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I am trying to use a if statement to determine if the roman numerals is 4, 9, 40, 90, 400, 900 but I am getting undefined [duplicate]

How can I convert integers into roman numerals?
function romanNumeralGenerator (int) {
}
For example, see the following sample inputs and outputs:
1 = "I"
5 = "V"
10 = "X"
20 = "XX"
3999 = "MMMCMXCIX"
Caveat: Only support numbers between 1 and 3999

There is a nice one here on this blog I found using google:
http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
function romanize (num) {
if (isNaN(num))
return NaN;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}

function romanize(num) {
var lookup = {M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) {
roman += i;
num -= lookup[i];
}
}
return roman;
}
Reposted from a 2008 comment located at: http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
VIEW DEMO

I don't understand why everyone's solution is so long and uses multiple for loops.
function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
};
var str = '';
for (var i of Object.keys(roman)) {
var q = Math.floor(num / roman[i]);
num -= q * roman[i];
str += i.repeat(q);
}
return str;
}

I've developed the recursive solution below. The function returns one letter and then calls itself to return the next letter. It does it until the number passed to the function is 0 which means that all letters have been found and we can exit the recursion.
var romanMatrix = [
[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']
];
function convertToRoman(num) {
if (num === 0) {
return '';
}
for (var i = 0; i < romanMatrix.length; i++) {
if (num >= romanMatrix[i][0]) {
return romanMatrix[i][1] + convertToRoman(num - romanMatrix[i][0]);
}
}
}

These functions convert any positive whole number to its equivalent Roman Numeral string; and any Roman Numeral to its number.
Number to Roman Numeral:
Number.prototype.toRoman= function () {
var num = Math.floor(this),
val, s= '', i= 0,
v = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
r = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
function toBigRoman(n) {
var ret = '', n1 = '', rem = n;
while (rem > 1000) {
var prefix = '', suffix = '', n = rem, s = '' + rem, magnitude = 1;
while (n > 1000) {
n /= 1000;
magnitude *= 1000;
prefix += '(';
suffix += ')';
}
n1 = Math.floor(n);
rem = s - (n1 * magnitude);
ret += prefix + n1.toRoman() + suffix;
}
return ret + rem.toRoman();
}
if (this - num || num < 1) num = 0;
if (num > 3999) return toBigRoman(num);
while (num) {
val = v[i];
while (num >= val) {
num -= val;
s += r[i];
}
++i;
}
return s;
};
Roman Numeral string to Number:
Number.fromRoman = function (roman, accept) {
var s = roman.toUpperCase().replace(/ +/g, ''),
L = s.length, sum = 0, i = 0, next, val,
R = { M: 1000, D: 500, C: 100, L: 50, X: 10, V: 5, I: 1 };
function fromBigRoman(rn) {
var n = 0, x, n1, S, rx =/(\(*)([MDCLXVI]+)/g;
while ((S = rx.exec(rn)) != null) {
x = S[1].length;
n1 = Number.fromRoman(S[2])
if (isNaN(n1)) return NaN;
if (x) n1 *= Math.pow(1000, x);
n += n1;
}
return n;
}
if (/^[MDCLXVI)(]+$/.test(s)) {
if (s.indexOf('(') == 0) return fromBigRoman(s);
while (i < L) {
val = R[s.charAt(i++)];
next = R[s.charAt(i)] || 0;
if (next - val > 0) val *= -1;
sum += val;
}
if (accept || sum.toRoman() === s) return sum;
}
return NaN;
};

I personally think the neatest way (not by any means the fastest) is with recursion.
function convert(num) {
if(num < 1){ return "";}
if(num >= 40){ return "XL" + convert(num - 40);}
if(num >= 10){ return "X" + convert(num - 10);}
if(num >= 9){ return "IX" + convert(num - 9);}
if(num >= 5){ return "V" + convert(num - 5);}
if(num >= 4){ return "IV" + convert(num - 4);}
if(num >= 1){ return "I" + convert(num - 1);}
}
console.log(convert(39));
//Output: XXXIX
This will only support numbers 1-40, but it can easily be extended by following the pattern.

This version does not require any hard coded logic for edge cases such as 4(IV),9(IX),40(XL),900(CM), etc. as the others do.
I have tested this code against a data set from 1-3999 and it works.
TLDR;
This also means this solution can handle numbers greater than the maximum roman scale could (3999).
It appears there is an alternating rule for deciding the next major roman numeral character. Starting with I multiply by 5 to get the next numeral V and then by 2 to get X, then by 5 to get L, and then by 2 to get C, etc to get the next major numeral character in the scale. In this case lets assume "T" gets added to the scale to allow for larger numbers than 3999 which the original roman scale allows. In order to maintain the same algorithm "T" would represent 5000.
I = 1
V = I * 5
X = V * 2
L = X * 5
C = L * 2
D = C * 5
M = D * 2
T = M * 5
This could then allow us to represent numbers from 4000 to 5000; MT = 4000 for example.
Code:
function convertToRoman(num) {
//create key:value pairs
var romanLookup = {M:1000, D:500, C:100, L:50, X:10, V:5, I:1};
var roman = [];
var romanKeys = Object.keys(romanLookup);
var curValue;
var index;
var count = 1;
for(var numeral in romanLookup){
curValue = romanLookup[numeral];
index = romanKeys.indexOf(numeral);
while(num >= curValue){
if(count < 4){
//push up to 3 of the same numeral
roman.push(numeral);
} else {
//else we had to push four, so we need to convert the numerals
//to the next highest denomination "coloring-up in poker speak"
//Note: We need to check previous index because it might be part of the current number.
//Example:(9) would attempt (VIIII) so we would need to remove the V as well as the I's
//otherwise removing just the last three III would be incorrect, because the swap
//would give us (VIX) instead of the correct answer (IX)
if(roman.indexOf(romanKeys[index - 1]) > -1){
//remove the previous numeral we worked with
//and everything after it since we will replace them
roman.splice(roman.indexOf(romanKeys[index - 1]));
//push the current numeral and the one that appeared two iterations ago;
//think (IX) where we skip (V)
roman.push(romanKeys[index], romanKeys[index - 2]);
} else {
//else Example:(4) would attemt (IIII) so remove three I's and replace with a V
//to get the correct answer of (IV)
//remove the last 3 numerals which are all the same
roman.splice(-3);
//push the current numeral and the one that appeared right before it; think (IV)
roman.push(romanKeys[index], romanKeys[index - 1]);
}
}
//reduce our number by the value we already converted to a numeral
num -= curValue;
count++;
}
count = 1;
}
return roman.join("");
}
convertToRoman(36);

