Java Palindrome always returns true - for-loop

I tried to create a palindrome java program with JOptionPane by using for loop, but it ends up returning true all the time no matter the input is really a palindrome or not. Can guys please help if you guys know what's wrong with the code below, thanks.
public class program {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
JOptionPane.showMessageDialog(null, "Welcome to The Palindrome!", "Hello", JOptionPane.INFORMATION_MESSAGE);
String str = JOptionPane.showInputDialog("Please input a string");
int len = str.length();
int j = len - 1;
int i = 0;
boolean result;
for(i = 0; i <= (len - 1)/2; i++);
{
if(str.charAt(i) != str.charAt(j))
result = false;
j--;
}
if(result = true)
JOptionPane.showMessageDialog(null, str + " is a palindrome.", "ByeBye", JOptionPane.INFORMATION_MESSAGE);
if(result = false)
JOptionPane.showMessageDialog(null, str + " is not a palindrome.", "ByeBye", JOptionPane.INFORMATION_MESSAGE);
}

Instead of using traditional way to check palindrome, just use the smart way. Here you go
boolean result = str.equalsIgnoreCase(new StringBuffer(str).reverse().toString());

When you check for the value of result, you are using =, which assigns that value to the result variable, and always evaluates to true.
To correct your code, you can either remove the equal sign, or use result == true (usually you use the former, as it is more concise).
However, this may result in an error, as you are not initialising the value of result. I recommend setting it's value to true as the default value.

Related

Filtering return on serial port

I have a CO2 sensor on my Arduino Mega and sometimes randomly when I'm reading the CO2 measurement, the sensor will return a "?". The question mark causes my program to crash and return "input string was not in a correct format".
I haven't tried anything because I don't know what approach would be the best for this. The CO2 sensor returns the measurement in the form of "Z 00000" but when this question mark appears it shows that all that returned was a "\n". Currently, I have the program just reading the 5 digits after the Z.
if (returnString != "")
{
val = Convert.ToDouble(returnString.Substring(returnString.LastIndexOf('Z')+ 1));
}
What I expect to return is the digits after Z which works but every so often I will get a random line return which crashes everything.
According to the C# documentation the ToDouble method throws FormatException whenever the input string is invalid. You should catch the exception to avoid further issues.
try {
val = Convert.ToDouble(returnString.Substring(returnString.LastIndexOf('Z')+ 1));
}
catch(FormatException e) {
//If you want to do anything in case of an error
//Otherwise you can leave it blank
}
Also I'd recommend using some sort of statemachine for parsing the data in your case, that could discard all invalid characters. Something like this:
bool z_received = false;
int digits = 0;
int value = 0;
//Called whenever you receive a byte from the serial port
void onCharacter(char input) {
if(input == 'Z') {
z_received = true;
}
else if(z_received && input <= '9' && input >= '0') {
value *= 10;
value += (input - '0');
digits++;
if(digits == 5) {
onData(value);
value = 0;
z_received = false;
digits = 0;
}
}
else {
value = 0;
z_received = false;
digits = 0;
}
}
void onData(int data) {
//do something with the data
}
This is just a mock-up, should work in your case if you can direct the COM port's byte stream into the onCharacter function.

