Develop Reference

How to encode chars in 2-bits? in java

DNA molecules are denoted by one of four values: A, C, G, or T. I need to convert a string of characters from A, C, G, and T to an array of bytes, encoding each of the characters
with two bits.A with bits 00, C with bits 01, G with 10, and T with 11. I don't understand how to convert characters to 2 bits. I was trying to shift and mask, but got wrong result.
At the very beginning, I check if there are characters in the line. Then i convert each character into a bit value and insert it into an array. When i insert ACGT, in the output i got 0 1 3 2. And here I have a problem, because I don’t understand how to convert the value to 2 bits.
Scanner text = new Scanner(System.in);
String str = text.nextLine();
if (str.contains("A") && str.contains("C") && str.contains("G") && str.contains("T")){
System.out.println("");
}
else
{
System.out.println("wrong command format");
}
byte mas[] = str.getBytes();
System.out.println("String in byte array : " + Arrays.toString(mas));
for (int i = 0; i < mas.length; i++){
byte mask = 3;
byte number = mas[i];
byte result = (byte)((number >> 1) & mask);
System.out.println(result);
}
}
}

It seems that you want to save the bits in a byte. The following example might give some ideas.
public class Main
{
private static final int A = 0x00; // b00
private static final int C = 0x01; // b01
private static final int G = 0x02; // b10
private static final int T = 0x03; // b11
public static void main(String[] args) throws Exception
{
byte store = 0;
store = setByte(store, 0, A);
store = setByte(store, 1, C);
store = setByte(store, 2, G);
store = setByte(store, 3, T);
System.out.println(Integer.toBinaryString(store));
//11111111111111111111111111100100
System.out.println(getByte(store, 0)); //0
System.out.println(getByte(store, 1)); //1
System.out.println(getByte(store, 2)); //2
System.out.println(getByte(store, 3)); //3
}
//Behavior :: Store "value" into "store".
//Reminder :: Valid index 0 - 3. Valid value 0 - 3.
private static byte setByte(byte store, int index, int value)
{
store = (byte)(store & ~(0x3 << (2 * index)));
return store |= (value & 0x3) << (2 * index);
}
private static byte getByte(byte store, int index)
{
return (byte)((store >> (2 * index)) & 0x3);
}
}

I haven't tested this, but it may help you.
byte test = 69;
byte insert = 0b01;
byte index = 2;
final byte ones = 0b00000011;
//Clear out the data at specified index
test = (byte) (test & ~(ones << index));
//Insert data
test |= (byte) (insert << index);
It works as follows:
Clear the 2 bits at the index in the byte (using bitwise AND).
Insert the 2 data bits at the index in the byte using bitwise OR).

You can "convert" the chars ACGT to 0, 1, 2, 3 using bit arithmetic.
byte[] bytes = str.getBytes();
for (int i = 0; i < bytes.length; i++) {
bytes[i] = (byte)(bytes[i] >> 1 & 3 ^ bytes[i] >> 2 & 1);
}
I suspect your initial check should be:
if (!str.matches("[ACGT]+") {
System.out.println("wrong command format");
return;
}

Java: convert byte[] to base36 String

I'm a bit lost. For a project, I need to convert the output of a hash-function (SHA256) - which is a byte array - to a String using base 36.
So In the end, I want to convert the (Hex-String representation of the) Hash, which is
43A718774C572BD8A25ADBEB1BFCD5C0256AE11CECF9F9C3F925D0E52BEAF89
to base36, so the example String from above would be:
3SKVHQTXPXTEINB0AT1P0G45M4KI8U0HR8PGB96DVXSTDJKI1
For the actual conversion to base36, I found some piece of code here on StackOverflow:
public static String toBase36(byte[] bytes) {
//can provide a (byte[], offset, length) method too
StringBuffer sb = new StringBuffer();
int bitsUsed = 0; //will point how many bits from the int are to be encoded
int temp = 0;
int tempBits = 0;
long swap;
int position = 0;
while((position < bytes.length) || (bitsUsed != 0)) {
swap = 0;
if(tempBits > 0) {
//there are bits left over from previous iteration
swap = temp;
bitsUsed = tempBits;
tempBits = 0;
}
//fill some bytes
while((position < bytes.length) && (bitsUsed < 36)) {
swap <<= 8;
swap |= bytes[position++];
bitsUsed += 8;
}
if(bitsUsed > 36) {
tempBits = bitsUsed - 36; //this is always 4
temp = (int)(swap & ((1 << tempBits) - 1)); //get low bits
swap >>= tempBits; //remove low bits
bitsUsed = 36;
}
sb.append(Long.toString(swap, 36));
bitsUsed = 0;
}
return sb.toString();
}
Now I'm doing this:
// this creates my hash, being a 256-bit byte array
byte[] hash = PBKDF2.deriveKey(key.getBytes(), salt.getBytes(), 2, 256);
System.out.println(hash.length); // outputs "256"
System.out.println(toBase36(hash)); // outputs total crap
the "total crap" is something like
-7-14-8-1q-5se81u0e-3-2v-24obre-73664-7-5-5cor1o9s-6h-4k6hr-5-4-rt2z0-30-8-2u-8-onz-4a2j-6-8-18-8trzza3-3-2x-6-4153to-4e3l01me-6-azz-2-k-4ckq-nav-gu-irqpxx-el-1j-6-rmf8hs-1bb5ax-3z25u-2-2r-t5-22-6-6w1v-1p
so it's not even close to what I want. I tried to find a solution now, but it seems I'm a bit lost here. How do I get the base36-encoded String representation of the Hash that I need?