I know this is an old question but I'm pretty proud of this solution :) It only handles numbers less than 1000 but could easily be expanded to include however large you'd need by adding on to the 'numeralCodes' 2D array.
var numeralCodes = [["","I","II","III","IV","V","VI","VII","VIII","IX"], // Ones
["","X","XX","XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"], // Tens
["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"]]; // Hundreds
function convert(num) {
var numeral = "";
var digits = num.toString().split('').reverse();
for (var i=0; i < digits.length; i++){
numeral = numeralCodes[i][parseInt(digits[i])] + numeral;
}
return numeral;
}
<input id="text-input" type="text">
<button id="convert-button" onClick="var n = parseInt(document.getElementById('text-input').value);document.getElementById('text-output').value = convert(n);">Convert!</button>
<input id="text-output" style="display:block" type="text">

Loops may be more elegant but I find them hard to read. Came up with a more or less hard coded version that's easy on the eyes. As long as you understand the very first line, the rest is a no-brainer.
function romanNumeralGenerator (int) {
let roman = '';
roman += 'M'.repeat(int / 1000); int %= 1000;
roman += 'CM'.repeat(int / 900); int %= 900;
roman += 'D'.repeat(int / 500); int %= 500;
roman += 'CD'.repeat(int / 400); int %= 400;
roman += 'C'.repeat(int / 100); int %= 100;
roman += 'XC'.repeat(int / 90); int %= 90;
roman += 'L'.repeat(int / 50); int %= 50;
roman += 'XL'.repeat(int / 40); int %= 40;
roman += 'X'.repeat(int / 10); int %= 10;
roman += 'IX'.repeat(int / 9); int %= 9;
roman += 'V'.repeat(int / 5); int %= 5;
roman += 'IV'.repeat(int / 4); int %= 4;
roman += 'I'.repeat(int);
return roman;
}

I created two convert functions.
The first function can convert numbers to roman using reduce.
And the second function is very similar to the first function, the function uses the same way to convert the value.
Everything that you need to change is the _roman property. Because you have to extend this const with scale what you want, I place there max number 1000 but you can put more.
Larger scale with roman numbers you can find here https://www.tuomas.salste.net/doc/roman/numeri-romani.html
const _roman = { M: 1000, CM: 900, D: 500, CD: 400, C: 100, XC: 90, L: 50, XL: 40, X: 10, IX: 9, V: 5, IV: 4, I: 1 };
// 1903 => MCMIII
function toRoman(number = 0) {
return Object.keys(_roman).reduce((acc, key) => {
while (number >= _roman[key]) {
acc += key;
number -= _roman[key];
}
return acc;
}, '');
}
// MCMIII => 1903
function fromRoman(roman = '') {
return Object.keys(_roman).reduce((acc, key) => {
while (roman.indexOf(key) === 0) {
acc += _roman[key];
roman = roman.substr(key.length);
}
return acc;
}, 0);
}
console.log(toRoman(1903)); // should return 'MCMIII
console.log(fromRoman('MCMIII')); // should return 1903

JavaScript
function romanize (num) {
if (!+num)
return false;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}
many other suggestions can be found at http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter

Here is the solution with recursion, that looks simple:
const toRoman = (num, result = '') => {
const map = {
M: 1000,
CM: 900, D: 500, CD: 400, C: 100,
XC: 90, L: 50, XL: 40, X: 10,
IX: 9, V: 5, IV: 4, I: 1,
};
for (const key in map) {
if (num >= map[key]) {
if (num !== 0) {
return toRoman(num - map[key], result + key);
}
}
}
return result;
};
console.log(toRoman(402)); // CDII
console.log(toRoman(3000)); // MMM
console.log(toRoman(93)); // XCIII
console.log(toRoman(4)); // IV

This function will convert any number smaller than 3,999,999 to roman. Notice that numbers bigger than 3999 will be inside a label with text-decoration set to overline, this will add the overline that is the correct representation for x1000 when the number is bigger than 3999.
Four million (4,000,000) would be IV with two overlines so, you would need to use some trick to represent that, maybe a DIV with border-top, or some background image with those two overlines... Each overline represents x1000.
function convert(num){
num = parseInt(num);
if (num > 3999999) { alert('Number is too big!'); return false; }
if (num < 1) { alert('Number is too small!'); return false; }
var result = '',
ref = ['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I'],
xis = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
if (num <= 3999999 && num >= 4000) {
num += ''; // need to convert to string for .substring()
result = '<label style="text-decoration: overline;">'+convert(num.substring(0,num.length-3))+'</label>';
num = num.substring(num.length-3);
}
for (x = 0; x < ref.length; x++){
while(num >= xis[x]){
result += ref[x];
num -= xis[x];
}
}
return result;
}