Implementing tokenize function with CString

For the sake of learning, I'm trying to implement my own simple Tokenize function with CStrings. I currently have this file:
11111
22222
(ENDWPT)
222222
333333
(ENDWPT)
6060606
ggggggg
hhhhhhh
(ENDWPT)
iiiiiii
jjjjjjj
kkkkkkk
lllllll
mmmmmmm
nnnnnnn
Which I would like to be tokenized with the delimiter (ENDWPT).
I coded the following function, which attempts to find the delimiter position, then add the delimiter length and extract the text to this position. After that, update a counter that is used so that the next time the function is called it begins searching for the delimiter from the previous index. The function looks like this:
bool MyTokenize(CString strText, CString& strOut, int& iCount)
{
CString strDelimiter = L"(ENDWPT)";
int iIndex = strText.Find(strDelimiter, iCount);
if (iIndex != -1)
{
iIndex += strDelimiter.GetLength();
strOut = strText.Mid(iCount, iIndex);
iCount = iIndex;
return true;
}
return false;
}
And is being called like so:
int nCount = 0;
while ((MyTokenize(strText, strToken, nCount)) == true)
{
// Handle tokenized strings here
}
Right now, the function is splitting the strings in the wrong way, I think it is because Find() may be returning the wrong index. I think it should be returning 12, but it is actually returning 14??.
I ran out of ideas, if anyone can figure this out I would really appreciate it.
If delimiter is found (iIndex) then read iIndex - iCount count, starting from (iCount). Then modify iCount
if(iIndex != -1)
{
strOut = strText.Mid(iCount, iIndex - iCount);
iCount = iIndex + strDelimiter.GetLength();
return true;
}
The source string may not end with delimiter, it needs a special case for that.
You can also pick better names to match the usage for CString::Mid(int nFirst, int nCount) to make it easier to understand. MFC uses camelCase coding style, with type identifiers in front of variables, which is unnecessary in C++, I'll avoid it in this example:
bool MyTokenize(CString &source, CString& token, int& first)
{
CString delimeter = L"(ENDWPT)";
int end = source.Find(delimeter, first);
if(end != -1)
{
int count = end - first;
token = source.Mid(first, count);
first = end + delimeter.GetLength();
return true;
}
else
{
int count = source.GetLength() - first;
if(count <= 0)
return false;
token = source.Mid(first, count);
first = source.GetLength();
return true;
}
}
...
int first = 0;
CString source = ...
CString token;
while(MyTokenize(source, token, first))
{
// Handle tokenized strings here
}

Methods, more than one return?

I have the following method:
From what I learned methods which are not voids need a return. For the following examples I can see two returns, once after if(), and one at the end.
For this example if String s is not a digit it will return the boolean as false. Which makes sense. If it is a digit then it will check whether it is in the interval. I guess I am confused regarding whether we can have multiple returns in such cases and what the limitations are, if there are any. thank you.
private boolean ElementBienFormat(String s) {
for (int i = 0; i < s.length(); i++) {
if (!Character.isDigit(s.charAt(i))) {
return false;
}
}
int n = Integer.valueOf(s);
return (n>=0 && n <=255);
A method will "quit" (return) when control reaches a return. So in this case as soon as a character is not a digit in the input String control will go back to the caller (with the appropriate value).
boolean success = ElementBienFormat( "a" ); // <-- control would go back here with the value of false.
Another quick note is that a void method can have multiple return statements as well
private void Method( int n )
{
if( n < 0 )
return;
//...
//implicit
//return;
}

Simple encryption algorithm for homework. not getting decryption working properly

