On 8/25/13 7:47 AM, Gilles wrote:
> 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)?
Thanks for the review!
Yes, I think below would work. The key is the base, as you note.
Or being less lazy and just actually directly implementing
lexicographic order check. I did not see the need to test with n >
10; but I would be fine with the change below to allow it.
Phil
>
>> + *
>> + * @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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