Develop Reference

Shuffle a string via swapping content to different index

I have an assignment in which we are instructed to create a method which takes a string, scrambles the content of the string, and then returns the scrambled string (ie "Hello" returns "elloH"). However, we are only allowed to do this via loops and basic string functions (no arrays can be used).
My Teacher has left the following suggestion for us:
The best way to shuffle is to generate 2 random numbers (index number) and swap the numbers(the content, not the index) based on the random index numbers.
Continue to do it for 100 times, they are shuffled and all characters remain the same but in different positions
How would this nessasarily be done? For reference here is my try at the solution, however it does not work and I'm not sure where to go from here:
public void shuffle(String s1) {
int i1 = 0;
int i2 = 0;
String s2 = "";
for(int a2 = 0; a2 < s1.length(); a2++) {
char c1 = s1.charAt(i1);
s2 += c1;
for(int a1 = 0; a1 < 100; a1++) {
double d1 = Math.random();
d1 *= (s1.length()-1);
i1 = (int) d1;
}
}
System.out.println(s2);
}

The problem with your code is that you don't swap characters in the String but rather select randomly across the String which causes some characters to be used many times and some not at all.
I would do something like this using StringBuilder that is suitable for String manipulations as long as the String itself is immutable (calling any method from it does not change the former object but returns a new one).
// it's a good practice to delcare formal parameters in a method as final
public void shuffle(final String string) {
// store the string length to a variable for sake of comfort
final int length = string.length();
// use mutable StringBuilder for easy String manipulations
final StringBuilder stringBuilder = new StringBuilder(string);
// initialize the Random object generator once and use many times
// as you don't want to initialize with each loop and throw it away
final Random random = new Random();
// define the number of iterations
// to repeat as many times as the string is long is more than enough
for (int i=0; i<length; i++) {
// pick two random indices to be swapped
final int firstIndex = random.nextInt(length);
final int secondIndex = random.nextInt(length);
// remember the swapped characters
// otherwise it would end up in a mess and duplicated characters
final char firstChar = stringBuilder.charAt(firstIndex);
final char secondChar = stringBuilder.charAt(secondIndex);
// perform the swap: basically set the characters to their positions
stringBuilder.setCharAt(firstIndex, secondChar);
stringBuilder.setCharAt(secondIndex, firstChar);
}
// and see the result
System.out.println(stringBuilder);
}
Run three times (it looks nicely swapped and the characters remain unchanged):
WoodeHl rl!l
Wer lldooHl!
HW! llrloeod

Hi here is the updated code with your requirement I tried not using the concepts out of string manipulation the swap is not using StringBuilder (that would be the best way to do that but if not needed here as a custom logic to do so)
and also used Math.random() to generate the random number if you want to see all 100 iterations just print the s1 in the for loop.
static String swap(String str, int i, int j) {
// if both indexes are the same change nothing
if (i == j)
return str;
// if the second index is last then there will be no substring from j+1 to last;
if (j == str.length() - 1)
return str.substring(0, i) + str.charAt(j) + str.substring(1 + i, j) + str.charAt(i);
return str.substring(0, i) + str.charAt(j) + str.substring(1 + i, j) + str.charAt(i)
+ str.substring(j + 1, str.length());
}
public static void shuffle(String s1) {
// iterate the same logic 100 times
for (int a1 = 0; a1 < 100; a1++) {
// generate 2 random numbers in the range of the length of string
int randomNumber1 = (int) (Math.random() * s1.length());
int randomNumber2 = (int) (Math.random() * s1.length());
// setting the lower random number to randomnumber1 (it is swapping two number without temp)
if (randomNumber1 > randomNumber2) {
randomNumber1 = randomNumber1 + randomNumber2;
randomNumber2 = randomNumber1 - randomNumber2;
randomNumber1 -= randomNumber2;
}
// calling the swap method to swap the chars in string.
s1 = swap(s1, randomNumber1, randomNumber2);
}
System.out.println(s1);
}

Since you cannot use arrays, try using StringBuilder.
private String shuffleUtil(String s) {
final StringBuilder result = new StringBuilder(s);
for (int rep = 1; rep <= 100; rep++) {
final int randomOne = ThreadLocalRandom.current().nextInt(0, s.length());
final int randomTwo = ThreadLocalRandom.current().nextInt(0, s.length());
final char c = result.charAt(randomOne);
result.setCharAt(randomOne, result.charAt(randomTwo));
result.setCharAt(randomTwo, c);
}
return result.toString();
}
public void shuffle(String s) {
final String result = shuffleUtil(s);
System.out.println(s);
}