This is a homework question that I can't get my head around at all
Its a very simple encryption algorithm. You start with a string of characters as your alphabet:
ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!, .
Then ask the user to enter there own string that will act as a map such as:
0987654321! .,POIUYTREWQASDFGHJKLMNBVCXZ
Then the program uses this to make a map and allows you to enter text that gets encrypted.
For example MY NAME IS JOSEPH would be encrypted as .AX,0.6X2YX1PY6O3
This is all very easy, however he said that its a one to one mapping and thus implied that if I enter .AX,0.6X2YX1PY6O3 back into the program I will get out MY NAME IS JOSEPH
This doesn't happen, because .AX,0.6X2YX1PY6O3 becomes Z0QCDZQGAQFOALDH
The mapping only works to decrypt when you go backwards but the question implies that the program just loops and runs the one algorithm every time.
Even if some could say that it is possible I would be happy, I have pages and pages of paper filled up with possible workings, but I came up with nothing, the only solution to run the algorithm backwards back I don't think we are allowed to do that.
Any ideas?
Edit:
Unfortunately I can't get this to work (Using the orbit computation idea) What am I doing wrong?
//import scanner class
import java.util.Scanner;
public class Encryption {
static Scanner inputString = new Scanner(System.in);
//define alphabet
private static String alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!, .";
private static String map;
private static int[] encryptionMap = new int[40];//mapping int array
private static boolean exit = false;
private static boolean valid = true;
public static void main(String[] args) {
String encrypt, userInput;
userInput = new String();
System.out.println("This program takes a large reordered string");
System.out.println("and uses it to encrypt your data");
System.out.println("Please enter a mapping string of 40 length and the same characters as below but in different order:");
System.out.println(alpha);
//getMap();//don't get user input for map, for testing!
map=".ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!, ";//forced input for testing only!
do{
if (valid == true){
System.out.println("Enter Q to quit, otherwise enter a string:");
userInput = getInput();
if (userInput.charAt(0) != 'Q' ){//&& userInput.length()<2){
encrypt = encrypt(userInput);
for (int x=0; x<39; x++){//here I am trying to get the orbit computation going
encrypt = encrypt(encrypt);
}
System.out.println("You entered: "+userInput);
System.out.println("Encrypted Version: "+encrypt);
}else if (userInput.charAt(0) == 'Q'){//&& userInput.length()<2){
exit = true;
}
}
else if (valid == false){
System.out.println("Error, your string for mapping is incorrect");
valid = true;//reset condition to repeat
}
}while(exit == false);
System.out.println("Good bye");
}
static String encrypt(String userInput){
//use mapping array to encypt data
String encrypt;
StringBuffer tmp = new StringBuffer();
char current;
int alphaPosition;
int temp;
//run through the user string
for (int x=0; x<userInput.length(); x++){
//get character
current = userInput.charAt(x);
//get location of current character in alphabet
alphaPosition = alpha.indexOf(current);
//encryptionMap.charAt(alphaPosition)
tmp.append(map.charAt(alphaPosition));
}
encrypt = tmp.toString();
return(encrypt);
}
static void getMap(){
//get a mapping string and validate from the user
map = getInput();
//validate code
if (map.length() != 40){
valid = false;
}
else{
for (int x=0; x<40; x++){
if (map.indexOf(alpha.charAt(x)) == -1){
valid = false;
}
}
}
if (valid == true){
for (int x=0; x<40; x++){
int a = (int)(alpha.charAt(x));
int y = (int)( map.charAt(x));
//create encryption map
encryptionMap[x]=(a-y);
}
}
}
static String getInput(){
//get input(this repeats)
String input = inputString.nextLine();
input = input.toUpperCase();
if ("QUIT".equals(input) || "END".equals(input) || "NO".equals(input) || "N".equals(input)){
StringBuffer tmp = new StringBuffer();
tmp.append('Q');
input = tmp.toString();
}
return(input);
}
}
You will (probably) not get your original string back if you apply that substitution again. I say probably because you can construct such inputs (they all do things like if A->B then B->A). But most inputs won't do that. You would have to construct the reverse map to decrypt.
However, there is a trick you can do if you're only allowed to go forward. Keep applying the mapping and you'll eventually return to your original input. The number of times you'll have to do that depends on your input. To figure out how many times, compute the orbit of each character, and take the least common multiple of all the orbit sizes. For your input the orbits are size 1 (T->T, W->W), 2 (B->9->B H->3->H U->R->U P->O->P), 4 (C->8->N->,->C), 9 (A->...->Y->A), and 17 (E->...->V->E). The LCM of all those is 612, so 611 forward mappings applied to the ciphertext will return you to the plaintext.
Well, you can get your string back this way only if you do reverse mapping. One to one mapping means that a single letter of your default alphabet maps to only one letter of your new alphabet and vice versa. I.e. you can't map ABCD to ABBA. It doesn't imply that you can get your initial string by doing a second round of encryption.
The thing you have described can be achieved if you use a finite alphabet and a displacement to encode your string. You can choose the displacement in such a way that after a number of rounds of encryption totalDisplacement mod alphabetSize == 0 Than you will get your string back going only forward.

How do I create a URL shortener? [closed]