I created two twin arrays one with arabic numbers the other with the roman characters.
function convert(num) {
var result = '';
var rom = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
var ara = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
Then I added a cycle which scan the roman elements, adding the biggest still comprised in NUM to RESULT, then we decrease NUM of the same amount.
It is like we map a part of NUM in roman numbers and then we decrease it of the same amount.
for (var x = 0; x < rom.length; x++) {
while (num >= ara[x]) {
result += rom[x];
num -= ara[x];
}
}
return result;
}

If you want to convert a big number with more symbols, maybe this algo could help.
The only premise for symbols is that must be odd and follow the same rule (1, 5, 10, 50,100 ...., 10^(N)/2, 10^(N)).
var rnumbers = ["I","V","X","L","C","D","M"];
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:1px solid black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border:1px solid black; border-bottom:1px none black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:3px double black; padding:1px;">'+n+'</span> '}));
String.prototype.repeat = function( num ) {
return new Array( num + 1 ).join( this );
};
function toRoman(n) {
if(!n) return "";
var strn = new String(n);
var strnlength = strn.length;
var ret = "";
for(var i = 0 ; i < strnlength; i++) {
var index = strnlength*2 -2 - i*2;
var str;
var m = +strn[i];
if(index > rnumbers.length -1) {
str = rnumbers[rnumbers.length-1].repeat(m*Math.pow(10,Math.ceil((index-rnumbers.length)/2)));
}else {
str = rnumbers[index].repeat(m);
if (rnumbers.length >= index + 2) {
var rnregexp = rnumbers[index]
.split("(").join('\\(')
.split(")").join('\\)');
str = str.replace(new RegExp('(' + rnregexp + '){9}'), rnumbers[index] + rnumbers[index + 2])
.replace(new RegExp('(' + rnregexp + '){5}'), rnumbers[index + 1])
.replace(new RegExp('(' + rnregexp + '){4}'), rnumbers[index] + rnumbers[index + 1])
}
}
ret +=str;
}
return ret;
}
<input type="text" value="" onkeyup="document.getElementById('result').innerHTML = toRoman(this.value)"/>
<br/><br/>
<div id="result"></div>

After testing some of the implementations in this post, I have created a new optimized one in order to execute faster. The time execution is really low comparing with the others, but obviously the code is uglier :).
It could be even faster with an indexed array with all the posibilities.
Just in case it helps someone.
function concatNumLetters(letter, num) {
var text = "";
for(var i=0; i<num; i++){
text += letter;
}
return text;
}
function arabicToRomanNumber(arabic) {
arabic = parseInt(arabic);
var roman = "";
if (arabic >= 1000) {
var thousands = ~~(arabic / 1000);
roman = concatNumLetters("M", thousands);
arabic -= thousands * 1000;
}
if (arabic >= 900) {
roman += "CM";
arabic -= 900;
}
if (arabic >= 500) {
roman += "D";
arabic -= 500;
}
if (arabic >= 400) {
roman += "CD";
arabic -= 400;
}
if (arabic >= 100) {
var hundreds = ~~(arabic / 100);
roman += concatNumLetters("C", hundreds);
arabic -= hundreds * 100;
}
if (arabic >= 90) {
roman += "XC";
arabic -= 90;
}
if (arabic >= 50) {
roman += "L";
arabic -= 50;
}
if (arabic >= 40) {
roman += "XL";
arabic -= 40;
}
if (arabic >= 10) {
var dozens = ~~(arabic / 10);
roman += concatNumLetters("X", dozens);
arabic -= dozens * 10;
}
if (arabic >= 9) {
roman += "IX";
arabic -= 9;
}
if (arabic >= 5) {
roman += "V";
arabic -= 5;
}
if (arabic >= 4) {
roman += "IV";
arabic -= 4;
}
if (arabic >= 1) {
roman += concatNumLetters("I", arabic);
}
return roman;
}

function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
}
var result = '';
for (var key in roman) {
if (num == roman[key]) {
return result +=key;
}
var check = num > roman[key];
if(check) {
result = result + key.repeat(parseInt(num/roman[key]));
num = num%roman[key];
}
}
return result;
}
console.log(convertToRoman(36));

I didn't see this posted already so here's an interesting solution using only string manipulation:
var numbers = [1, 4, 5, 7, 9, 14, 15, 19, 20, 44, 50, 94, 100, 444, 500, 659, 999, 1000, 1024];
var romanNumeralGenerator = function (number) {
return 'I'
.repeat(number)
.replace(/I{5}/g, 'V')
.replace(/V{2}/g, 'X')
.replace(/X{5}/g, 'L')
.replace(/L{2}/g, 'C')
.replace(/C{5}/g, 'D')
.replace(/D{2}/g, 'M')
.replace(/DC{4}/g, 'CM')
.replace(/C{4}/g, 'CD')
.replace(/LX{4}/g, 'XC')
.replace(/X{4}/g, 'XL')
.replace(/VI{4}/g, 'IX')
.replace(/I{4}/g, 'IV')
};
console.log(numbers.map(romanNumeralGenerator))