Adding two numbers stored as arrays of chars

I'm trying to write an algorithm which adds two numbers that are stored as chars in two arrays. Unfortunately, it doesn't work. When I try to debug it, I see that the variables a and b get the value -1 which makes no sense. Any idea what might be the problem?
public class rechner2 {
public static void main(String[] args) {
final char[] zahl1 = {1, 2, 3};
final char[] zahl2 = {7, 8, 9};
//Add arrays zahl1 and zahl2.
char [] zwischenarray = add(zahl1, zahl2);
for (int i = 0; i < zwischenarray.length; i++) {
System.out.println(zwischenarray[i]);
}
}
private static char[] add(char[] zahl1, char[] zahl2) {
int len;
if (zahl1.length < zahl2.length) {
len = zahl2.length;
} else {
len = zahl1.length;
}
char[] finalresult = new char [len + 1];
int carryover = 0;
for (int i = 0; i < len; i++) {
int a = Character.getNumericValue(zahl1[i]);
int b = Character.getNumericValue(zahl2[i]);
int c = a + b + carryover;
if (c > 9) {
carryover = 1;
c = c - 10;
} else {
carryover = 0;
}
finalresult[i] = (char)c;
}
if (carryover == 1) {
finalresult[len + 1] = 1;
}
return finalresult;
}
}

in this code I believe 2 bug
instead of char , i guess better to us int
length of the array
here is the code:
public class rechner2 {
public static void main(String[] args) {
int[] zahl1 = {1,2,3};
int[] zahl2 = {7,8,9};
//Add arrays zahl1 and zahl2.
int [] zwischenarray = add(zahl1, zahl2);
for (int i = 0; i < zwischenarray.length; i++) {
System.out.println(zwischenarray[i]);
}
}
private static int[] add(int[] zahl1, int[] zahl2) {
int len;
if (zahl1.length < zahl2.length) {
len = zahl2.length;
} else {
len = zahl1.length;
}
int[] finalresult = new int [len + 1];
int carryover = 0;
for (int i = 0; i <= len-1; i++) {
int a = (zahl1[i]);
int b = (zahl2[i]);
int c = a + b + carryover;
if (c > 9) {
carryover = 1;
c = c - 10;
} else {
carryover = 0;
}
finalresult[i] = c;
}
if (carryover == 1) {
finalresult[len] = 1;
}
return finalresult;
}
}

Your code is conflicted: The numbers / characters in your array are actually integers, not "printable" or "human readable" characters. But, parts of your code are treating them as if they are "printable".
Let's go back decades, and use ASCII for the beginning of this explanation. ASCII has "Printable" and "Nonprintable" characters. The "Nonprintable" characters are known as "Control codes."
Control codes include codes that move the cursor on a display terminal or print head on a printing terminal. They include thing like CR (Carriage Return), LF (Line Feed), HT (Horizontal tab), and BS (Backspace). Others are used by data communications hardware to control the flow of data, or to report status.
Printable characters correspond to what you see on a terminal screen or printout. They include uppercase alphabetic, lower case alphabetic, digits, punctuation, and the space character. They are "human readable."
Look at the list of printable characters in the Wikipedia article. Take 5 as an example. It's represented as '53' in base ten, which corresponds to '35' in base sixteen, or '011 0101' in binary. Note that it is not the same as the binary number five, which would be '0000 0101'.
Java uses 16 bit Unicode, not ASCII, for its char type. The Java compiler allows arithmetic to be done on char data, as if it was the same as short.
These lines in your code expect your char variables and constants are printable characters:
int a = Character.getNumericValue(zahl1[i]);
int b = Character.getNumericValue(zahl2[i]);
In addition, that you specified zwischenarray as char tells the compiler to handle the contents as printable characters in this line:
System.out.println(zwischenarray[i]);
But, the rest of your code treats your char data as integer data types.
You have a bug in this line: finalresult[len + 1] = 1;. After that bug is fixed, how do you fix the rest of your code? There are different ways, and which is best depends on your intention.
For demonstration purpose, try this: Replace the following
int a = Character.getNumericValue(zahl1[i]);
int b = Character.getNumericValue(zahl2[i]);
int c = a + b + carryover;
with
int c = zahl1[i] + zahl2 [i] + carryover;
Also, put a cast in your output line:
System.out.println((short)zwischenarray[i]);
And run it. That will demonstrate you can do arithmetic on Java char data.
Now, remove the (short) cast in output line, and change all occurrences of char to short (or int). Your program still works.
This is because of the way you entered the values for zahl1 and zahl2. Your source code consists of printable characters and white space. By omitting the single quotes, you told the compiler to convert the values to binary integers. For example, your source code 9 became binary 0000 1001 in the runtime code. If you wanted your 9 to remain as a printable character, you needed to enclose it in single quote marks: '9' .
By enclosing all the values in zahl1 and zahl2 in single quote marks, the use of Character.getNumericValue is appropriate. But, you would still need the (short) or (int) cast in your System.out. line to see your output.
Character.getNumericValue is returning -1 because the values passed are outside of the range it was designed to work with.
Here are two ways to convert a base 10 digit represented as a binary integer to the equivalent printable character:
finalresult[i] = (char) (c + '0');
But, my preference is for this:
final String digit = "0123456789";
finalresult[i] = digit.charAt (c);