Try using BigInteger:
String hash = "43A718774C572BD8A25ADBEB1BFCD5C0256AE11CECF9F9C3F925D0E52BEAF89";
//use a radix of 16, default would be 10
String base36 = new BigInteger( hash, 16 ).toString( 36 ).toUpperCase();

This might work:
BigInteger big = new BigInteger(your_byte_array_to_hex_string, 16);
big.toString(36);

How do I convert a large binary String to byte array java?

I have a large binary string "101101110...", and I am trying to store it into a byte array. what is the best way of doing it?
Lets say I have largeString = "0100111010111011011000000001000110101"
Result that I'm looking for:
[78, 187, 96, 17, 21]
01001110 10111011 01100000 00010001 10101
What i've tried:
byte[] b= new BigInteger(largeString,2).toByteArray();
however it did not give me the result I'm looking for...

You can easily build an ArrayList on which you can call toArray if you want an actual array;
List<Integer> list = new ArrayList<>();
for(String str : largeString.split("(?<=\\G.{8})"))
list.add(Integer.parseInt(str, 2));
System.out.println(list); // Outputs [78, 187, 96, 17, 21]

Assuming that your binary string module 8 equals 0 binString.lenght()%8==0
/**
* Get an byte array by binary string
* #param binaryString the string representing a byte
* #return an byte array
*/
public static byte[] getByteByString(String binaryString) {
int splitSize = 8;
if(binaryString.length() % splitSize == 0){
int index = 0;
int position = 0;
byte[] resultByteArray = new byte[binaryString.length()/splitSize];
StringBuilder text = new StringBuilder(binaryString);
while (index < text.length()) {
String binaryStringChunk = text.substring(index, Math.min(index + splitSize, text.length()));
Integer byteAsInt = Integer.parseInt(binaryStringChunk, 2);
resultByteArray[position] = byteAsInt.byteValue();
index += splitSize;
position ++;
}
return resultByteArray;
}
else{
System.out.println("Cannot convert binary string to byte[], because of the input length. '" +binaryString+"' % 8 != 0");
return null;
}
}

Do it in a loop. Split the string at 8-character chunks and convert them separately. In "pseudocode" it's something like:
byte[] result = new byte[subs.size()];
int i = 0;
int j = 0;
while(i+8 <= s.length){
result[j] = new Byte.valueOf(largeString.substring(i, i+8), 2);
i+=8;
j++;
}
result[j] = new Byte.valueOf(largeString.substring(i, largeString.length));

convert hex string to base64

Please help me on converting a hexstring to base64
here is the cede where I'm getting the exception
String hexString = "bf940165bcc3bca12321a5cc4c753220129337b48ad129d880f718d147a2cd1bfa79de92239ef1bc06c2f05886b0cd5d";
private static String ConvertHexStringToBase64(String hexString) {
System.out.println(hexString);
if ((hexString.length()) % 2 > 0)
throw new NumberFormatException("Input string was not in a correct format.");
byte[] buffer = new byte[hexString.length() / 2];
int i = 0;
while (i < hexString.length())
{
buffer[i / 2] = Byte.parseByte(hexString.substring(i, 2));
i += 2;
}
System.out.println("hexSring"+hexString+"afterconverttobase64"+Base64.encodeBase64String(buffer));
return Base64.encodeBase64String(buffer);
}
I'm getting an exception here::bf940165bcc3bca12321a5cc4c753220129337b48ad129d880f718d147a2cd1bfa79de92239ef1bc06c2f05886b0cd5d
Exception in thread "main" java.lang.NumberFormatException: For input string: "bf"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:449)
at java.lang.Byte.parseByte(Byte.java:151)
at java.lang.Byte.parseByte(Byte.java:108)
at com.motorola.gst.DecryptTest3.ConvertHexStringToBase64(DecryptTest3.java:38)
at com.motorola.gst.DecryptTest3.main(DecryptTest3.java:16)