This function works on the the different character sets in each digit. To add another digit add the roman numeral string the 1 place, 5 place and next 1 place. This is nice because you update it with only knowing the next set of characters used.
function toRoman(n){
var d=0,o="",v,k="IVXLCDM".split("");
while(n!=0){
v=n%10,x=k[d],y=k[d+1],z=k[d+2];
o=["",x,x+x,x+x+x,x+y,y,y+x,y+x+x,y+x+x+x,x+z][v]+o;
n=(n-v)/10,d+=2;
}
return o
}
var out = "";
for (var i = 0; i < 100; i++) {
out += toRoman(i) + "\n";
}
document.getElementById("output").innerHTML = out;
<pre id="output"></pre>

function convertToRoman(num) {
var romans = {
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'
};
var popped, rem, roman = '',
keys = Object.keys(romans);
while (num > 0) {
popped = keys.pop();
m = Math.floor(num / popped);
num = num % popped;
console.log('popped:', popped, ' m:', m, ' num:', num, ' roman:', roman);
while (m-- > 0) {
roman += romans[popped];
}
while (num / popped === 0) {
popped = keys.pop();
delete romans[popped];
}
}
return roman;
}
var result = convertToRoman(3999);
console.log(result);
document.getElementById('roman').innerHTML = 'Roman: ' + result;
p {
color: darkblue;
}
<p>Decimal: 3999</p>
<p id="roman">Roman:</p>

I just made this at freecodecamp. It can easily be expanded.
function convertToRoman(num) {
var roman ="";
var values = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
var literals = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"];
for(i=0;i<values.length;i++){
if(num>=values[i]){
if(5<=num && num<=8) num -= 5;
else if(1<=num && num<=3) num -= 1;
else num -= values[i];
roman += literals[i];
i--;
}
}
return roman;
}

Here's a regular expression solution:
function deromanize(roman) {
var r = 0;
// regular expressions to check if valid Roman Number.
if (!/^M*(?:D?C{0,3}|C[MD])(?:L?X{0,3}|X[CL])(?:V?I{0,3}|I[XV])$/.test(roman))
throw new Error('Invalid Roman Numeral.');
roman.replace(/[MDLV]|C[MD]?|X[CL]?|I[XV]?/g, function(i) {
r += {M:1000, CM:900, D:500, CD:400, C:100, XC:90, L:50, XL:40, X:10, IX:9, V:5, IV:4, I:1}[i];
});
return r;
}

I really liked the solution by jaggedsoft but I couldn't reply because my rep is TOO LOW :( :(
I broke it down to explain it a little bit for those that don't understand it. Hopefully it helps someone.
function convertToRoman(num) {
var lookup =
{M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) { //while input is BIGGGER than lookup #..1000, 900, 500, etc.
roman += i; //roman is set to whatever i is (M, CM, D, CD...)
num -= lookup[i]; //takes away the first num it hits that is less than the input
//in this case, it found X:10, added X to roman, then took away 10 from input
//input lowered to 26, X added to roman, repeats and chips away at input number
//repeats until num gets down to 0. This triggers 'while' loop to stop.
}
}
return roman;
}
console.log(convertToRoman(36));

IF this number in HTMLElement (such as span), we recommend to Add HTML attribute data-format :
Number.prototype.toRoman = function() {
var e = Math.floor(this),
t, n = "",
i = 3999,
s = 0;
v = [1e3, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1], r = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
if (e < 1 || e > i) return "";
while (s < 13) {
t = v[s];
while (e >= t) {
e -= t;
n += r[s]
}
if (e == 0) return n;
++s
}
return ""
};
var fnrom = function(e) {
if (parseInt(e.innerHTML)) {
e.innerHTML = parseInt(e.innerHTML).toRoman()
}
};
setTimeout(function() {
[].forEach.call(document.querySelectorAll("[data-format=roman]"), fnrom)
}, 10)
Phase <span data-format="roman">4</span> Sales

function convertToRoman(num) {
let roman = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
let arabic = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
let index = 0;
let result = "";
while (num > 0) {
if (num >= arabic[index]) {
result += roman[index];
num -= arabic[index];
} else index++;
}
return result;
}

/*my beginner-nooby solution for numbers 1-999 :)*/
function convert(num) {
var RomNumDig = [['','I','II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX'],['X','XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC'], ['C','CC','CCC','CD','D','DC','DCC','DCCC','CM']];
var lastDig = num%10;
var ourNumb1 = RomNumDig[0][lastDig]||'';
if(num>=10) {
var decNum = (num - lastDig)/10;
if(decNum>9)decNum%=10;
var ourNumb2 = RomNumDig[1][decNum-1]||'';}
if(num>=100) {
var hundNum = ((num-num%100)/100);
var ourNumb3 = RomNumDig[2][hundNum-1]||'';}
return ourNumb3+ourNumb2+ourNumb1;
}
console.log(convert(950));//CML
/*2nd my beginner-nooby solution for numbers 1-10, but it can be easy transformed for larger numbers :)*/
function convert(num) {
var ourNumb = '';
var romNumDig = ['I','IV','V','IX','X'];
var decNum = [1,4,5,9,10];
for (var i=decNum.length-1; i>0; i--) {
while(num>=decNum[i]) {
ourNumb += romNumDig[i];
num -= decNum[i];
}
}
return ourNumb;
}
console.log(convert(9));//IX

function toRoman(n) {
var decimals = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
var roman = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
for (var i = 0; i < decimals.length; i++) {
if(n < 1)
return "";
if(n >= decimals[i]) {
return roman[i] + toRoman(n - decimals[i]);
}
}
}

This works for all numbers only in need of roman numerals M and below.
function convert(num) {
var code = [
[1000, "M"], [900, "CM"], [800, "DCCC"], [700, "DCC"], [600, "DC"],
[500, "D"], [400, "CD"], [300, "CCC"], [200, "CC"],
[100, "C"], [90, "XC"], [80, "LXXX"], [70, "LXX"], [60, "LX"],
[50, "L"], [40, "XL"], [30, "XXX"], [20, "XX"],
[10, "X"], [9, "IX"], [8, "VIII"], [7, "VII"], [6, "VI"],
[5, "V"], [4, "IV"], [3, "III"], [2, "II"], [1, "I"],
];
var rom = "";
for(var i=0; i<code.length; i++) {
while(num >= code[i][0]) {
rom += code[i][1];
num -= code[i][0];
}
}
return rom;
}

