Coding Challenge Of The Day

[wizardzz]

I didn't see you award him any points. Go on, 5 him!

[Chris Maunder]

Actually it's a personality test to see who among you are worthy of being labelled a Brogrammer.

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

[Steve Mayfield]

Don't you mean 'V' him?

Steve _________________ I C(++) therefore I am

[S Houghtelin]

Thanks wizardzz, I suppose he meant like virtual points, or maybe pints, I'll definitely takes pints! :-D

It was broke, so I fixed it.

[Bassam Abdul Baki]

You may want to snag CodeBroject.com just in case (CodePreject.com too).

Web - BM - RSS - Math - LinkedIn

[Sentenryu]

as a nerd, i can't resist... between, it's only a quick n' dirty solution:

public static int RomanToArabic(string romans) {
if (string.IsNullOrWhiteSpace(romans))
{
return 0;
}
//remove whitespace...
romans = romans.Replace(" ", string.Empty);
//make sure no one pass out an "A" for example...
if (!Regex.IsMatch(romans, "(I|V|X|C|M|D|L)+"))
{
throw new ArgumentException("Invalid numeral!!");
}

        //actual validation that matters (based on http://en.wikipedia.org/wiki/Roman\_numerals): 
        //The symbols "I", "X", "C", and "M" can be repeated three times in succession, but no more. "D", "L", and "V" can never be repeated.
        string invalidSequences = "(XXXX+)|(IIII+)|(CCCC+)|(MMMM+)|(DD+)|(LL+)|(VV+)";

        if (Regex.IsMatch(romans, invalidSequences))
        {
            throw new ArgumentException("Invalid sequence!!");
        }

        int result = 0;

        result += Regex.Matches(romans, "IV").Count \* 4;
        romans = Regex.Replace(romans, "IV", string.Empty);

        result += Regex.Matches(romans, "IX").Count \* 9;
        romans = Regex.Replace(romans, "IX", string.Empty);

        result += Regex.Matches(romans, "XL").Count \* 40;
        romans = Regex.Replace(romans, "XL", string.Empty);

        result += Regex.Matches(romans, "XC").Count \* 90;
        romans = Regex.Replace(romans, "XC", string.Empty);

        result += Regex.Matches(romans, "CD").Count \* 400;
        romans = Regex.Replace(romans, "CD", string.Empty);

        result += Regex.Matches(romans, "CM").Count \* 900;
        romans = Regex.Replace(romans, "CM", string.Empty);

        result += romans.Count(x => x == 'I');
        romans = romans.Replace("I", string.Empty);

        result += romans.Count(x => x == 'V') \* 5;
        romans = romans.Replace("V", string.Empty);

        result += romans.Count(x => x == 'X') \* 10;
        romans = romans.Replace("X", string.Empty);

        result += romans.Count(x => x == 'L') \* 50;
        romans = romans.Replace("L", string.Empty);

        result += romans.Count(x => x == 'C') \* 100;
        romans = romans.Replace("C", string.Empty);

        result += romans.Count(x => x == 'D') \* 500;
        romans = romans.Replace("D", strin

[S Houghtelin]

Just like positive reenforcment in the schools today, everybody gets a trophy, even when they're wrong. Thanks! :thumbsup: :-D

It was broke, so I fixed it.

[AspDotNetDev]

Good catch. Live long and prosper.

Thou mewling ill-breeding pignut!

[Karl Sanford]

Chris Maunder wrote:

Roman numerals to Arabic numbers

totally read this backward... at least now I have a solution for Arabic to Roman Numerals :doh:

Be The Noise

[Lost User]

#!/usr/bin/python

import sys

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

a=0
b=0
c=0

for d in map(eval,sys.argv[1]):
if d > a:
b += d - c
c = 0
else:
b += c
c = d
a = d
print b + c

[Corporal Agarn]

Tried to leave a similar message earlier but CP locked. I was going to add this is not QA. :)

[jesarg]

private int ToInt(string rom)
{
int length = rom.Length;
int result = 0;
for(int loop1=0; loop1 < length; loop1++)
{
char current = rom[loop1];
char next = loop1 < length - 1 ? rom[loop1 + 1] : ' ';
switch(current)
{
case 'I':
result += next == 'V' || next == 'X' ? -1 : 1;
break;
case 'V':
result += 5;
break;
case 'X':
result += next == 'L' || next == 'C' ? -10 : 10;
break;
case 'L':
result += 50;
break;
case 'C':
result += next == 'D' || next == 'M' ? -100 : 100;
break;
case 'D':
result += 500;
break;
case 'M':
result += 1000;
break;
}
}
return result;
}

[Lost User]

Or, somewhat more tersely,

def r2d3(r,(s,M,C,X,I,z,V,L,D)=["%%s%d"%v for v in[1001,1000,100,10,1,0,5,50,500]]):
return eval("".join([
p%"+-"[eval( p%"%s>"%d%"")]
for d,p in zip(map(eval,'%sz'%r),map(eval,'s%s'%r))
][1:]))

[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