First of all you have to specify the specify the radix(16 in your case) in the parseByte method to avoid the numberFormat exception :
buffer[i / 2] = Byte.parseByte(hexString.substring(i, 2),16);
However your code seems broken, take a look at the corrected one :
if ((hexString.length()) % 2 > 0)
throw new NumberFormatException("Input string was not in a correct format.");
int[] buffer = new int[hexString.length() / 2];
int i = 2;
while (i < hexString.length())
{
buffer[i / 2] = Integer.parseInt(hexString.substring(i, i + 2),16);
i += 2;
}
Your loop was wrong and you have to parse as Integer because you have some values inside your input string that overflow the byte capability ...
If you need byte you could cast the parsed int to byte in this way :
byte[] buffer = new byte[hexString.length() / 2];
int i = 2;
while (i < hexString.length())
{
buffer[i / 2] = (byte)Integer.parseInt(hexString.substring(i, i + 2),16);
i += 2;
}

Found a similar solution, thought it might be good to share:
public static string convertHexToBase64String(String hexString)
{
string base64 = "";
//--Important: remove "0x" groups from hexidecimal string--
hexString = hexString.Replace("0x", "");
byte[] buffer = new byte[hexString.Length / 2];
for (int i = 0; i < hexString.Length; i++)
{
try
{
buffer[i / 2] = Convert.ToByte(Convert.ToInt32(hexString.Substring(i, 2), 16));
}
catch (Exception ex) { }
i += 1;
}
base64 = Convert.ToBase64String(buffer);
return base64;
}
Hope it helps someone else.

Store binary sequence in byte array?

I need to store a couple binary sequences that are 16 bits in length into a byte array (of length 2). The one or two binary numbers don't change, so a function that does conversion might be overkill. Say for example the 16 bit binary sequence is 1111000011110001. How do I store that in a byte array of length two?

String val = "1111000011110001";
byte[] bval = new BigInteger(val, 2).toByteArray();
There are other options, but I found it best to use BigInteger class, that has conversion to byte array, for this kind of problems. I prefer if, because I can instantiate class from String, that can represent various bases like 8, 16, etc. and also output it as such.
Edit: Mondays ... :P
public static byte[] getRoger(String val) throws NumberFormatException,
NullPointerException {
byte[] result = new byte[2];
byte[] holder = new BigInteger(val, 2).toByteArray();
if (holder.length == 1) result[0] = holder[0];
else if (holder.length > 1) {
result[1] = holder[holder.length - 2];
result[0] = holder[holder.length - 1];
}
return result;
}
Example:
int bitarray = 12321;
String val = Integer.toString(bitarray, 2);
System.out.println(new StringBuilder().append(bitarray).append(':').append(val)
.append(':').append(Arrays.toString(getRoger(val))).append('\n'));