This is the first time I really got stuck on freecodecamp. I perused through some solutions here and was amazed at how different they all were. Here is what ended up working for me.
function convertToRoman(num) {
var roman = "";
var lookupObj = {
1000:"M",
900:"CM",
500:"D",
400:"CD",
100:"C",
90:"XC",
50:"L",
40:"XL",
10:"X",
9:"IX",
4:"IV",
5:"V",
1:"I",
};
var arrayLen = Object.keys(lookupObj).length;
while(num>0){
for (i=arrayLen-1 ; i>=0 ; i--){
if(num >= Object.keys(lookupObj)[i]){
roman = roman + lookupObj[Object.keys(lookupObj)[i]];
num = num - Object.keys(lookupObj)[i];
break;
}
}
}
return roman;
}
convertToRoman(1231);

function convertToRoman(num) {
var romNumerals = [["M", 1000], ["CM", 900], ["D", 500], ["CD", 400], ["C", 100], ["XC", 90], ["L", 50], ["XL", 40], ["X", 10], ["IX", 9], ["V", 5], ["IV", 4], ["I", 1]];
var runningTotal = 0;
var roman = "";
for (var i = 0; i < romNumerals.length; i++) {
while (runningTotal + romNumerals[i][1] <= num) {
runningTotal += romNumerals[i][1];
roman += romNumerals[i][0];
}
}
return roman;
}

Convert roman number to arabic using javascript

Hi I am trying to convert roman numerals to arabic using javascript. I wrote a code but it is failing.
The rules I am trying to follow are :
if Larger number is before smaller number then addition and if smaller number is before larger number then subtraction.
Along with that I have few other rules as well like 'D','L' and 'V' can't be repeated at all and 'M' can be repeated only twice (Not sure how to implement this, can I use regex for it and how?)
Code :
function romanToArabic(roman){
if(roman == null)
return -1;
var value;
for(var i=0;i<roman.length;i++){
current = char_to_int(roman.charAt(i));
next = char_to_int(roman.charAt(i+1));
console.log("Current",current);
console.log("Next",next);
if(current >= next){
value = current + next;
console.log(value);
}
else {
console.log(value);
value = next - current;
}
}
return value;
}
function char_to_int(character) {
switch(character){
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return -1;
}
}
console.log(romanToArabic('IIX'));
Can somebody help? Would appreciate it!
Added screenshots :

To those, who might need to translate conventional roman numbers as opposed to irregular subtractive notation (e.g. 'IIX' instead of 'VIII' for 8), I might suggest my own, slightly shorter method:
const test = ['XIV'/*14*/, 'MXMVI'/*1996*/, 'CII'/*102*/, 'CDI'/*401*/];
const roman2arabic = s => {
const map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000};
return [...s].reduce((r,c,i,s) => map[s[i+1]] > map[c] ? r-map[c] : r+map[c], 0);
};
console.log(test.map(roman2arabic));
.as-console-wrapper {min-height: 100%}
Though, it can be modified to follow unconventional logic:
const test = ['IIV'/*3*/,'XXMMII'/*1982*/, 'IIIXV'/*12*/, 'XII'/*conventional 12*/];
const roman2arabic = s => {
const map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000};
return [...s]
.reduceRight(({sum,order},c,i,s) =>
Object.keys(map).indexOf(c) < order ?
{sum: sum-map[c], order} :
{sum: sum+map[c], order: Object.keys(map).indexOf(c)},
{sum:0,order:Object.keys(map).indexOf(s[s.length-1])})
.sum;
};
console.log(test.map(roman2arabic));
.as-console-wrapper {min-height: 100%}

The issue is that your code only subtracts the value corresponding to one character, while in IIX you need to subtract twice (although that kind of representation for the number 8 is quite unconventional -- 8 would normally be represented as VIII).
The solution is to keep collecting a separate sum for when the symbol is the same, so that after reading the first two "I", you have two separate sums:
total: 2
value of all "I": 2
Then when you encounter the "X" and detect that a subtraction is needed, you first undo the addition already done for the grand total, and then perform the subtraction with the value you collected for the "I":
total: -2
After this, you start with a reset value for "X":
total: 10 + -2 = 8
value for all "X": 10
Here is your code adapted for that to happen:
function romanToArabic(roman){
if(roman == null)
return -1;
var totalValue = 0,
value = 0, // Initialise!
prev = 0;
for(var i=0;i<roman.length;i++){
var current = char_to_int(roman.charAt(i));
if (current > prev) {
// Undo the addition that was done, turn it into subtraction
totalValue -= 2 * value;
}
if (current !== prev) { // Different symbol?
value = 0; // reset the sum for the new symbol
}
value += current; // keep adding same symbols
totalValue += current;
prev = current;
}
return totalValue;
}
function char_to_int(character) {
switch(character){
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return -1;
}
}
console.log(romanToArabic('IIX'));
As for your additional question to limit the number of consecutive "I" to at most two, "D" at most one, ... you could use a regular expression test at the start of your function:
if (/III|XXX|CCC|MMM|VV|LL|DD|[^IVXLCDM]/.test(roman))
return -1;
You can just append other invalid sub-sequences separated by |. For instance, if you would not want an "I" to appear directly in front of "L", "C", "D" or "M", then extend to:
if (/III|XXX|CCC|MMM|VV|LL|DD|[^IVXLCDM]|I[LCDM]/.test(roman))
return -1;

