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Open HTML to PDF is a CSS 2.1 renderer written in Java. This artifact contains the core rendering and layout code.
package com.openhtmltopdf.util;
import java.math.BigInteger;
/**
* The PermutationGenerator Java class systematically generates permutations. It relies on the fact that any set with n
* elements can be placed in one-to-one correspondence with the set {1, 2, 3, ..., n}. The algorithm is described by
* Kenneth H. Rosen, Discrete Mathematics and Its Applications, 2nd edition (NY: McGraw-Hill, 1991), pp. 282-284.
*
* The class is very easy to use. Suppose that you wish to generate all permutations of the strings "a", "b", "c", and
* "d". Put them into an array. Keep calling the permutation generator's {@link #getNext ()} method until there are no
* more permutations left. The {@link #getNext ()} method returns an array of integers, which tell you the order in
* which to arrange your original array of strings. Here is a snippet of code which illustrates how to use the
* PermutationGenerator class.
*
* int[] indices;
* String[] elements = {"a", "b", "c", "d"};
* PermutationGenerator x = new PermutationGenerator (elements.length);
* StringBuffer permutation;
* while (x.hasMore ()) {
* permutation = new StringBuffer ();
* indices = x.getNext ();
* for (int i = 0; i < indices.length; i++) {
* permutation.append (elements[indices[i]]);
* }
* System.out.println (permutation.toString ());
* }
*
* One caveat. Don't use this class on large sets. Recall that the number of permutations of a set containing n elements
* is n factorial, which is a very large number even when n is as small as 20. 20! is 2,432,902,008,176,640,000.
*
* NOTE: This class was taken from the internet, as posted by Michael Gilleland on this website. The code was posted with the following comment: "The
* source code is free for you to use in whatever way you wish."
*
* @author Michael Gilleland, Merriam Park Software (http://www.merriampark.com/index.htm)
*/
public class PermutationGenerator {
private int[] a;
private BigInteger numLeft;
private BigInteger total;
//-----------------------------------------------------------
// Constructor. WARNING: Don't make n too large.
// Recall that the number of permutations is n!
// which can be very large, even when n is as small as 20 --
// 20! = 2,432,902,008,176,640,000 and
// 21! is too big to fit into a Java long, which is
// why we use BigInteger instead.
//----------------------------------------------------------
public PermutationGenerator(int n) {
if (n < 1) {
throw new IllegalArgumentException("Min 1");
}
a = new int[n];
total = getFactorial(n);
reset();
}
//------
// Reset
//------
public void reset() {
for (int i = 0; i < a.length; i++) {
a[i] = i;
}
numLeft = new BigInteger(total.toString());
}
//------------------------------------------------
// Return number of permutations not yet generated
//------------------------------------------------
public BigInteger getNumLeft() {
return numLeft;
}
//------------------------------------
// Return total number of permutations
//------------------------------------
public BigInteger getTotal() {
return total;
}
//-----------------------------
// Are there more permutations?
//-----------------------------
public boolean hasMore() {
return numLeft.compareTo(BigInteger.ZERO) == 1;
}
//------------------
// Compute factorial
//------------------
private static BigInteger getFactorial(int n) {
BigInteger fact = BigInteger.ONE;
for (int i = n; i > 1; i--) {
fact = fact.multiply(new BigInteger(Integer.toString(i)));
}
return fact;
}
//--------------------------------------------------------
// Generate next permutation (algorithm from Rosen p. 284)
//--------------------------------------------------------
public int[] getNext() {
if (numLeft.equals(total)) {
numLeft = numLeft.subtract(BigInteger.ONE);
return ArrayUtil.cloneOrEmpty(a);
}
int temp;
// Find largest index j with a[j] < a[j+1]
int j = a.length - 2;
while (a[j] > a[j + 1]) {
j--;
}
// Find index k such that a[k] is smallest integer
// greater than a[j] to the right of a[j]
int k = a.length - 1;
while (a[j] > a[k]) {
k--;
}
// Interchange a[j] and a[k]
temp = a[k];
a[k] = a[j];
a[j] = temp;
// Put tail end of permutation after jth position in increasing order
int r = a.length - 1;
int s = j + 1;
while (r > s) {
temp = a[s];
a[s] = a[r];
a[r] = temp;
r--;
s++;
}
numLeft = numLeft.subtract(BigInteger.ONE);
return ArrayUtil.cloneOrEmpty(a);
}
} // end class