Convert Binary to Hexadecimal in Java without methods

I am currently a beginner in programming and I am trying to write a program in java to convert binary in hexadecimal numbers.
I know that the program will have to divide the number in groups of 4 and convert them to hexadecimal.
Ex: 11101111 (b2) --> E + F --- EF
However, since I used ints to do the conversion of the numbers, I'm stuck when I need to print a letter because it is a String.
Can someone point me to the right way? What am I doing wrong? I've also tried another version with an auxiliary array to store each group of 4 digits but I can't manage to insert a proper dimension to the array.
Unfortunately I am not allowed to use any function other than Scanner and Math, the method lenght and charAt and the basic stuff. I can't modify the public static line either.
EDIT: So after your inputs and so many tries, I managed to get this code. However it gives me an error if I insert too many numbers, eg: 0111011010101111. I've tried to change int to double but that didn't fix the problem.
import java.util.Scanner;
public class Bin2HexString {
public static void main(String[] args) {
Scanner keyb = new Scanner(System.in);
System.out.println("Valor?");
int vlr = keyb.nextInt();
String num = "";
int aux = vlr;
// Hexadecimal numbers
String arr[] = {"0","1","2","3","4","5","6","7","8","9","A", "B", "C", "D", "E", "F"};
String bits[] = {"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
String letters = "";
//Divide in groups of 4
int r;
for (; aux > 0; ) {
r = aux % 10000;
aux = aux / 10000;
num = "" + r;
for (;num.length() < 4;) { //add missing zeros
String zero = "0";
num = zero + num;
}
int charint = 0,bitint = 0;
for (int i = 0; i < arr.length;i++) {
String aux2 = bits[i];
String aux3 = arr[i];
for (int j = 0; j < num.length();j++) { // compare each group with arr[i]
char charvl = num.charAt(j);
char bitsvl = aux2.charAt(j);
charint = ((int) (charvl)-'0');
bitint = ((int) (bitsvl) - '0');
if (bitint != charint)
break;
}
if (bitint == charint)
letters = aux3 + "" + letters;
}
}
System.out.println(letters);
}
}

Having thought about this for a while to determine the most effective and useful way to do this is to write methods which convert a string from any base between 2 and 16 to an int and back to a string again.
This way you have useful methods for other things. And note that they methods can be easily changed and names to simply hard code the desired radix into the method to limit it to binary and hex methods.
The indexOf utility method was written to avoid using the builtin String method.
final static String hex = "0123456789ABCDEF";
static int stringToInt(String str, int radix) {
if (radix < 2 || radix > 16) {
System.out.println("Base must be between 2 and 16 inclusive");
return -1;
}
int v = 0;
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
int idx = indexOf(hex, c);
if (idx < 0 || idx > radix) {
System.out.println("Illegal character in string (" + c + ")");
}
v = v * radix + idx;
}
return v;
}
static String intToBase(int v, int radix) {
if (radix < 2 || radix > 16) {
System.out.println("Base must be between 2 and 16 inclusive");
return null;
}
String s = "";
while (v > 0) {
int idx = v % radix;
s = hex.charAt(idx) + s;
v /= radix;
}
return s;
}
static int indexOf(String str, char c) {
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == c) {
return i;
}
}
return -1;
}
And here is an example of their use.
// generate some test data
Random r = new Random(23);
String[] bitStrings =
r.ints(20, 20, 4000).mapToObj(Integer::toBinaryString).toArray(
String[]::new);
for (String bitstr : bitStrings) {
int v = baseToInt(bitstr, 2);
String hex = intToBase(v, 16);
System.out.printf("%12s = %s%n", bitstr, hex);
}
Which prints the following:
101110000011 = B83
111001111100 = E7C
10001110111 = 477
100110001111 = 98F
111001010 = 1CA
111001001111 = E4F
111000011010 = E1A
100001010010 = 852
11011001101 = 6CD
111010010111 = E97

Just some quick notes:
First this is wrong:
//Divide in groups of 4
for (; aux > 0; ) {
r = aux % 10000;
aux = aux / 10000;
Not at all what you want to do. Try it by hand and see what happens. Take a simple number that you know the answer to, and try it. You won't get the right answer. A good test is 17, which is 11 hex.
Try this instead: convert directly to the base you want. Hex is base 16 (its radix is 16), so you use 16 instead.
//Divide in groups of 4
for (; aux > 0; ) {
r = aux % 16;
aux = aux / 16;
Try those numbers with the test case, which is 17, and see what you get. That will get you much closer.