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Closed 1 year ago.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
I want to create a URL shortener service where you can write a long URL into an input field and the service shortens the URL to "http://www.example.org/abcdef".
Instead of "abcdef" there can be any other string with six characters containing a-z, A-Z and 0-9. That makes 56~57 billion possible strings.
My approach:
I have a database table with three columns:
id, integer, auto-increment
long, string, the long URL the user entered
short, string, the shortened URL (or just the six characters)
I would then insert the long URL into the table. Then I would select the auto-increment value for "id" and build a hash of it. This hash should then be inserted as "short". But what sort of hash should I build? Hash algorithms like MD5 create too long strings. I don't use these algorithms, I think. A self-built algorithm will work, too.
My idea:
For "http://www.google.de/" I get the auto-increment id 239472. Then I do the following steps:
short = '';
if divisible by 2, add "a"+the result to short
if divisible by 3, add "b"+the result to short
... until I have divisors for a-z and A-Z.
That could be repeated until the number isn't divisible any more. Do you think this is a good approach? Do you have a better idea?
Due to the ongoing interest in this topic, I've published an efficient solution to GitHub, with implementations for JavaScript, PHP, Python and Java. Add your solutions if you like :)
I would continue your "convert number to string" approach. However, you will realize that your proposed algorithm fails if your ID is a prime and greater than 52.
Theoretical background
You need a Bijective Function f. This is necessary so that you can find a inverse function g('abc') = 123 for your f(123) = 'abc' function. This means:
There must be no x1, x2 (with x1 ≠ x2) that will make f(x1) = f(x2),
and for every y you must be able to find an x so that f(x) = y.
How to convert the ID to a shortened URL
Think of an alphabet we want to use. In your case, that's [a-zA-Z0-9]. It contains 62 letters.
Take an auto-generated, unique numerical key (the auto-incremented id of a MySQL table for example).
For this example, I will use 12510 (125 with a base of 10).
Now you have to convert 12510 to X62 (base 62).
12510 = 2×621 + 1×620 = [2,1]
This requires the use of integer division and modulo. A pseudo-code example:
digits = []
while num > 0
remainder = modulo(num, 62)
digits.push(remainder)
num = divide(num, 62)
digits = digits.reverse
Now map the indices 2 and 1 to your alphabet. This is how your mapping (with an array for example) could look like:
0 → a
1 → b
...
25 → z
...
52 → 0
61 → 9
With 2 → c and 1 → b, you will receive cb62 as the shortened URL.
http://shor.ty/cb
How to resolve a shortened URL to the initial ID
The reverse is even easier. You just do a reverse lookup in your alphabet.
e9a62 will be resolved to "4th, 61st, and 0th letter in the alphabet".
e9a62 = [4,61,0] = 4×622 + 61×621 + 0×620 = 1915810
Now find your database-record with WHERE id = 19158 and do the redirect.
Example implementations (provided by commenters)
C++
Python
Ruby
Haskell
C#
CoffeeScript
Perl
Why would you want to use a hash?
You can just use a simple translation of your auto-increment value to an alphanumeric value. You can do that easily by using some base conversion. Say you character space (A-Z, a-z, 0-9, etc.) has 62 characters, convert the id to a base-40 number and use the characters as the digits.
public class UrlShortener {
private static final String ALPHABET = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
private static final int BASE = ALPHABET.length();
public static String encode(int num) {
StringBuilder sb = new StringBuilder();
while ( num > 0 ) {
sb.append( ALPHABET.charAt( num % BASE ) );
num /= BASE;
}
return sb.reverse().toString();
}
public static int decode(String str) {
int num = 0;
for ( int i = 0; i < str.length(); i++ )
num = num * BASE + ALPHABET.indexOf(str.charAt(i));
return num;
}
}
Not an answer to your question, but I wouldn't use case-sensitive shortened URLs. They are hard to remember, usually unreadable (many fonts render 1 and l, 0 and O and other characters very very similar that they are near impossible to tell the difference) and downright error prone. Try to use lower or upper case only.