const romans = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
};
// MXMIV
function roman2arabic(nums){
let sum = 0;
const numsArr = nums.split('');
const isSimpleRoman = (num) => num in romans;
const arabicCompare = (current, prev) => romans[current] < romans[prev];
const orderNums = (acc, current) => {
const prev = acc[acc.length - 1] || null;
const arabCurrent = romans[current];
if (prev && isSimpleRoman(prev) && arabicCompare(current, prev)) {
sum -= arabCurrent;
acc.pop() && acc.push(current + prev);
} else {
sum += arabCurrent;
acc.push(current);
}
return acc;
};
return numsArr.reduceRight(orderNums, []) && sum;
}

const romans = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
};
function roman2arabicRecursion(nums){
const numsArr = nums.split('');
const recursion = (arr, index, sum) => {
const current = arr[index];
const prev = arr[index + 1] || null;
if(prev && romans[current] < romans[prev]){
sum -= romans[current];
} else {
sum += romans[current];
}
if(index === 0) return sum;
return recursion(arr, index - 1, sum);
}
return recursion(numsArr, numsArr.length - 1, 0);
};

Alternatively || operator could be used isntead of ??
const toArabic = (romanNumber) => {
const map = {
M: 1000,
D: 500,
C: 100,
L: 50,
X: 10,
V: 5,
I: 1,
};
const nums = romanNumber.split('');
let result = 0;
for (let i = 0; i < nums.length; i += 1) {
const first = map[nums[i]];
const second = map[nums[i + 1]] ?? 0;
if (first < second) {
result += second - first;
i += 1;
} else {
result += first;
}
}
return result;
};
console.log(toArabic('CMXI')); // 911
console.log(toArabic('MXXIV')); // 1024

We are so used to reading from Left to Right, we overlook the Right to Left alternatives.
The whole point of Roman notation is you want to check if V becomes before X
That is much easier when you reverse the Roman string.
Or in JavaScript, use the hardly ever used reduceRight method
(code optimized for better GZIP/Brotli compression)
const romanToArabic = (input) => [...input].reduceRight((
acc,
letter,
idx,
arr,
value = {m:1000, d:500, c:100, l:50, x:10, v:5, i:1}[letter.toLowerCase()],
doubleSubtraction = letter == arr[idx + 1] // ignore IIX notation
) => {
if (value < acc.high && !doubleSubtraction)
acc.Arabic -= value;
else
acc.Arabic += acc.high = value;
//console.log(idx, letter, acc, 'value:', value, acc.high, arr[idx + 1]);
return acc;
}, { high:0, Arabic:0 }).Arabic; // return Arabic value
//TESTS
Object.entries({
"cxxiv": 124,
"ix": 9,
"iix": 10,
"xL": 40,
"MMMDXLIX": 3549,
"MMMMCMXCIX": 4999}
).map(([roman,value])=>{
let converted = romanToArabic(roman);
console.log(roman, "=", converted);
console.assert(converted == value, "wrong conversion,", roman, "must be", value)
})

I resolved this exercise like this:
function arabic(num) {
let ch;
let sum = 0;
for (let i = 0; i < num.length; i++) {
ch = num[i];
switch (ch) {
case 'I':
if (num[i + 1] === 'V' || num[i + 1] === 'X') {
continue;
}
sum = sum + 1;
break;
case 'V':
if (num[i - 1] === 'I') {
sum = sum + 4;
break;
}
sum = sum + 5;
break;
case 'X':
if (num[i - 1] === 'I') {
sum = sum + 9;
break;
}
if (num[i + 1] === 'C') {
continue;
}
sum = sum + 10;
break;
case 'L':
sum = sum + 50;
break;
case 'C':
if (num[i + 1] === 'D' || num[i + 1] === 'M') {
continue;
}
if (num[i - 1] === 'X') {
sum = sum + 90;
break;
}
sum = sum + 100;
break;
case 'D':
if (num[i - 1] === 'C') {
sum = sum + 400;
break;
}
sum = sum + 500;
break;
case 'M':
if (num[i - 1] === 'C') {
sum = sum + 900;
break;
}
sum = sum + 1000;
break;
}
}
return sum;
}

Converting to Roman Numerals in Javascript - Weird bug

function convertToRoman(num) {
var thisMap = {
1:[1],
2:[1, 1],
3:[1, 1, 1],
4:[1, 5],
5:[5],
6:[5, 1],
7:[5, 1, 1],
8:[5, 1, 1, 1],
9:[1, 10],
0:[0]
};
var numMap = {
1000:"M",
500:"D",
100:"C",
50:"L",
10:"X",
5:"V",
1:"I"
};
numArr = num.toString().split("");
var thisIndex = 1;
var tallyArr = [];
for (var i = numArr.length - 1; i >= 0; i--) {
tallyArr.unshift(thisMap[numArr[i]]);
}
thisIndex = Math.pow(10, tallyArr.length - 1);
checkArr = [];
<<<BUG HERE>>>
for (var x = 0; x < tallyArr.length; x++) {
for (var y = 0; y < tallyArr[x].length; y++) {
tallyArr[x][y] *= thisIndex;
}
thisIndex = thisIndex / 10;
}
<<</BUG HERE>>>
var finalArr = [];
for (var a = 0; a < tallyArr.length; a++) {
for (var b = 0; b < tallyArr[a].length; b++) {
finalArr.push(numMap[tallyArr[a][b]]);
}
}
finalAnswer = finalArr.join("");
return finalAnswer;
}
convertToRoman(88);
So this is my function for converting a number into a Roman Numeral in Javascript. It basically formats every number into the right format using thisMap, then uses thisIndex to multiply by either 1000, 100 or 10, and then compares to numMap to get the correct Roman Numeral.
It seems to work in most of the test cases, except with 44, 99, or 3999.
In these cases, it seems to multiply the numbers by the wrong amount, so 44 becomes XLXL, when it should be XLIV.
I think the bug is between the <<>> tags I've inserted, because that is where the numbers seem to be multiplied wrong.
However, I can't spot the problem.
Thanks.