I'm assuming by "without methods" in the title, you are attempting to write your own integer parsing method instead of using Scanner.nextInt(int radix). In that case, my first advice would be work with a string instead of an integer - you'll be able to handle larger numbers and you can simply make an array of substrings (length 4) to convert to letters.
So, if you use the string approach - first scan in a string, not an int. Then I'd recommend a hash table with the 4-bit strings as keys and the hexadecimal equivalents as values. That should make calculation quite fast.
e.g.
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class HashMapBin2Hex
{
public static void main(String[] args)
{
//Read the string in
Scanner sc = new Scanner(System.in);
System.out.println("Binary number?");
String bin = sc.nextLine();
//Pad the bitstring with leading zeros to make a multiple of four
String zeros = "";
int i;
if (bin.length() % 4 != 0)
{
for (i = 0; i < 4 - (bin.length() % 4); i++)
{
zeros += "0";
}
}
bin = zeros + bin;
//Split the padded string into 4-bit chunks
String[] chunks = new String[bin.length() / 4];
for (i = 0; (i * 4) < bin.length() - 1; i++)
{
chunks[i] = bin.substring(i * 4, (i * 4) + 4);
}
//Convert the chunks to hexadecimal
String hex = "";
Map<String, String> bin2hex = new HashMap<>();
bin2hex.put("0000", "0");
bin2hex.put("0001", "1");
bin2hex.put("0010", "2");
bin2hex.put("0011", "3");
bin2hex.put("0100", "4");
bin2hex.put("0101", "5");
bin2hex.put("0110", "6");
bin2hex.put("0111", "7");
bin2hex.put("1000", "8");
bin2hex.put("1001", "9");
bin2hex.put("1010", "A");
bin2hex.put("1011", "B");
bin2hex.put("1100", "C");
bin2hex.put("1101", "D");
bin2hex.put("1110", "E");
bin2hex.put("1111", "F");
for (String s : chunks)
{
hex += bin2hex.get(s);
}
System.out.println("Hexadecimal: " + hex);
sc.close();
}
}
Further iterations could have some error checking to prevent catastrophic failure in the case of characters other than 0 or 1.
And of course, if you're fine with the other way (builtins), the following is far easier and more robust (ie will throw an exception if the string contains anything other than 0s and 1s):
import java.util.Scanner;
public class BuiltinBin2Hex
{
public static void main(String[] args)
{
//Read the binary number in
Scanner sc = new Scanner(System.in);
System.out.println("Binary number?");
int bin = sc.nextInt(2);
//And print as hexadecimal
System.out.println("Hexadecimal: " + Integer.toString(bin, 16));
sc.close();
}
}

Generate all possible string from a given length

I would like to be able to generate all possible strings from a given length, and I frankly don't know how to code that. So for further explanation, I and a friend would like to demonstrate some basic hacking techniques, so bruteforcing comes up. Of course, he will be my victim, no illegal thing there.
However, the only thing he told me is that his PW will be 4-char-long, but I'm pretty sure his PW won't be in any dictionnary, that would be toi easy.
So I came up with the idea of generating EVERY 4-char-long-string possible, containing a-z characters (no caps).
Would someone have a lead to follow to code such an algorithm ? I don't really bother with performances, if it takes 1 night to generate all PW, that's no problem.
Don't forget, that's only on demonstration purposes.

You can do it just how you'd do it with numbers. Start with aaaa. Then increment the 'least significant' part, so aaab. Keep going until you get to aaaz. Then increment to aaba. Repeat until you get to zzzz.
So all you need to do is implement is
String getNext(String current)
To expand on this; It possibly isnt the quickest way of doing things, but it is the simplest to get right.
As the old adage goes - 'first make it right, then make it fast'. Getting a working implementation that passes all your tests (you do have tests, right?) is what you do first. You then rewrite it to make it fast, using your tests as reassurance you're not breaking the core functionality.