Also, try to have a format where you mix the numbers and characters in a predefined form. There are studies that show that people tend to remember one form better than others (think phone numbers, where the numbers are grouped in a specific form). Try something like num-char-char-num-char-char. I know this will lower the combinations, especially if you don't have upper and lower case, but it would be more usable and therefore useful.
My approach: Take the Database ID, then Base36 Encode it. I would NOT use both Upper AND Lowercase letters, because that makes transmitting those URLs over the telephone a nightmare, but you could of course easily extend the function to be a base 62 en/decoder.
Here is my PHP 5 class.
<?php
class Bijective
{
public $dictionary = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
public function __construct()
{
$this->dictionary = str_split($this->dictionary);
}
public function encode($i)
{
if ($i == 0)
return $this->dictionary[0];
$result = '';
$base = count($this->dictionary);
while ($i > 0)
{
$result[] = $this->dictionary[($i % $base)];
$i = floor($i / $base);
}
$result = array_reverse($result);
return join("", $result);
}
public function decode($input)
{
$i = 0;
$base = count($this->dictionary);
$input = str_split($input);
foreach($input as $char)
{
$pos = array_search($char, $this->dictionary);
$i = $i * $base + $pos;
}
return $i;
}
}
A Node.js and MongoDB solution
Since we know the format that MongoDB uses to create a new ObjectId with 12 bytes.
a 4-byte value representing the seconds since the Unix epoch,
a 3-byte machine identifier,
a 2-byte process id
a 3-byte counter (in your machine), starting with a random value.
Example (I choose a random sequence)
a1b2c3d4e5f6g7h8i9j1k2l3
a1b2c3d4 represents the seconds since the Unix epoch,
4e5f6g7 represents machine identifier,
h8i9 represents process id
j1k2l3 represents the counter, starting with a random value.
Since the counter will be unique if we are storing the data in the same machine we can get it with no doubts that it will be duplicate.
So the short URL will be the counter and here is a code snippet assuming that your server is running properly.
const mongoose = require('mongoose');
const Schema = mongoose.Schema;
// Create a schema
const shortUrl = new Schema({
long_url: { type: String, required: true },
short_url: { type: String, required: true, unique: true },
});
const ShortUrl = mongoose.model('ShortUrl', shortUrl);
// The user can request to get a short URL by providing a long URL using a form
app.post('/shorten', function(req ,res){
// Create a new shortUrl */
// The submit form has an input with longURL as its name attribute.
const longUrl = req.body["longURL"];
const newUrl = ShortUrl({
long_url : longUrl,
short_url : "",
});
const shortUrl = newUrl._id.toString().slice(-6);
newUrl.short_url = shortUrl;
console.log(newUrl);
newUrl.save(function(err){
console.log("the new URL is added");
})
});
I keep incrementing an integer sequence per domain in the database and use Hashids to encode the integer into a URL path.
static hashids = Hashids(salt = "my app rocks", minSize = 6)
I ran a script to see how long it takes until it exhausts the character length. For six characters it can do 164,916,224 links and then goes up to seven characters. Bitly uses seven characters. Under five characters looks weird to me.
Hashids can decode the URL path back to a integer but a simpler solution is to use the entire short link sho.rt/ka8ds3 as a primary key.
Here is the full concept:
function addDomain(domain) {
table("domains").insert("domain", domain, "seq", 0)
}
function addURL(domain, longURL) {
seq = table("domains").where("domain = ?", domain).increment("seq")
shortURL = domain + "/" + hashids.encode(seq)
table("links").insert("short", shortURL, "long", longURL)
return shortURL
}
// GET /:hashcode
function handleRequest(req, res) {
shortURL = req.host + "/" + req.param("hashcode")
longURL = table("links").where("short = ?", shortURL).get("long")
res.redirect(301, longURL)
}
C# version:
public class UrlShortener
{
private static String ALPHABET = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
private static int BASE = 62;
public static String encode(int num)
{
StringBuilder sb = new StringBuilder();
while ( num > 0 )
{
sb.Append( ALPHABET[( num % BASE )] );
num /= BASE;
}
StringBuilder builder = new StringBuilder();
for (int i = sb.Length - 1; i >= 0; i--)
{
builder.Append(sb[i]);
}
return builder.ToString();
}
public static int decode(String str)
{
int num = 0;
for ( int i = 0, len = str.Length; i < len; i++ )
{
num = num * BASE + ALPHABET.IndexOf( str[(i)] );
}
return num;
}
}
You could hash the entire URL, but if you just want to shorten the id, do as marcel suggested. I wrote this Python implementation:
https://gist.github.com/778542
Take a look at https://hashids.org/ it is open source and in many languages.
Their page outlines some of the pitfalls of other approaches.
If you don't want re-invent the wheel ... http://lilurl.sourceforge.net/
// simple approach
$original_id = 56789;
$shortened_id = base_convert($original_id, 10, 36);
$un_shortened_id = base_convert($shortened_id, 36, 10);
alphabet = map(chr, range(97,123)+range(65,91)) + map(str,range(0,10))
def lookup(k, a=alphabet):
if type(k) == int:
return a[k]
elif type(k) == str:
return a.index(k)
def encode(i, a=alphabet):
'''Takes an integer and returns it in the given base with mappings for upper/lower case letters and numbers 0-9.'''
try:
i = int(i)
except Exception:
raise TypeError("Input must be an integer.")
def incode(i=i, p=1, a=a):
# Here to protect p.
if i <= 61:
return lookup(i)
else:
pval = pow(62,p)
nval = i/pval
remainder = i % pval
if nval <= 61:
return lookup(nval) + incode(i % pval)
else:
return incode(i, p+1)
return incode()
def decode(s, a=alphabet):
'''Takes a base 62 string in our alphabet and returns it in base10.'''
try:
s = str(s)
except Exception:
raise TypeError("Input must be a string.")
return sum([lookup(i) * pow(62,p) for p,i in enumerate(list(reversed(s)))])a
Here's my version for whomever needs it.
Why not just translate your id to a string? You just need a function that maps a digit between, say, 0 and 61 to a single letter (upper/lower case) or digit. Then apply this to create, say, 4-letter codes, and you've got 14.7 million URLs covered.
Here is a decent URL encoding function for PHP...
// From http://snipplr.com/view/22246/base62-encode--decode/
private function base_encode($val, $base=62, $chars='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ') {
$str = '';
do {
$i = fmod($val, $base);
$str = $chars[$i] . $str;
$val = ($val - $i) / $base;
} while($val > 0);
return $str;
}
Don't know if anyone will find this useful - it is more of a 'hack n slash' method, yet is simple and works nicely if you want only specific chars.
$dictionary = "abcdfghjklmnpqrstvwxyz23456789";
$dictionary = str_split($dictionary);
// Encode
$str_id = '';
$base = count($dictionary);
while($id > 0) {
$rem = $id % $base;
$id = ($id - $rem) / $base;
$str_id .= $dictionary[$rem];
}
// Decode
$id_ar = str_split($str_id);
$id = 0;
for($i = count($id_ar); $i > 0; $i--) {
$id += array_search($id_ar[$i-1], $dictionary) * pow($base, $i - 1);
}
Did you omit O, 0, and i on purpose?
I just created a PHP class based on Ryan's solution.
<?php
$shorty = new App_Shorty();
echo 'ID: ' . 1000;
echo '<br/> Short link: ' . $shorty->encode(1000);
echo '<br/> Decoded Short Link: ' . $shorty->decode($shorty->encode(1000));
/**
* A nice shorting class based on Ryan Charmley's suggestion see the link on Stack Overflow below.
* #author Svetoslav Marinov (Slavi) | http://WebWeb.ca
* #see http://stackoverflow.com/questions/742013/how-to-code-a-url-shortener/10386945#10386945
*/
class App_Shorty {
/**
* Explicitly omitted: i, o, 1, 0 because they are confusing. Also use only lowercase ... as
* dictating this over the phone might be tough.
* #var string
*/
private $dictionary = "abcdfghjklmnpqrstvwxyz23456789";
private $dictionary_array = array();
public function __construct() {
$this->dictionary_array = str_split($this->dictionary);
}
/**
* Gets ID and converts it into a string.
* #param int $id
*/
public function encode($id) {
$str_id = '';
$base = count($this->dictionary_array);
while ($id > 0) {
$rem = $id % $base;
$id = ($id - $rem) / $base;
$str_id .= $this->dictionary_array[$rem];
}
return $str_id;
}
/**
* Converts /abc into an integer ID
* #param string
* #return int $id
*/
public function decode($str_id) {
$id = 0;
$id_ar = str_split($str_id);
$base = count($this->dictionary_array);
for ($i = count($id_ar); $i > 0; $i--) {
$id += array_search($id_ar[$i - 1], $this->dictionary_array) * pow($base, $i - 1);
}
return $id;
}
}
?>
public class TinyUrl {
private final String characterMap = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
private final int charBase = characterMap.length();
public String covertToCharacter(int num){
StringBuilder sb = new StringBuilder();
while (num > 0){
sb.append(characterMap.charAt(num % charBase));
num /= charBase;
}
return sb.reverse().toString();
}
public int covertToInteger(String str){
int num = 0;
for(int i = 0 ; i< str.length(); i++)
num += characterMap.indexOf(str.charAt(i)) * Math.pow(charBase , (str.length() - (i + 1)));
return num;
}
}
class TinyUrlTest{
public static void main(String[] args) {