I tried a different approach.
function convertToRoman(num) {
let arabicArray = [ 1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000, 4000, 5000, 5001]
let romanArray = ['I', 'IV', 'V', 'IX', 'X', 'XL', 'L', 'XC', 'C', 'CD', 'D', 'CM', 'M', 'MV', 'V', 'limit of 5000']
let roman = ""
loop()
function loop() {
for (let i = 0; i < arabicArray.length; i++) {
if (num < arabicArray[i]) {
roman += romanArray[i - 1]
num -= arabicArray[i - 1]
while (num != 0) {loop()} break;
}
}
}
return roman
}
console.log(convertToRoman(3))
However this gives you a limit to 5000.

Try this: x loop should run through all the length of tallyArr except the last one.
function convertToRoman(num) {
// ... code ...
for (var x = 0; x < tallyArr.length - 1; x++) {
for (var y = 0; y < tallyArr[x].length; y++) {
tallyArr[x][y] *= thisIndex;
}
thisIndex = thisIndex / 10;
}
// ... more code ...
}

Your solution seems a bit over engineered and overly complicated sometimes simpler is better and what might seem like a clever answer and looking for the overly eloquent solution can trip you up.
function convertToRoman(num) {
var output = "";
var numMap = [
{ limit: 1000, value: "M" },
{ limit: 900, value: "CM" },
{ limit: 500, value: "D" },
{ limit: 400, value: "CD" },
{ limit: 100, value: "C" },
{ limit: 90, value: "XC" },
{ limit: 50, value: "L" },
{ limit: 40, value: "XL" },
{ limit: 10, value: "X" },
{ limit: 9, value: "IX" },
{ limit: 5, value: "V" },
{ limit: 4, value: "IV" },
{ limit: 1, value: "I" }
];
for(var index = 0; index < numMap.length; index++) {
var value = numMap[index].value,
limit = numMap[index].limit;
while(num >= limit) {
output += value;
num -= limit;
}
}
return output;
}
alert(convertToRoman(1));
alert(convertToRoman(4));
alert(convertToRoman(5));
alert(convertToRoman(9));
alert(convertToRoman(10));
alert(convertToRoman(88));
alert(convertToRoman(2016));
JSFiddle

var romanNumber = [
[1, 'I'], [2, 'II'], [3, 'III'],[4, 'IV'],
[5, 'V'], [6, 'VI'],[7, 'VII'], [8, 'VIII'],
[9, 'IX'],[10, 'X']
];
function convertToRoman(number) {
if (number === 0) {
return '';
}
for (var i = 0; i < romanNumber.length; i++) {
if (number === romanNumber[i][0]) {
return romanNumber[i][1];
}
}
}
console.log(convertToRoman(1));
console.log(convertToRoman(2));
console.log(convertToRoman(3));
console.log(convertToRoman(4));
console.log(convertToRoman(5));
console.log(convertToRoman(6));
console.log(convertToRoman(7));
console.log(convertToRoman(8));
console.log(convertToRoman(9));
console.log(convertToRoman(10));

Algorithm to convert Roman Numerals with Javascript [duplicate]

This question already has answers here:
Convert a number into a Roman numeral in JavaScript
(93 answers)
Closed 6 years ago.
I want to implement Algorithm which converts Roman numeral to Arabic with Javascript. Using the following suggested methods.
Array.prototype.splice()
Array.prototype.indexOf()
Array.prototype.join()
I have already found the algorithm which solves this task
function convertToRoman(num) {
var numeric = [ 5000,4000,1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ];
var roman = [ 'V\u0305','I\u0305V\u0305','M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I' ];
var output = '', i, len = numeric.length;
for (i = 0; i < len; i++) {
while (numeric[i] <= num) {
output += roman[i];
num -= numeric[i];
}
}
return output;
}
convertToRoman(4999);
Nevertheless i'm curious how to implement an algorithm with above mentioned methods.
Thanks, please do not judge me harshly, i'm a beginner programmer.

I think the question has already been answered:
https://stackoverflow.com/a/9083857/4269495
Number to Roman Numeral:
Number.prototype.toRoman= function () {
var num = Math.floor(this),
val, s= '', i= 0,
v = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
r = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
function toBigRoman(n) {
var ret = '', n1 = '', rem = n;
while (rem > 1000) {
var prefix = '', suffix = '', n = rem, s = '' + rem, magnitude = 1;
while (n > 1000) {
n /= 1000;
magnitude *= 1000;
prefix += '(';
suffix += ')';
}
n1 = Math.floor(n);
rem = s - (n1 * magnitude);
ret += prefix + n1.toRoman() + suffix;
}
return ret + rem.toRoman();
}
if (this - num || num < 1) num = 0;
if (num > 3999) return toBigRoman(num);
while (num) {
val = v[i];
while (num >= val) {
num -= val;
s += r[i];
}
++i;
}
return s;
};

Linear arrangement algorithm

I am not sure if title is correct.
I have few labels which have set their positions in y scale in range:
range = [0, 100px]
for example: 5 labels in positions:
positions = [5px, 6px, 8px, 72px, 76px]
Now I want my algorithm to correct these positions to not let them be closer than 10px to each other and do minimal corrections.
I am expecting calling my function like this:
result = calculateNewPositions(range, positions, min(10px, 100px / positions.length))
and result in this case should be:
[0px, 10px, 20px, 69px, 79px]
What is name of this alghoritm or how to implement that?