The absolutely simplest way is to use four nested loops:
char[] pw = new char[4];
for (pw[0] = 'a' ; pw[0] <= 'z' ; pw[0]++)
for (pw[1] = 'a' ; pw[1] <= 'z' ; pw[1]++)
for (pw[2] = 'a' ; pw[2] <= 'z' ; pw[2]++)
for (pw[3] = 'a' ; pw[3] <= 'z' ; pw[3]++)
System.out.println(new String(pw));
This does not scale well, because adding extra characters requires adding a level of nesting. Recursive approach is more flexible, but it is harder to understand:
void findPwd(char[] pw, int pos) {
if (pos < 0) {
System.out.println(new String(pwd));
return;
}
for (pw[pos] = 'a' ; pw[pos] <= 'z' ; pw[pos]++)
findPwd(pw, pos-1);
}
Call recursive method like this:
char[] pw = new char[4];
findPwd(pw, 3);

private static void printAllStringsOfLength(int len) {
char[] guess = new char[len];
Arrays.fill(guess, 'a');
do {
System.out.println("Current guess: " + new String(guess));
int incrementIndex = guess.length - 1;
while (incrementIndex >= 0) {
guess[incrementIndex]++;
if (guess[incrementIndex] > 'z') {
if (incrementIndex > 0) {
guess[incrementIndex] = 'a';
}
incrementIndex--;
}
else {
break;
}
}
} while (guess[0] <= 'z');
}

public class GenerateCombinations {
public static void main(String[] args) {
List<Character> characters = new ArrayList<Character>();
for (char c = 'a'; c <= 'z'; c++) {
characters.add(c);
}
List<String> allStrings = new ArrayList<String>();
for (Character c : characters) {
for (Character d : characters) {
for (Character e : characters) {
for (Character f : characters) {
String s = "" + c + d + e + f;
allStrings.add(s);
}
}
}
}
System.out.println(allStrings.size()); // 456 976 combinations
}
}

This is something you can do recursively.
Lets define every (n)-character password the set of all (n-1)-character passwords, prefixed with each of the letters a thru z. So there are 26 times as many (n)-character passwords as there are (n-1)-character passwords. Keep in mind that this is for passwords consisting of lower-case letters. Obviously, you can increase the range of each letter quite easily.
Now that you've defined the recursive relationship, you just need the terminating condition.
That would be the fact that there is only one (0)-character password, that being the empty string.
So here's the recursive function:
def printNCharacterPasswords (prefix, num):
if num == 0:
print prefix
return
foreach letter in 'a'..'z':
printNCharacterPasswords (prefix + letter, num - 1)
to be called with:
printNCharacterPasswords ("", 4)
And, since Python is such a wonderful pseudo-code language, you can see it in action with only the first five letters:
def printNCharacterPasswords (prefix, num):
if num == 0:
print prefix
return
for letter in ('a', 'b', 'c', 'd', 'e'):
printNCharacterPasswords (prefix + letter, num - 1)
printNCharacterPasswords ("", 2)
which outputs:
aa
ab
ac
ad
ae
ba
bb
bc
bd
be
ca
cb
cc
cd
ce
da
db
dc
dd
de
ea
eb
ec
ed
ee

A aroth points out, using a digit counter approach is faster. To make this even faster, you can use a combination of an inner loop for the last digit and a counter for the rest (so the number of digits can be variable)
public static void main(String... args) {
long start = System.nanoTime();
int letters = 26;
int count = 6;
final int combinations = (int) Math.pow(letters, count);
char[] chars = new char[count];
Arrays.fill(chars, 'a');
final int last = count - 1;
OUTER:
while (true) {
for (chars[last] = 'a'; chars[last] <= 'z'; chars[last]+=2) {
newComination(chars);
chars[last]++;
newComination(chars);
}
UPDATED:
{
for (int i = last - 1; i >= 0; i--) {
if (chars[i]++ >= 'z')
chars[i] = 'a';
else
break UPDATED;
}
// overflow;
break OUTER;
}
}
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to generate %,d combinations%n", time / 1e9, combinations);
}
private static void newComination(char[] chars) {
}
prints
Took 0.115 seconds to generate 308,915,776 combinations
Note: the loop is so simple, its highly likely that the JIT can eliminate key pieces of code (after in-lining newCombination) and the reason its so fast is its not really calculating every combination.
A simpler way to generate combinations.
long start = System.nanoTime();
int letters = 26;
int count = 6;
final int combinations = (int) Math.pow(letters, count);
StringBuilder sb = new StringBuilder(count);
for (int i = 0; i < combinations; i++) {
sb.setLength(0);
for (int j = 0, i2 = i; j < count; j++, i2 /= letters)
sb.insert(0, (char) ('a' + i2 % letters));
// System.out.println(sb);
}
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to generate %,d combinations%n", time / 1e9, combinations);
prints
aaaa
aaab
aaac
....
zzzx
zzzy
zzzz
Took 0.785 seconds to generate 456,976 combinations
It spends most of its time waiting for the screen to update. ;)
If you comment out the line which prints the combinations, and increase the count to 5 and 6
Took 0.671 seconds to generate 11,881,376 combinations
Took 15.653 seconds to generate 308,915,776 combinations

