Talk:Multiplicative order

Clarification

There's more than one way to determine the multiplicative order. While my J is next to non-existant, I can tell you're not using one of the naive ones :-) As the point of Rosetta is to compare equal algorithms in different languages (and not different algorithms in different languages), please state in the task which algorithm you'd like to be used. Also, IMHO the tasks should be chosen to allow a meaningful comparison of languages, and I am not sure if this is a good task for this purpose, because it's more centered on the algorithm than on the possible different ways different languages work. Dirkt 03:57, 8 December 2007 (MST)

The J solution implements the algorithm described in Bach & Shallit, Algorithmic Number Theory I, exercise 5.8, page 118.

If you implement a specific algorithm, it would be nice to describe the algorithm for those who don't happen to have this book lying around.

Judging by the other programming tasks, the submissions in the various languages don't always implement the same algorithm, but use algorithms that are "natural" to the language. That too is a meaningful comparison. Roger Hui 08:28, 8 December 2007 (MST)

But they should implement the same algorithm as far as mathematical principles, say, are concerned. (And if they don't, it's bad, and it should be changed.) Using different mathematical principles to arrive at the same result has nothing to do with what's "natural" for one language. And that's exactly my criticism here: I don't see how this task brings out features that are "natural" to a language. Instead it highlights features of a different algorithmic approach. It's of course tempting to show off with a clever "deep" algorithm, but it misses somehow the point.

And J has enough interesting features that require a different "natural" approach than in other languages that it would be really worthwhile to pick tasks which highlight this. And if you'd add a bit of explanation to really point this out in the implementation, for those not familiar with J, all the better.

One thing this Wiki is really good for is to try to get up to speed in an unfamiliar language, by comparing the approaches with those in a familiar language. This doesn't work if all similarity is suppressed, because the algorithms are essentially incomparable.Dirkt 13:46, 8 December 2007 (MST)

Java solution

There is currently a mismatched left paren in the line:

for(;x.modPow(retVal, y) != BigInteger.ONE;retVal = retVal.add(BigInteger.ONE);

Roger Hui 21:27, 7 December 2007 (MST)

You're allowed to fix it too if you know how. No one will be angry. --Mwn3d 09:43, 8 December 2007 (MST)