I'm stuck, and I really don't know what to do ...arff...
so this is the code I found and I added the line you gave me. But it only writes the last permutation (4 3 2 1)
int[] indices;
String[] elements = {
"1 ", "2 ", "3 ", "4 "
};
PermutationGenerator x = new PermutationGenerator (elements.length);
StringBuffer permutation;
void setup() {
while (x.hasMore()) {
permutation = new StringBuffer ();
indices = x.getNext ();
for (int i = 0; i < indices.length; i++) {
permutation.append (elements[indices[i]]);
}
println (permutation.toString ());
saveStrings("data.txt", new String[ ] {permutation.toString()});
}
}
import java.math.BigInteger;
class PermutationGenerator {
int[] a;
BigInteger numLeft;
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.
//----------------------------------------------------------
PermutationGenerator (int n) {
if (n < 1) {
throw new IllegalArgumentException ("Min 1");
}
a = new int[n];
total = getFactorial (n);
reset ();
}
//------
// Reset
//------
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
//------------------------------------------------
BigInteger getNumLeft () {
return numLeft;
}
//------------------------------------
// Return total number of permutations
//------------------------------------
BigInteger getTotal () {
return total;
}
//-----------------------------
// Are there more permutations?
//-----------------------------
boolean hasMore () {
return numLeft.compareTo (BigInteger.ZERO) == 1;
}
//------------------
// Compute factorial
//------------------
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)
//--------------------------------------------------------
int[] getNext () {
if (numLeft.equals (total)) {
numLeft = numLeft.subtract (BigInteger.ONE);
return 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 a;
}
}