public class AnagramEngine {
private static int[] anagramIndex;
public AnagramEngine(String str) {
AnagramEngine.generate(str);
}
private static void generate(String str) {
java.util.Map<Integer, Character> anagram = new java.util.HashMap<Integer, Character>();
for(int i = 0; i < str.length(); i++) {
anagram.put((i+1), str.charAt(i));
}
anagramIndex = new int[size(str.length())];
StringBuffer rev = new StringBuffer(AnagramEngine.start(str)+"").reverse();
int end = Integer.parseInt(rev.toString());
for(int i = AnagramEngine.start(str), index = 0; i <= end; i++){
if(AnagramEngine.isOrder(i))
anagramIndex[index++] = i;
}
for(int i = 0; i < anagramIndex.length; i++) {
StringBuffer toGet = new StringBuffer(anagramIndex[i] + "");
for(int j = 0; j < str.length(); j++) {
System.out.print(anagram.get(Integer.parseInt(Character.toString(toGet.charAt(j)))));
}
System.out.print("\n");
}
System.out.print(size(str.length()) + " iterations");
}
private static boolean isOrder(int num) {
java.util.Vector<Integer> list = new java.util.Vector<Integer>();
String str = Integer.toString(num);
char[] digits = str.toCharArray();
for(char vecDigits : digits)
list.add(Integer.parseInt(Character.toString(vecDigits)));
int[] nums = new int[str.length()];
for(int i = 0; i < nums.length; i++)
nums[i] = i+1;
for(int i = 0; i < nums.length; i++) {
if(!list.contains(nums[i]))
return false;
}
return true;
}
private static int start(String str) {
StringBuffer num = new StringBuffer("");
for(int i = 1; i <= str.length(); i++)
num.append(Integer.toString(i));
return Integer.parseInt(num.toString());
}
private static int size(int num) {
int size;
if(num == 1) {
return 1;
}
else {
size = num * size(num - 1);
}
return size;
}
public static void main(final String[] args) {
final java.util.Scanner sc = new java.util.Scanner(System.in);
System.out.printf("\n%s\t", "Entered word:");
String word = sc.nextLine();
System.out.printf("\n");
new AnagramEngine(word);
}
}

Put all the characters you expect the password to contain into an array. Write a stub function to test if your algorithm finds the correct password. Start with passwords of length 1, work your way up to 4 and see if your fake password is found on each iteration.

you can use the following code for getting random string. It will return you a string of 32 chars. you can get string of desired length by using substring(). Like if you want a string with 10 chars then:
import java.security.SecureRandom;
import java.math.BigInteger;
SecureRandom srandom = new SecureRandom();
String rand = new BigInteger(176, srandom).toString(32);
rand.substring(0,7);

Is there functionality to generate a random character in Java?

Does Java have any functionality to generate random characters or strings? Or must one simply pick a random integer and convert that integer's ascii code to a character?

To generate a random char in a-z:
Random r = new Random();
char c = (char)(r.nextInt(26) + 'a');

There are many ways to do this, but yes, it involves generating a random int (using e.g. java.util.Random.nextInt) and then using that to map to a char. If you have a specific alphabet, then something like this is nifty:
import java.util.Random;
//...
Random r = new Random();
String alphabet = "123xyz";
for (int i = 0; i < 50; i++) {
System.out.println(alphabet.charAt(r.nextInt(alphabet.length())));
} // prints 50 random characters from alphabet
Do note that java.util.Random is actually a pseudo-random number generator based on the rather weak linear congruence formula. You mentioned the need for cryptography; you may want to investigate the use of a much stronger cryptographically secure pseudorandom number generator in that case (e.g. java.security.SecureRandom).

You could also use the RandomStringUtils from the Apache Commons project:
Dependency:
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>3.8.1</version>
</dependency>
Usages:
RandomStringUtils.randomAlphabetic(stringLength);
RandomStringUtils.randomAlphanumeric(stringLength);

private static char rndChar () {
int rnd = (int) (Math.random() * 52); // or use Random or whatever
char base = (rnd < 26) ? 'A' : 'a';
return (char) (base + rnd % 26);
}
Generates values in the ranges a-z, A-Z.

String abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char letter = abc.charAt(rd.nextInt(abc.length()));
This one works as well.

In following 97 ascii value of small "a".
public static char randomSeriesForThreeCharacter() {
Random r = new Random();
char random_3_Char = (char) (97 + r.nextInt(3));
return random_3_Char;
}
in above 3 number for a , b , c or d and if u want all character like a to z then you replace 3 number to 25.

