Ant is a freelance Java architect and developer. He has been writing a blog and white papers since 2004 and writes about anything he finds interesting, related to Java or software. Most recently he has been working on enterprise systems involving Eclipse RCP, Google GWT, Hibernate, Spring and J2ME. He believes very strongly in being involved in all parts of the software life cycle. Ant is a DZone MVB and is not an employee of DZone and has posted 27 posts at DZone. You can read more from them at their website. View Full User Profile

Base X Encoding

  • submit to reddit

Ever needed to shorten a number so that its easier to remember? Or provide someone with a temporary PIN which is short enough to remember, but long enough to pretty much ensure it wont be randomly guessed by someone else? Converting a binary number into a hexadecimal is exactly the process used in such cases. But hexadecimal only has 16 characters in its "dictionary". Base64 is the next step up, with a bigger dictionary containing all alphanumerics (upper and lower case) as well as "/" and "+".

I need a solution which didn't contain certain characters. For example, its easy to mix up an O with a 0. Or an I,l and a 1. I wanted a solution whereby I could encode a number, but using my own definition of the dictionary. So I built just such a solution. You can see the source code below. It contains a main method which runs a simple test, the output of which is:

    Original: 123456789012345678901234567890
    encoded: 2aYls9bkamJJSwhr0
    decoded: 123456789012345678901234567890
    Passed! decoded value is the same as the original.

As you can see, the encoded version is only half as long as the input. Using an 89 character dictionary, it gets even shorter:

    encoded: "9Kgbz])M.w8KgK

The implementation uses the BigInteger class from Java, so you can encode REALLY big numbers. My phone number is now only 5 characters long and really easy to remember:


* Copyright (c) 2010 Ant Kutschera, maxant
* The code below is free software: you can redistribute it and/or modify
* it under the terms of the Lesser GNU General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
* The code in this file is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* Lesser GNU General Public License for more details.
* You should have received a copy of the Lesser GNU General Public License
* along with Foobar. If not, see


import java.math.BigInteger;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

* allows you to convert a whole number into a compacted representation of that number,
* based upon the dictionary you provide. very similar to base64 encoding, or indeed hex
* encoding.
public class BaseX {

* contains hexadecimals 0-F only.
public static final char[] DICTIONARY_16 =
new char[]{'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};

* contains only alphanumerics, in capitals and excludes letters/numbers which can be confused,
* eg. 0 and O or L and I and 1.
public static final char[] DICTIONARY_32 =
new char[]{'1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','J','K','M','N','P','Q','R','S','T','U','V','W','X','Y','Z'};

* contains only alphanumerics, including both capitals and smalls.
public static final char[] DICTIONARY_62 =
new char[]{'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

* contains alphanumerics, including both capitals and smalls, and the following special chars:
* +"@*#%&/|()=?'~[!]{}-_:.,; (you might not be able to read all those using a browser!
public static final char[] DICTIONARY_89 =
new char[]{'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','+','"','@','*','#','%','&','/','|','(',')','=','?','~','[',']','{','}','$','-','_','.',':',',',';','<','>'};

protected char[] dictionary;

* create an encoder with the given dictionary.
* @param dictionary the dictionary to use when encoding and decoding.
public BaseX(char[] dictionary){
this.dictionary = dictionary;

* creates an encoder with the {@link #DICTIONARY_62} dictionary.
* @param dictionary the dictionary to use when encoding and decoding.
public BaseX(){
this.dictionary = DICTIONARY_62;

* tester method.
public static void main(String[] args) {
String original = "123456789012345678901234567890";
System.out.println("Original: " + original);
BaseX bx = new BaseX(DICTIONARY_62);
String encoded = bx.encode(new BigInteger(original));
System.out.println("encoded: " + encoded);
BigInteger decoded = bx.decode(encoded);
System.out.println("decoded: " + decoded);
System.out.println("Passed! decoded value is the same as the original.");
System.err.println("FAILED! decoded value is NOT the same as the original!!");

* encodes the given string into the base of the dictionary provided in the constructor.
* @param value the number to encode.
* @return the encoded string.
public String encode(BigInteger value) {

List<Character> result = new ArrayList<Character>();
BigInteger base = new BigInteger("" + dictionary.length);
int exponent = 1;
BigInteger remaining = value;
BigInteger a = base.pow(exponent); //16^1 = 16
BigInteger b = remaining.mod(a); //119 % 16 = 7 | 112 % 256 = 112
BigInteger c = base.pow(exponent - 1);
BigInteger d = b.divide(c);

//if d > dictionary.length, we have a problem. but BigInteger doesnt have
//a greater than method :-( hope for the best. theoretically, d is always
//an index of the dictionary!
remaining = remaining.subtract(b); //119 - 7 = 112 | 112 - 112 = 0



//need to reverse it, since the start of the list contains the least significant values
StringBuffer sb = new StringBuffer();
for(int i = result.size()-1; i >= 0; i--){
return sb.toString();

* decodes the given string from the base of the dictionary provided in the constructor.
* @param str the string to decode.
* @return the decoded number.
public BigInteger decode(String str) {

//reverse it, coz its already reversed!
char[] chars = new char[str.length()];
str.getChars(0, str.length(), chars, 0);

char[] chars2 = new char[str.length()];
int i = chars2.length -1;
for(char c : chars){
chars2[i--] = c;

//for efficiency, make a map
Map<Character, BigInteger> dictMap = new HashMap<Character, BigInteger>();
int j = 0;
for(char c : dictionary){
dictMap.put(c, new BigInteger("" + j++));

BigInteger bi = BigInteger.ZERO;
BigInteger base = new BigInteger("" + dictionary.length);
int exponent = 0;
for(char c : chars2){
BigInteger a = dictMap.get(c);
BigInteger b = base.pow(exponent).multiply(a);
bi = bi.add(new BigInteger("" + b));

return bi;





Published at DZone with permission of Ant Kutschera, author and DZone MVB.

(Note: Opinions expressed in this article and its replies are the opinions of their respective authors and not those of DZone, Inc.)



Dain Nilsson replied on Tue, 2010/02/23 - 3:20pm

Here's a slightly improved version:

import java.math.BigInteger;
import java.util.Map;
import java.util.HashMap;

public class BaseX {
	private final char[] dictionary;
	private final Map charMap = new HashMap();
	private final BigInteger base;

	public BaseX(char[] dictionary) {
		this.dictionary = dictionary;
		int i = 0;
		for (char c : dictionary)
			charMap.put(c, i++);
		base = BigInteger.valueOf(i);

	public String encode(BigInteger value) {
		StringBuilder s = new StringBuilder();
		BigInteger[] parts = { value, BigInteger.ZERO };
		while (parts[0].compareTo(base) >= 0) {
			parts = parts[0].divideAndRemainder(base);

		return s.reverse().toString();

	public BigInteger decode(String encoded) {
		BigInteger result = BigInteger.ZERO;
		int len = encoded.length();
		//Sum up each "digit" multiplied by the base raised to its position.
		for (int i = 0; i < len; i++) {
			BigInteger digit = BigInteger.valueOf(charMap
			result = result.add(digit.multiply(base.pow((len - i) - 1)));

		return result;

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.