Can you encode to less bits when you don't need to preserve order?

问题

Say you have a List of 32-bit Integers and the same collection of 32-bit Integers in a Multiset (a set that allows duplicate members)

Since Sets don't preserve order but List do, does this mean we can encode a Multiset in less bits than the List?

If so how would you encode the Multiset?

If this is true what other examples are there where not needing to preserve order saves bits?

Note, I just used 32-bit Integers as an example. Does the data type matter in the encoding? Does the data type need to be fixed length and comparable for you to get the savings?

EDIT

Any solution should work well for collections that have low duplication as well as high duplication. Its obvious with high duplication encoding a Multiset by just simply counting duplicates is very easy, but this takes more space if there is no duplication in the collection.

回答1:

In the multiset, each entry would be a pair of numbers: The integer value, and a count of how many times it is used in the set. This means additional repeats of each value in the multiset do not cost any more to store (you just increment the counter).

However (assuming both values are ints) this would only be more efficient storage than a simple list if each list item is repeated twice or more on average - There could be more efficient or higher performance ways of implementing this, depending on the ranges, sparsity, and repetitive of the numbers being stored. (For example, if you know there won't be more than 255 repeats of any value, you could use a byte rather than an int to store the counter)

This approach would work with any types of data, as you are just storing the count of how many repeats there are of each data item. Each data item needs to be comparable (but only to the point where you know that two items are the same or different). There is no need for the items to take the same amount of storage each.

回答2:

If there are duplicates in the multiset, it could be compressed to a smaller size than a naive list. You may want to have a look at Run-length encoding, which can be used to efficiently store duplicates (a very simple algorithm).

Hope that is what you meant...

回答3:

Data compression is a fairly complicated subject, and there are redundancies in data that are hard to use for compression.

It is fundamentally ad hoc, since a non-lossy scheme (one where you can recover the input data) that shrinks some data sets has to enlarge others. A collection of integers with lots of repeats will do very well in a multimap, but if there's no repetition you're using a lot of space on repeating counts of 1. You can test this by running compression utilities on different files. Text files have a lot of redundancy, and can typically be compressed a lot. Files of random numbers will tend to grow when compressed.

I don't know that there really is an exploitable advantage in losing the order information. It depends on what the actual numbers are, primarily if there's a lot of duplication or not.

回答4:

In principle, this is the equivalent of sorting the values and storing the first entry and the ordered differences between subsequent entries.

In other words, for a sparsely populated set, only little saving can be had, but for a more dense set, or one with clustered entries - more significant compression is possible (i.e. less bits need to be stored per entry, possibly less than one in the case of many duplicates). I.e. compression is possible but the level depends on the actual data.