You could use generators from the Quickcheck specification-based test framework.
To create a random string use anyString method.
String x = anyString();
You could create strings from a more restricted set of characters or with min/max size restrictions.
Normally you would run tests with multiple values:
#Test
public void myTest() {
for (List<Integer> any : someLists(integers())) {
//A test executed with integer lists
}
}

using dollar:
Iterable<Character> chars = $('a', 'z'); // 'a', 'b', c, d .. z
given chars you can build a "shuffled" range of characters:
Iterable<Character> shuffledChars = $('a', 'z').shuffle();
then taking the first n chars, you get a random string of length n. The final code is simply:
public String randomString(int n) {
return $('a', 'z').shuffle().slice(n).toString();
}
NB: the condition n > 0 is cheched by slice
EDIT
as Steve correctly pointed out, randomString uses at most once each letter. As workaround
you can repeat the alphabet m times before call shuffle:
public String randomStringWithRepetitions(int n) {
return $('a', 'z').repeat(10).shuffle().slice(n).toString();
}
or just provide your alphabet as String:
public String randomStringFromAlphabet(String alphabet, int n) {
return $(alphabet).shuffle().slice(n).toString();
}
String s = randomStringFromAlphabet("00001111", 4);

This is a simple but useful discovery. It defines a class named RandomCharacter with 5 overloaded methods to get a certain type of character randomly. You can use these methods in your future projects.
public class RandomCharacter {
/** Generate a random character between ch1 and ch2 */
public static char getRandomCharacter(char ch1, char ch2) {
return (char) (ch1 + Math.random() * (ch2 - ch1 + 1));
}
/** Generate a random lowercase letter */
public static char getRandomLowerCaseLetter() {
return getRandomCharacter('a', 'z');
}
/** Generate a random uppercase letter */
public static char getRandomUpperCaseLetter() {
return getRandomCharacter('A', 'Z');
}
/** Generate a random digit character */
public static char getRandomDigitCharacter() {
return getRandomCharacter('0', '9');
}
/** Generate a random character */
public static char getRandomCharacter() {
return getRandomCharacter('\u0000', '\uFFFF');
}
}
To demonstrate how it works let's have a look at the following test program displaying 175 random lowercase letters.
public class TestRandomCharacter {
/** Main method */
public static void main(String[] args) {
final int NUMBER_OF_CHARS = 175;
final int CHARS_PER_LINE = 25;
// Print random characters between 'a' and 'z', 25 chars per line
for (int i = 0; i < NUMBER_OF_CHARS; i++) {
char ch = RandomCharacter.getRandomLowerCaseLetter();
if ((i + 1) % CHARS_PER_LINE == 0)
System.out.println(ch);
else
System.out.print(ch);
}
}
}
and the output is:
if you run one more time again:
I am giving credit to Y.Daniel Liang for his book Introduction to Java Programming, Comprehensive Version, 10th Edition, where I cited this knowledge from and use in my projects.
Note: If you are unfamiliar with overloaded methhods, in a nutshell Method Overloading is a feature that allows a class to have more than one method having the same name, if their argument lists are different.

Take a look at Java Randomizer class.
I think you can randomize a character using the randomize(char[] array) method.

My propose for generating random string with mixed case like: "DthJwMvsTyu".
This algorithm based on ASCII codes of letters when its codes a-z (97 to 122) and A-Z (65 to 90) differs in 5th bit (2^5 or 1 << 5 or 32).
random.nextInt(2): result is 0 or 1.
random.nextInt(2) << 5: result is 0 or 32.
Upper A is 65 and lower a is 97. Difference is only on 5th bit (32) so for generating random char we do binary OR '|' random charCaseBit (0 or 32) and random code from A to Z (65 to 90).
public String fastestRandomStringWithMixedCase(int length) {
Random random = new Random();
final int alphabetLength = 'Z' - 'A' + 1;
StringBuilder result = new StringBuilder(length);
while (result.length() < length) {
final char charCaseBit = (char) (random.nextInt(2) << 5);
result.append((char) (charCaseBit | ('A' + random.nextInt(alphabetLength))));
}
return result.toString();
}

