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| /**
* Get the power set of {0, 1, ..., n}.
*
* @param n the upper bound in the index set {0, 1, ..., n} to get the power set of
* @return the powerset including the empty set and the set itself as the first and last element, respectively
* @author Mikkel Meyer Andersen, scienco [at] mikl [dot] dk
*/
public static int[][] getPowerset(int n) {
if (n <= 0) {
throw new IllegalArgumentException("n must be be at least 1");
}
// stop at (2^n) - 1, so calculate 2^n
final int stop = 1 << n;
int i0 = 0, i1;
int[][] powerset = new int[stop][];
// Starting at i = 0 gives an empty first pair, so start at i = 1
for (int i = 0; i < stop; ++i) {
int j = i;
int k = i;
int size = 0;
// First calculate size of the array needed
for (int r = n - 1; r >= 0; --r) {
if ((j & 1) == 1) {
size++;
}
j >>>= 1;
}
i1 = 0;
int[] set = new int[size];
for (int r = n - 1; r >= 0; --r) {
if ((k & 1) == 1) {
set[i1++] = r;
}
k >>>= 1;
}
powerset[i0++] = set;
}
return powerset;
}
/**
* Partition an index set {0, 1, ..., n} into two parts A and B such that A
* union B is the whole set and A and B are disjoint. If you use this in
* commercial software, do contact me first. If not, feel free to use this
* code as long as you don't take credit for it yourself.
*
* @param n the upper bound in the index set {0, 1, ..., n} to get pairs of
* @return the pairs
* @author Mikkel Meyer Andersen, scienco [at] mikl [dot] dk
*/
public static int[][][] getPairs(int n) {
if (n <= 1) {
throw new IllegalArgumentException(
"n must be be at least 2 (it takes at least two elements to create a pair)");
}
// stop at 2^(n-1) - 1, so calculate 2^(n-1)
final int stop = 1 << (n - 1);
int i0 = 0, i1, i2;
int[][][] pairs = new int[stop - 1][][];
// Starting at i = 0 gives an empty first pair, so start at i = 1
for (int i = 1; i < stop; ++i) {
int j = i;
int k = i;
int size = 0;
// First calculate size of the array needed
for (int r = n - 1; r >= 0; --r) {
if ((j & 1) == 1) {
size++;
}
j >>>= 1;
}
i1 = 0;
i2 = 0;
int[] set1 = new int[size];
int[] set2 = new int[n - size];
for (int r = n - 1; r >= 0; --r) {
if ((k & 1) == 1) {
set1[i1++] = r;
} else {
set2[i2++] = r;
}
k >>>= 1;
}
pairs[i0++] = new int[][] {
set1, set2
};
}
return pairs;
}
/**
* Formats the power set of {0, 1, ..., n} obtained by {@link getPowerset}
* in a nicely looking string.
*
* @param powerset the powerset to format nicely
* @author Mikkel Meyer Andersen, scienco [at] mikl [dot] dk
*/
public static String powersetToString(int[][] powerset) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < powerset.length; ++i) {
int[] set = powerset[i];
for (int j = 0; j < set.length; ++j) {
sb.append(set[j]);
if (j < set.length - 1) {
sb.append(", ");
}
}
if (i < powerset.length - 1) {
sb.append("\n");
}
}
return sb.toString();
}
/**
* Formats the different pairs partitioning the index set of {0, 1, ..., n}
* obtained by {@link getPairs} in a nicely looking string.
*
* @param pairs the pairs to format nicely
* @author Mikkel Meyer Andersen, scienco [at] mikl [dot] dk
*/
public static String pairsToString(int[][][] pairs) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < pairs.length; ++i) {
int[] set1 = pairs[i][0];
int[] set2 = pairs[i][1];
sb.append("(");
for (int j = 0; j < set1.length; ++j) {
sb.append(set1[j]);
if (j < set1.length - 1) {
sb.append(", ");
}
}
sb.append("), (");
for (int j = 0; j < set2.length; ++j) {
sb.append(set2[j]);
if (j < set2.length - 1) {
sb.append(", ");
}
}
sb.append(")");
if (i < pairs.length - 1) {
sb.append("\n");
}
}
return sb.toString();
} |
Mikkel Meyer Andersen |
July 7, 2011 at 13:47 (UTC) |