Here's an algorithm that should work pretty well for most cases, and tries to make as minimal amount of adjustments as necessary from the original values.
Iterate through each pair of elements.
If the space is not large enough, move them apart from each other by 1, making sure not to violate the range.
Repeat until all the elements have enough space between them.
And here is a sample implementation:
function calculateNewPositions(positions, minSpacing, rangeMin, rangeMax) {
var temp = positions.slice(0);
var madeChange;
do {
madeChange = false;
for (var i = 0; i < temp.length - 1; i++)
if (temp[i + 1] - temp[i] < minSpacing) {
if (temp[i] > rangeMin) { temp[i]--; madeChange = true; }
if (temp[i + 1] < rangeMax) { temp[i + 1]++; madeChange = true; }
}
} while (madeChange);
return temp;
}
Demo: https://jsfiddle.net/aaxmuw2t/
Example Result: [0, 10, 20, 69, 79]
Note that this algorithm is very simplistic and may not always yield the best result for really complex arrays with lots of close numbers. For example, if you input [33, 34, 35, 36], you get [19, 29, 40, 50], which has an extra unnecessary space.

calculateNewPositions = function(positions, minDelta) {
var newPositions = [0]
positions.slice(1).forEach(function(pos, index) {
var delta = positions[index + 1] - positions[index]
newPositions.push(newPositions[index] + Math.max(delta, minDelta))
})
return newPositions
}
https://tonicdev.com/lipp/pos-diff

I finally did something like this:
var fixPositions = function(range, pos, delta, strict) {
var i;
var leftSpaces = [];
var halfDelta = strict ? delta / 2 : 0;
delta = Math.min(delta, (range[1] - range[0] / (pos.length + (strict ? 0 : 1))));
// calculate all spaces that are greater than delta
leftSpaces.push(Math.max(pos[0] - range[0] - halfDelta, 0));
for (i = 1; i < pos.length; i++) {
leftSpaces.push(Math.max(pos[i] - pos[i-1] - delta, 0));
}
leftSpaces.push(Math.max(range[1] - pos[pos.length-1] - halfDelta, 0));
// save indexes of big spaces
var nonZeroSpacesIdx = [];
leftSpaces.map(function(space, i) {
if (space > 0) {
nonZeroSpacesIdx.push(i);
}
});
// sort indexes by spaces sizes (start from smaller)
nonZeroSpacesIdx.sort(function(a, b) {
return leftSpaces[a] - leftSpaces[b];
});
// loop until spaces sum are greater than range
var spacesSum = Infinity;
while (nonZeroSpacesIdx.length > 0 && spacesSum > 0) {
spacesSum = 0;
for (i = 0; i < nonZeroSpacesIdx.length; i++) {
spacesSum += leftSpaces[nonZeroSpacesIdx[i]];
}
var missingDiff = (spacesSum + (pos.length - 1) * delta + halfDelta * 2) - (range[1] - range[0]);
if (missingDiff <= 0) {
break;
}
// find min diff which can be substracted from all spaces
var minDiff = Math.min(missingDiff / nonZeroSpacesIdx.length, leftSpaces[nonZeroSpacesIdx[0]]);
for (i = 0; i < nonZeroSpacesIdx.length; i++) {
leftSpaces[nonZeroSpacesIdx[i]] -= minDiff;
}
// remove index of first space if its equal zero
if (leftSpaces[nonZeroSpacesIdx[0]] <= 0) {
nonZeroSpacesIdx.shift();
}
}
// reconstruct new positions
var newPos = [];
newPos.push(range[0] + leftSpaces[0] + halfDelta);
for (i = 1; i < leftSpaces.length - 1; i++) {
newPos[i] = newPos[i-1] + leftSpaces[i] + delta;
}
return newPos;
};
// result should be from range: [5, 95]
console.log(fixPositions([0, 100], [5, 6, 8, 72, 76], 10, true));
// result should be from range: [0, 100]
console.log(fixPositions([0, 100], [5, 6, 8, 72, 76], 10, false));
https://jsfiddle.net/fcwu1oyu/14/
Its not giving exact same values for my input, but it does the job for my pie charts:

A work in progress solution
The code pushes two too close couples appart with one on each side. This is symetrically and results in sometimes to far pushed values, which can be corrected.
function disperse(values, threshold, range) {
var delta = Array.apply(null, { length: values.length }).map(function () { return 0; }),
converged = false;
while (!converged) {
converged = true;
delta = delta.map(function (d, i, dd) {
if (i < dd.length - 1 && dd.length > 1 && values[i + 1] + dd[i + 1] - values[i] - d < threshold) {
converged = false;
dd[i + 1] += 1;
return d - 1;
}
return d;
});
}
converged = false;
// try to minimise difference
while (!converged) {
converged = true;
delta = delta.map(function (d, i) {
var space;
if (i < delta.length - 2) {
space = values[i + 1] + delta[i + 1] - values[i] - d;
if (d < 0 && space > threshold) {
converged = false;
return d + space - threshold;
}
}
return d;
});
}
// respect lower range
delta.reduce(function (r, d, i, dd) {
if (values[i] + d < r) {
dd[i] = r - values[i];
return r + threshold;
}
return values[i] + threshold + d;
}, range[0]);
// respect upper range
delta.reduceRight(function (r, d, i, dd) {
if (values[i] + d > r) {
dd[i] = r - values[i];
return r - threshold;
}
return values[i] + d;
}, range[1]);
return values.map(function (v, i) {
return v + delta[i];
});
}
document.write('<pre>' + JSON.stringify(disperse([5, 6, 8, 72, 76], 10, [0, 100]), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(disperse([5, 6, 7, 8, 72, 76], 10, [0, 100]), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(disperse([24, 28, 92, 94, 95], 10, [0, 100]), 0, 4) + '</pre>');