Here is the code to generate random alphanumeric code. First you have to declare a string of allowed characters what you want to include in random number.and also define max length of string
SecureRandom secureRandom = new SecureRandom();
String CHARACTERS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789";
StringBuilder generatedString= new StringBuilder();
for (int i = 0; i < MAXIMUM_LENGTH; i++) {
int randonSequence = secureRandom .nextInt(CHARACTERS.length());
generatedString.append(CHARACTERS.charAt(randonSequence));
}
Use toString() method to get String from StringBuilder

polygenelubricants' answer is also a good solution if you only want to generate Hex values:
/** A list of all valid hexadecimal characters. */
private static char[] HEX_VALUES = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '0', 'A', 'B', 'C', 'D', 'E', 'F' };
/** Random number generator to be used to create random chars. */
private static Random RANDOM = new SecureRandom();
/**
* Creates a number of random hexadecimal characters.
*
* #param nValues the amount of characters to generate
*
* #return an array containing <code>nValues</code> hex chars
*/
public static char[] createRandomHexValues(int nValues) {
char[] ret = new char[nValues];
for (int i = 0; i < nValues; i++) {
ret[i] = HEX_VALUES[RANDOM.nextInt(HEX_VALUES.length)];
}
return ret;
}

In fact mentioned methods don't generate real random char. To generate real random char you should give it a random seed! in example time in millisecond. this code generate 10 random char and then Convert it to String:
import java.util.Random;
public class MyClass {
public static void main() {
String randomKey;
char[] tempArray={0,0,0,0,0,0,0,0,0,0}; //ten characters
long seed=System.currentTimeMillis();
Random random=new Random(seed);
for (int aux=0; aux<10;aux++){
tempArray[aux]=(char) random.nextInt(255);
System.out.println(tempArray[aux]);
}
randomKey=String.copyValueOf(tempArray);
System.out.println(randomKey);
}
}

I use this:
char uppercaseChar = (char) ((int)(Math.random()*100)%26+65);
char lowercaseChar = (char) ((int)(Math.random()*1000)%26+97);

java.util.Random is the more effective one I have tried out so far, having a precision of 98.65% uniqueness.
I have provided bellow some tests which generate 10000 batches of a hundred 2 alphanumeric chars strings and calculates the average.
Other random tools were RandomStringUtils from commons.lang3 and java.util.Math.
public static void main(String[] args) {
int unitPrintMarksTotal = 0;
for (int i = 0; i < 10000; i++) {
unitPrintMarksTotal += generateBatchOfUniquePrintMarks(i);
}
System.out.println("The precision across 10000 runs with 100 item batches is: " + (float) unitPrintMarksTotal / 10000);
}
private static int generateBatchOfUniquePrintMarks(int batch) {
Set<String> printMarks = new HashSet<>();
for (int i = 0; i < 100; i++) {
printMarks.add(generatePrintMarkWithJavaUtil());
}
System.out.println("Batch " + batch + " Unique number of elements is " + printMarks.size());
return printMarks.size();
}
// the best so far => 98.65
// with 3 chars => 99.98
// with 4 chars => 99.9997
private static String generatePrintMarkWithJavaUtil() {
int leftLimit = 48; // numeral '0'
int rightLimit = 122; // letter 'z'
int targetStringLength = 2;
String printMark;
do {
printMark = new Random().ints(leftLimit, rightLimit + 1)
.filter(i -> (i <= 57 || i >= 65) && (i <= 90 || i >= 97))
.limit(targetStringLength)
.collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
.toString();
} while (!isValid(printMark));
return printMark;
}
// 95.46
private static String generatePrintMarkWithCommonsLang3() {
String printMark;
do {
printMark = RandomStringUtils.randomAlphanumeric(2).toUpperCase();
} while (!isValid(printMark));
return printMark;
}
// 95.92
private static String generatePrintMarkWithMathRandom() {
final String ALPHA_NUMERIC_STRING = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
StringBuilder printMark;
do {
printMark = new StringBuilder();
int i = (int) (Math.random() * ALPHA_NUMERIC_STRING.length());
printMark.append(ALPHA_NUMERIC_STRING.charAt(i));
int j = (int) (Math.random() * ALPHA_NUMERIC_STRING.length());
printMark.append(ALPHA_NUMERIC_STRING.charAt(j));
} while (!isValid(printMark.toString()));
return printMark.toString();
}
private static boolean isValid(final String printMark) {
return true;
}

I simply want to tell you to use UUID.randomUUID() and toString() as its simplest way i also used

If you don't mind adding a new library in your code you can generate characters with MockNeat (disclaimer: I am one of the authors).
MockNeat mock = MockNeat.threadLocal();
Character chr = mock.chars().val();
Character lowerLetter = mock.chars().lowerLetters().val();
Character upperLetter = mock.chars().upperLetters().val();
Character digit = mock.chars().digits().val();
Character hex = mock.chars().hex().val();

public static void main(String[] args) {
// System.out.println("Enter a number to changeit at char ");
Random random = new Random();
int x = random.nextInt(26)+65; //0 to 25
System.out.println((char)x);
}

Random randomGenerator = new Random();
int i = randomGenerator.nextInt(256);
System.out.println((char)i);
Should take care of what you want, assuming you consider '0,'1','2'.. as characters.