I have been disappointed with all of the solutions I have found to converting strings of bits to byte arrays and vice versa -- all have been buggy (even the BigInteger solution above), and very few are as efficient as they should be.
I realize the OP was only concerned with a bit string to an array of two bytes, which the BitInteger approach seems to work fine for. However, since this post is currently the first search result when searching "bit string to byte array java" in Google, I am going to post my general solution here for people dealing with huge strings and/or huge byte arrays.
Note that my solution below is the only solution I have ran that passes all of my test cases -- many online solutions to this relatively simple problem simply do not work.
Code
/**
* Zips (compresses) bit strings to byte arrays and unzips (decompresses)
* byte arrays to bit strings.
*
* #author ryan
*
*/
public class BitZip {
private static final byte[] BIT_MASKS = new byte[] {1, 2, 4, 8, 16, 32, 64, -128};
private static final int BITS_PER_BYTE = 8;
private static final int MAX_BIT_INDEX_IN_BYTE = BITS_PER_BYTE - 1;
/**
* Decompress the specified byte array to a string.
* <p>
* This function does not pad with zeros for any bit-string result
* with a length indivisible by 8.
*
* #param bytes The bytes to convert into a string of bits, with byte[0]
* consisting of the least significant bits in the byte array.
* #return The string of bits representing the byte array.
*/
public static final String unzip(final byte[] bytes) {
int byteCount = bytes.length;
int bitCount = byteCount * BITS_PER_BYTE;
char[] bits = new char[bitCount];
{
int bytesIndex = 0;
int iLeft = Math.max(bitCount - BITS_PER_BYTE, 0);
while (bytesIndex < byteCount) {
byte value = bytes[bytesIndex];
for (int b = MAX_BIT_INDEX_IN_BYTE; b >= 0; --b) {
bits[iLeft + b] = ((value % 2) == 0 ? '0' : '1');
value >>= 1;
}
iLeft = Math.max(iLeft - BITS_PER_BYTE, 0);
++bytesIndex;
}
}
return new String(bits).replaceFirst("^0+(?!$)", "");
}
/**
* Compresses the specified bit string to a byte array, ignoring trailing
* zeros past the most significant set bit.
*
* #param bits The string of bits (composed strictly of '0' and '1' characters)
* to convert into an array of bytes.
* #return The bits, as a byte array with byte[0] containing the least
* significant bits.
*/
public static final byte[] zip(final String bits) {
if ((bits == null) || bits.isEmpty()) {
// No observations -- return nothing.
return new byte[0];
}
char[] bitChars = bits.toCharArray();
int bitCount = bitChars.length;
int left;
for (left = 0; left < bitCount; ++left) {
// Ignore leading zeros.
if (bitChars[left] == '1') {
break;
}
}
if (bitCount == left) {
// Only '0's in the string.
return new byte[] {0};
}
int cBits = bitCount - left;
byte[] bytes = new byte[((cBits) / BITS_PER_BYTE) + (((cBits % BITS_PER_BYTE) > 0) ? 1 : 0)];
{
int iRight = bitCount - 1;
int iLeft = Math.max(bitCount - BITS_PER_BYTE, left);
int bytesIndex = 0;
byte _byte = 0;
while (bytesIndex < bytes.length) {
while (iLeft <= iRight) {
if (bitChars[iLeft] == '1') {
_byte |= BIT_MASKS[iRight - iLeft];
}
++iLeft;
}
bytes[bytesIndex++] = _byte;
iRight = Math.max(iRight - BITS_PER_BYTE, left);
iLeft = Math.max((1 + iRight) - BITS_PER_BYTE, left);
_byte = 0;
}
}
return bytes;
}
}
Performance
I was bored at work so I did some performance testing comparing against the accepted answer here for when N is large. (Pretending to ignore the fact that the BigInteger approach posted above doesn't even work properly as a general approach.)
This is running with a random bit string of size 5M and a random byte array of size 1M:
String -> byte[] -- BigInteger result: 39098ms
String -> byte[] -- BitZip result: 29ms
byte[] -> String -- Integer result: 138ms
byte[] -> String -- BitZip result: 71ms
And the code:
public static void main(String[] argv) {
int testByteLength = 1000000;
int testStringLength = 5000000;
// Independently random.
final byte[] randomBytes = new byte[testByteLength];
final String randomBitString;
{
StringBuilder sb = new StringBuilder();
Random rand = new Random();
for (int i = 0; i < testStringLength; ++i) {
int value = rand.nextInt(1 + i);
sb.append((value % 2) == 0 ? '0' : '1');
randomBytes[i % testByteLength] = (byte) value;
}
randomBitString = sb.toString();
}
byte[] resultCompress;
String resultDecompress;
{
Stopwatch s = new Stopwatch();
TimeUnit ms = TimeUnit.MILLISECONDS;
{
s.start();
{
resultCompress = compressFromBigIntegerToByteArray(randomBitString);
}
s.stop();
{
System.out.println("String -> byte[] -- BigInteger result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultCompress = zip(randomBitString);
}
s.stop();
{
System.out.println("String -> byte[] -- BitZip result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultDecompress = decompressFromIntegerParseInt(randomBytes);
}
s.stop();
{
System.out.println("byte[] -> String -- Integer result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultDecompress = unzip(randomBytes);
}
s.stop();
{
System.out.println("byte[] -> String -- BitZip result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
}
}