Coding Challenge Of The Day

[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

[Chris Maunder]

Sweet.

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[nrkn]

Better:

static int RomanToInt( string roman ) {
return roman
.ToLower()
.Select(
n =>
new Dictionary {
{'i', 1},
{'v', 5},
{'x', 10},
{'l', 50},
{'c', 100},
{'d', 500},
{'m', 1000},
}[ n ]
)
.Aggregate(
new[] {0, 0, 0},
( c, d ) =>
new[] {
c[ 0 ] + ( d > c[ 2 ] ? d - c[ 1 ] : c[ 1 ] ),
d > c[ 2 ] ? 0 : d,
d
}
)
.Take( 2 )
.Sum();
}

[AnonimityPreferred]

My previous code after some Herbalife:-

FUNC rr(c)
r=0;p=0;b=len(c)
WHILE b>0
v=at(c[b],'IVXLCDM')-1
f=10^int(v/2)*(v%2*4+1)
r+=if(fp,p:=f,)
b--
END
RETURN r

Compiles with xHarbour.

[VadimAlk]

private Dictionary R2A_dictionary = new Dictionary()
{
{'I', 1},
{'V', 5},
{'X', 10},
{'L', 50},
{'C', 100},
{'D', 500},
{'M', 1000},
};

private int ConvertRomanToArabic(string roman)
{
int result = 0;
int previous = Int32.MaxValue, current = 0;
for (int i = 0; i < roman.Length; i++)
{
current = R2A_dictionary[roman[i]];
result += current;
if (current > previous)
result -= previous << 1;
previous = current;
}
return result;
}

[jsc42]

If there are bonus points for using hated constructs, how about this one-liner using JavaScript's 'eval' function?

function RtoA(r)
{ return eval(r.toUpperCase()
.replace(/CM/g, '+900').replace(/CD/g, '+400').replace(/XC/g, '+90').replace(/XL/g, '+40')
.replace(/IX/g, '+9').replace(/M/g, '+1000').replace(/D/g, '+500').replace(/C/g, '+100')
.replace(/L/g, '+50').replace(/X/g, '+10').replace(/V/g, '+5').replace(/I/g, '+1'));
}

or for obscure use of Regular Expression matching and using string length as scale factors:

function RtoA(r) // r is roman numerals in any case
{
var m = r.match(/^(M*)(CM)?(D)?(CD)?(C*)(XC)?(XL)?(L)?(X*)(IX)?(IV)?(V)?(I*)$/i);

return  (   m\[1\].length \* 1000 +    // M
        m\[2\].length \* 450 + // CM
        m\[3\].length \* 500 + // D
        m\[4\].length \* 200 + // CD
        m\[5\].length \* 100 + // C
        m\[6\].length \* 45 +  // XC
        m\[7\].length \* 25 +  // XL
        m\[8\].length \* 50 +  // L
        m\[9\].length \* 10 +  // X
        m\[10\].length \* 4.5 +    // IX
        m\[11\].length \* 2 +  // IV
        m\[12\].length \* 5 +  // V
        m\[13\].length        // I
    );

} // RtoA

Of the three submissions that I have made, this is the only one that enforces the ordering of numerals (e.g. all Cs must be before all Is) and the fact that some character sequences can only appear once (e.g. D can only appear once, except when following CD which also can only appear once). The odd scale multiplier factors (e.g. 4.5) come from the fact that some letter combinations have 2 characters and the substring lengths are thus twice as long (IX = 2 chars = value of 9, so each char is worth 9 / 2 = 4.5).

[Chris Maunder]

Oh wow. I think it's too early in the morning for that sort of shenanigans, especially a Monday morning. I feel queasy.

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP