Hi.
[...]
+
+ /**
+ * Verifies that the iterator generates a lexicographically
+ * increasing sequence of b(n,k) arrays, each having length k
+ * and each array itself increasing.
+ *
+ * Note: the lexicographic order check only works for n < 10.
What about not using a fixed base (cf. below)?
+ *
+ * @param iterator
+ * @param n size of universe
+ * @param k size of subsets
+ */
+ private void checkIterator(Iterator<int[]> iterator, int n, int
k) {
+ long lastLex = -1;
+ long length = 0;
+ while (iterator.hasNext()) {
+ final int[] iterate = iterator.next();
+ Assert.assertEquals(k, iterate.length);
+ final long curLex = lexNorm(iterate);
^^^^^^^^^^^^^^^
Replace with
final long curLex = lexNorm(iterate, n);
+ Assert.assertTrue(curLex > lastLex);
+ lastLex = curLex;
+ length++;
+ for (int i = 1; i < iterate.length; i++) {
+ Assert.assertTrue(iterate[i] > iterate[i - 1]);
+ }
+ }
[...]
+
+ /**
+ * Returns the value represented by the digits in the input
array in reverse order.
+ * For example [3,2,1] returns 123.
+ *
+ * @param iterate input array
+ * @return lexicographic norm
+ */
+ private long lexNorm(int[] iterate) {
^^^
Replace with
private long lexNorm(int[] iterate, int n)
+ long ret = 0;
+ for (int i = iterate.length - 1; i >= 0; i--) {
+ ret += iterate[i] * ArithmeticUtils.pow(10l, (long) i);
^^^^^^^^^^^^^^
Replace with
ret += iterate[i] * ArithmeticUtils.pow(n, (long) i);
+ }
+ return ret;
+ }
+}
[...]
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