ([1 “I”]) , `recur`

will then be called with (- 1 1) and (str “V” “I”) .

In the final round in the loop, `i`

will be zero, and `s`

will be returned, resulting in the string `"VI"`

, which is the correct roman numeral for 6.

## Steve Miner

6 Aug 2012Just for reference, there’s also the built-in way:

(defn roman-cl [n]

(clojure.pprint/cl-format nil “~@R” n))

## Steve Miner

6 Aug 2012Another version:

http://fnclojure.posterous.com/roman-numerals-in-clojure

## Ulrik Sandberg

6 Aug 2012Thanks for sharing your versions. The Common Lisp format replica looks pretty nice.

Your optimized version seems to be around five times faster than both my version and cl-format.

Pingback: A Decimal to Roman Numeral converter in Clojure | Japila :: verba docent, exempla trahunt

## kawas

12 Aug 2012Hi Ulrik,

I like your very elegant solution.

Speaking of optimization, I see couple of tricks that speed things up 100 fold.

– Since you’re only using the first value of filter you may use ‘some’ and thus less destructuring :

(let [[n c] (some #(when (>= i (first %)) %) mapping)] …)

– Since your accumulating strings you may concat them all at the end using a vector for ‘s’, using ‘conj’ in ‘recur’ and applying ‘str’ to get the result.

Then you got something still elegant and quick enough for the job

(defn roman-numeral2 [i]

(loop [i i s []]

(if (zero? i)

(apply str s)

(let [[n c]

(some #(when (>= i (first %)) %) mapping)]

(recur (- i n) (conj s c))))))

user=> (time (count (roman-numeral 98765432)))

“Elapsed time: 12094.501546 msecs”

98772

user=> (time (count (roman-numeral2 98765432)))

“Elapsed time: 124.735959 msecs”

98772

Cheers

## Ulrik Sandberg

12 Aug 2012Very nice improvements. Thank you.

## Steve Miner

26 Feb 2013My post has moved to a new URL:

http://fnclojure.wordpress.com/2012/08/06/roman-numerals-in-clojure/