Coding Challenge Of The Day

[Luc Pattyn]

that would be where most Romans are, don't you think? :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

[Luc Pattyn]

no arrays, no dictionaries, no obscure language, just some original coding:

public static bool TryParseRoman(string wellFormedRoman, out int val) {
val=0;
int prevDigit=0;
foreach (char c in wellFormedRoman) {
int index="IVXLCDM".IndexOf(c);
if (index<0) {val=0; return false;}
int digit=1;
int factor=5;
while (--index>=0) { digit*=factor; factor=7-factor; }
if (prevDigit

BTW: is it Friday already?

:)

Luc Pattyn [My Articles] Nil Volentibus Arduum

[PIEBALDconsult]

I'm unsure when I wrote this, but it must have been the 90s, in VAX C or DEC C. I'm shocked (shocked! I say) at the inconsistent braces. :wtf:

size_t
roman2dec
(
char *subject
)
{
size_t result = 0 , current , high = 0 ;
int index ;

for ( index = strlen(subject)-1 ; index >= 0 ; index-- ) {
    switch ( (char) \*(subject + index) )
    {
        case 'I' :
        case 'i' : current =    1 ; break ;

        case 'V' :
        case 'v' : current =    5 ; break ;

        case 'X' :
        case 'x' : current =   10 ; break ;

        case 'L' :
        case 'l' : current =   50 ; break ;

        case 'C' :
        case 'c' : current =  100 ; break ;

        case 'D' :
        case 'd' : current =  500 ; break ;

        case 'M' :
        case 'm' : current = 1000 ; break ;

        default  : current =    0 ; break ;
    }

    if ( current < high )
        result -= current ;
    else
    {
       result += current ;
       high = current ;
    }
}

return ( result ) ;

}

[PIEBALDconsult]

That's dreadful. :wtf:

[PIEBALDconsult]

Seems a waste of RAM.

[arcosupportus]

did you get here number?

[jsc42]

Apologies if other respondents have used a similar technique, but I have deliberately not looked at the other replies before attempting this challenge so that I am not influenced by other people's ideas. The following is in JavaScript, not obfuscated. The tricks are: Using an object literal as a lookup table and treating the double chars as subtractors from the second single chars (e.g. CM = [CM] + [M] = -100 + 1000 = 900). The only awkward part is preventing the last char being counted twice - once as a single char and once as a double char with no second char (this is done by appending '?').

// Roman to Arabic character translations
function RtoA(r) // r is roman numerals in any case
{
// Roman to Arabic conversion table
var Rch =
{
// Single chars: Add each value separately to the total
I: 1, V: 5, X: 10, L: 50, C: 100, D: 500, M: 1000,
// Double chars: Modify the sum of the second char as subtractors
CM: -100, CD: -100, XC: -10, XL: -10, IX: -1, IV: -1
};

var	R   = r.toUpperCase();  // Ignore case
var	A   = 0;    // Arabic equivalent

// Parse the text converting valid single chars and double chars
// Set any invalid single or double char combinations as translating to zero
for (var i = 0; i < R.length; i++)
{
	var	ch	= R.charAt(i);
	A	+= 
		Rch\[ch + (R.charAt(i + 1) || '?')\]	// Double char conversion
		||	Rch\[ch\]	// Single char conversion
		||	0;	// No conversion - invalid char
};	// for

return	A;

}

alert(RtoA('MDC')); // 1600
alert(RtoA('mcmxcix')); // 1999

[RobertHarris]

First time post so hope this looks OK; 1 CRT "NUMERALS: ": ; INPUT NUM 2 CRT "NUMBER : ":ICONV(NUM, 'NR') 3 END Written in UniVerse BASIC.

[yiangos]

Or, somewhere in between:

def a6(roman):
if len(roman)==0:
return 0
mapping={"I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000}
values=[mapping[digit] for digit in roman.upper()]
for i in range(len(values)):
if i

also, added capitalization tolerance, so that "MCM" and "mcm" evaluate the same. I haven't added a check that the argument is a valid roman numeral, but that is left as an exercise to the reader ;P

Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων!
(Alas! We're devoured by lamb-guised wolves!)

[Pascal Ganaye]

I hope it will be obscure enough. ;)

class Program
class Program
{
static int RomanToArabic(string r)
{
int x, y, z = x = y = 0;
foreach (char c in r)
y += ((z = (int)Math.Pow(10, (z = "ivxlcdm".IndexOf((char)(c | ' '))) / 2) * (1 + 4 * (z & 1))) > x ? -2 * x + (x = z) : x = z);
return y;
}

    static void Main(string\[\] args)
    {
        Console.WriteLine(RomanToArabic("MDC"));	// 1600
        Console.WriteLine(RomanToArabic("mcmxcix"));	// 1999
        Console.ReadKey();
    }
    
}

}

[Pascal Ganaye]

This is definitely better than my attempt.