TinyUrl tinyUrl = new TinyUrl();
int num = 122312215;
String url = tinyUrl.covertToCharacter(num);
System.out.println("Tiny url: " + url);
System.out.println("Id: " + tinyUrl.covertToInteger(url));
}
}
This is what I use:
# Generate a [0-9a-zA-Z] string
ALPHABET = map(str,range(0, 10)) + map(chr, range(97, 123) + range(65, 91))
def encode_id(id_number, alphabet=ALPHABET):
"""Convert an integer to a string."""
if id_number == 0:
return alphabet[0]
alphabet_len = len(alphabet) # Cache
result = ''
while id_number > 0:
id_number, mod = divmod(id_number, alphabet_len)
result = alphabet[mod] + result
return result
def decode_id(id_string, alphabet=ALPHABET):
"""Convert a string to an integer."""
alphabet_len = len(alphabet) # Cache
return sum([alphabet.index(char) * pow(alphabet_len, power) for power, char in enumerate(reversed(id_string))])
It's very fast and can take long integers.
For a similar project, to get a new key, I make a wrapper function around a random string generator that calls the generator until I get a string that hasn't already been used in my hashtable. This method will slow down once your name space starts to get full, but as you have said, even with only 6 characters, you have plenty of namespace to work with.
I have a variant of the problem, in that I store web pages from many different authors and need to prevent discovery of pages by guesswork. So my short URLs add a couple of extra digits to the Base-62 string for the page number. These extra digits are generated from information in the page record itself and they ensure that only 1 in 3844 URLs are valid (assuming 2-digit Base-62). You can see an outline description at http://mgscan.com/MBWL.
Very good answer, I have created a Golang implementation of the bjf:
package bjf
import (
"math"
"strings"
"strconv"
)
const alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
func Encode(num string) string {
n, _ := strconv.ParseUint(num, 10, 64)
t := make([]byte, 0)
/* Special case */
if n == 0 {
return string(alphabet[0])
}
/* Map */
for n > 0 {
r := n % uint64(len(alphabet))
t = append(t, alphabet[r])
n = n / uint64(len(alphabet))
}
/* Reverse */
for i, j := 0, len(t) - 1; i < j; i, j = i + 1, j - 1 {
t[i], t[j] = t[j], t[i]
}
return string(t)
}
func Decode(token string) int {
r := int(0)
p := float64(len(token)) - 1
for i := 0; i < len(token); i++ {
r += strings.Index(alphabet, string(token[i])) * int(math.Pow(float64(len(alphabet)), p))
p--
}
return r
}
Hosted at github: https://github.com/xor-gate/go-bjf
Implementation in Scala:
class Encoder(alphabet: String) extends (Long => String) {
val Base = alphabet.size
override def apply(number: Long) = {
def encode(current: Long): List[Int] = {
if (current == 0) Nil
else (current % Base).toInt :: encode(current / Base)
}
encode(number).reverse
.map(current => alphabet.charAt(current)).mkString
}
}
class Decoder(alphabet: String) extends (String => Long) {
val Base = alphabet.size
override def apply(string: String) = {
def decode(current: Long, encodedPart: String): Long = {
if (encodedPart.size == 0) current
else decode(current * Base + alphabet.indexOf(encodedPart.head),encodedPart.tail)
}
decode(0,string)
}
}
Test example with Scala test:
import org.scalatest.{FlatSpec, Matchers}
class DecoderAndEncoderTest extends FlatSpec with Matchers {
val Alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
"A number with base 10" should "be correctly encoded into base 62 string" in {
val encoder = new Encoder(Alphabet)
encoder(127) should be ("cd")
encoder(543513414) should be ("KWGPy")
}
"A base 62 string" should "be correctly decoded into a number with base 10" in {
val decoder = new Decoder(Alphabet)
decoder("cd") should be (127)
decoder("KWGPy") should be (543513414)
}
}
Function based in Xeoncross Class
function shortly($input){
$dictionary = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','0','1','2','3','4','5','6','7','8','9'];
if($input===0)
return $dictionary[0];
$base = count($dictionary);
if(is_numeric($input)){
$result = [];
while($input > 0){
$result[] = $dictionary[($input % $base)];
$input = floor($input / $base);
}
return join("", array_reverse($result));
}
$i = 0;
$input = str_split($input);
foreach($input as $char){
$pos = array_search($char, $dictionary);
$i = $i * $base + $pos;
}
return $i;
}
Here is a Node.js implementation that is likely to bit.ly. generate a highly random seven-character string.
It uses Node.js crypto to generate a highly random 25 charset rather than randomly selecting seven characters.
var crypto = require("crypto");