[Luc Pattyn]

Nice. I like the % ' ' bits, and appreciate you used literal VII. :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

[Pascal Ganaye]

oops I removed the % ' '... Ok I put back a:

"ivxlcdm".IndexOf((char)(c | ' '))

[Pascal Ganaye]

This is my final answer

static int RomanToArabic(string roman, int x = 0, int y=0)
{
return roman.Length == 0 ? 0 : (y = (int)Math.Pow(10, (y = "ivxlcdm".IndexOf((char)(roman[0] | ' '))) / 2) * (1 + 4 * (y & 1))) + (y > x ? -2 * x : 0) + RomanToArabic(roman.Substring(1), y);
}

[Alan Balkany]

Geez, I thought it was going to be something hard.

"Microsoft -- Adding unnecessary complexity to your work since 1987!"

[Reese Currie]

OK, I admit it isn't very small. Scala is number 46 on the Tiobe Index this month, so perhaps it is obscure enough to qualify. Being a "functional" language there are actually 3 functions but I nested them in one.

def convertRomanToArabic(romanNumeral: String): Int = {
  
  var previous = 0
  
  def convertSingleRomanNumeral(numeral: Char): Int = {
    numeral match {
      case 'I' => 1
      case 'V' => 5
      case 'X' => 10
      case 'L' => 50
      case 'C' => 100
      case 'D' => 500
      case 'M' => 1000
    }
  }
  
  def addRomans(next: Int, accumulator: Int): Int = {
    if (previous == 0) previous = accumulator
    var addto = if (previous > next) next \* -1 else next
    previous = next
    accumulator + addto
  }
  
  val values = romanNumeral.toList.map(n => convertSingleRomanNumeral(n))
  values.reduceRight(addRomans)
}

[AnonimityPreferred]

FUNCTION Roman2Dec(cRoman)

LOCAL nPointer := len(cRoman)  
LOCAL nPos := 0,nFaceVal := 0,nLastVal := 0, nReturn := 0

WHILE nPointer > 0
    nPos     := at(substr(cRoman, nPointer, 1),'IVXLCDM') -1 
    nFaceVal := (10 ^ int((nPos)/2)) \* (1+((nPos) % 2 \* 4))
    nReturn  += iif(nFaceVal < nLastVal, -nFaceVal, nFaceVal)
    IF nFaceVal > nLastVal
        nLastVal := nFaceVal
    ENDIF
    nPointer--    
ENDDO

RETURN nReturn

[firegryphon]

These aren't the roman numerals you're looking for.  You should go about your business.  Move along move along.

ragnaroknrol: Yes, but comparing a rabid wolverine gnawing on your face while stabbing you with a fountain pen to Vista is likely to make the wolverine look good, so it isn't exactly that big of a compliment.

[User 8912811]

In C#:

using System;
using System.Collections;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace RomanNumerConverter
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}

    private void button1\_Click(object sender, EventArgs e)
    {
        Hashtable ht = new Hashtable(); // Create a look-up table to look up Roman numerals and get there associated values.
        ht.Add('M', 1000);
        ht.Add('D', 500);
        ht.Add('C', 100);
        ht.Add('L', 50);
        ht.Add('X', 10);
        ht.Add('V', 5);
        ht.Add('I', 1);

        String RomanNumber = textBox1.Text;
        int value = 0;
        for (int index = 0; index < RomanNumber.Length; index++)  // Go through the entered Roman numeral a character at a time
        {
            int nextVal;
            int currentVal = Convert.ToInt32(ht\[RomanNumber\[index\]\]); // Get the value of the current character.
            if (index + 1 < RomanNumber.Length) // if you are not at the end of the string
            {
                nextVal = Convert.ToInt32(ht\[RomanNumber\[index + 1\]\]);// Peek at the next number to see you need to use it to combine characters for a single value(e.g. IV equates to 4)
            }
            else
            {
                nextVal = 0; // No more characters so just set this to 0 to simplfy the algorithm.
            }
            if (nextVal > currentVal)// If the value of nextVal is greater than currentVal, we have a subtractive situation (e.g. IV)
            {
                currentVal = nextVal - currentVal;  // To determine the value you have to subtract the value of the first number of the pair from the second number(e.g. IV = 5 - 1 = 4)
                index++;// Used nextVal to help determine the number so increment the index to skip over that character in the next iteration.
            }
            value += currentVal; // Add the values of the Roman numerals up.
        }
        label1.Text = value.ToString();  //Convert the sum to string to use as a text label.  This is the converted number in Arabic numerals.
    }
}

}

[nrkn]

Hideous C# one-liner* with whitespace added to ahahahaha enhance readability.

static double RomanToInt( string roman ) {
return (
Enumerable.Range( 0, 3 )
.Select(
i =>
( i + 1 ) * 2 - 2
)
.Aggregate(
roman.ToLower(),
( current, i ) =>
current
.Replace(
"ivxlcdm".Substring( i, 2 ),
new string( "ivxlcdm"[ i ], 4 )
)
.Replace(
"ivxlcdm"[ i ].ToString() + "ivxlcdm"[ i + 2 ].ToString(),
"ivxlcdm"[ i + 1 ].ToString() + new string( "ivxlcdm"[ i ], 4 )
)
)
.Select(
i =>
"ivxlcdm".IndexOf( i )
)
.Select(
x =>
x % 2 == 0 ?
Math.Pow( 10, x / 2 ) :
Math.Pow( 10, ( x + 1 ) / 2 ) / 2
)
.Sum()
);
}

*Not counting the function declaration - trivial to convert to a true a one-liner but you asked for a function