exports.shortURL = new function () {
this.getShortURL = function () {
var sURL = '',
_rand = crypto.randomBytes(25).toString('hex'),
_base = _rand.length;
for (var i = 0; i < 7; i++)
sURL += _rand.charAt(Math.floor(Math.random() * _rand.length));
return sURL;
};
}
My Python 3 version
base_list = list("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
base = len(base_list)
def encode(num: int):
result = []
if num == 0:
result.append(base_list[0])
while num > 0:
result.append(base_list[num % base])
num //= base
print("".join(reversed(result)))
def decode(code: str):
num = 0
code_list = list(code)
for index, code in enumerate(reversed(code_list)):
num += base_list.index(code) * base ** index
print(num)
if __name__ == '__main__':
encode(341413134141)
decode("60FoItT")
For a quality Node.js / JavaScript solution, see the id-shortener module, which is thoroughly tested and has been used in production for months.
It provides an efficient id / URL shortener backed by pluggable storage defaulting to Redis, and you can even customize your short id character set and whether or not shortening is idempotent. This is an important distinction that not all URL shorteners take into account.
In relation to other answers here, this module implements the Marcel Jackwerth's excellent accepted answer above.
The core of the solution is provided by the following Redis Lua snippet:
local sequence = redis.call('incr', KEYS[1])
local chars = '0123456789ABCDEFGHJKLMNPQRSTUVWXYZ_abcdefghijkmnopqrstuvwxyz'
local remaining = sequence
local slug = ''
while (remaining > 0) do
local d = (remaining % 60)
local character = string.sub(chars, d + 1, d + 1)
slug = character .. slug
remaining = (remaining - d) / 60
end
redis.call('hset', KEYS[2], slug, ARGV[1])
return slug
Why not just generate a random string and append it to the base URL? This is a very simplified version of doing this in C#.
static string chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
static string baseUrl = "https://google.com/";
private static string RandomString(int length)
{
char[] s = new char[length];
Random rnd = new Random();
for (int x = 0; x < length; x++)
{
s[x] = chars[rnd.Next(chars.Length)];
}
Thread.Sleep(10);
return new String(s);
}
Then just add the append the random string to the baseURL:
string tinyURL = baseUrl + RandomString(5);
Remember this is a very simplified version of doing this and it's possible the RandomString method could create duplicate strings. In production you would want to take in account for duplicate strings to ensure you will always have a unique URL. I have some code that takes account for duplicate strings by querying a database table I could share if anyone is interested.
This is my initial thoughts, and more thinking can be done, or some simulation can be made to see if it works well or any improvement is needed:
My answer is to remember the long URL in the database, and use the ID 0 to 9999999999999999 (or however large the number is needed).
But the ID 0 to 9999999999999999 can be an issue, because
it can be shorter if we use hexadecimal, or even base62 or base64. (base64 just like YouTube using A-Z a-z 0-9 _ and -)
if it increases from 0 to 9999999999999999 uniformly, then hackers can visit them in that order and know what URLs people are sending each other, so it can be a privacy issue
We can do this:
have one server allocate 0 to 999 to one server, Server A, so now Server A has 1000 of such IDs. So if there are 20 or 200 servers constantly wanting new IDs, it doesn't have to keep asking for each new ID, but rather asking once for 1000 IDs
for the ID 1, for example, reverse the bits. So 000...00000001 becomes 10000...000, so that when converted to base64, it will be non-uniformly increasing IDs each time.
use XOR to flip the bits for the final IDs. For example, XOR with 0xD5AA96...2373 (like a secret key), and the some bits will be flipped. (whenever the secret key has the 1 bit on, it will flip the bit of the ID). This will make the IDs even harder to guess and appear more random
Following this scheme, the single server that allocates the IDs can form the IDs, and so can the 20 or 200 servers requesting the allocation of IDs. The allocating server has to use a lock / semaphore to prevent two requesting servers from getting the same batch (or if it is accepting one connection at a time, this already solves the problem). So we don't want the line (queue) to be too long for waiting to get an allocation. So that's why allocating 1000 or 10000 at a time can solve the issue.

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