Friday, February 28, 2014

FSE encoding : Mapping sub-ranges in a memory efficient way

 With last blog post, we were left with a description of the FSE compression algorithm. However, the code associated was a bit cryptic. So let's spend some time understanding it.

Here are the lines of code :

nbBitsOut = (state + symbolTT.deltaNbBits) >> 16; flushBits(bitStream, state, nbBitsOut);
subrangeID = state >> nbBitsOut; state = stateTable[subrangeID + symbolTT.deltaFindState];

As suggested in an earlier blog post, the first task is to determine the number of bits to flush. This is basically one of 2 values, n or n+1, depending on state crossing a threshold.
symbolTT.deltaNbBits stores a value which, when added with state, makes the result of >> 16 produces either n or n+1, as required. It is technically equivalent to :

nbBitsOut = n;
if (state >= threshold) nbBitsOut += 1;

but as can be guessed, it's much faster, because it avoids a test, hence a branch.

The 2nd line just flushes the required nb of bits.

So we are left with the last 2 lines, which are more complex to grasp. It realises the conversion from newState to oldState (since we are encoding in backward direction). Let's describe how it works.

A naive way to do this conversion would be to create conversion tables, one per symbol, providing the destination state for each origin state. It works. It's just memory wasteful.
Consider for example a 4096 states table, for a 256 alphabet. Each state value uses 2 bytes. It results into 4K * 256 * 2 = 2 MB of memory. This is way too large for L1 cache, with immediate consequences on performance.

So we need a trick to reduce that amount of memory.
Let's have another look at a sub-range map :

Remember, all origin state values within a given sub-range have the same destination state. So what seems clear here is that we can simply reduce all origin state values by 9 bits, and get a much smaller map, with essentially the same information.

It's simple yet very effective. We now have a much smaller 8-state map for a symbol of probability 5/4096. This trick can be achieved with all other symbols, reducing the sum of all sub-range maps to a variable total between number of state and number of states x 2.

But we can do even better. Notice that the blue sub-ranges occupy 2 slots, providing the same destination state.
Remember that the red area corresponds to n=9 bits, and the blue area corresponds to n+1=10 bits.  What we just have to do then is to shift origin state by this amount of bits. Looks complex ? not really : we already have calculated this number of bits. Let's just use it now.

subrangeID = state >> nbBitsOut;

For this trick to work properly, state values must be scaled not from 0 to 4095, but from 4096 to 8191. Doing this, it results in the following sub-range map :

A few important properties to this transformation :
- There are as many cells as symbol probability.
- The first subrangeID is the same as symbol probability (in this example, 5).
- The sub-ranges are now stored in order (from 1 to 5). This is desirable, as it will simplify the creation of the map : we will just store the 5 destination states in order.
- Since sub-range maps have same size as symbol probability, and since the sum of probabilities is equal to the size of state table, the sum of all sub-ranges map is the size of state table ! We can now pack all sub-range maps into a single table, of size number of states.

Using again the previous example, of a 4096 states table, for an alphabet of 256 symbols. Each state value uses 2 bytes. We essentially now disregard the alphabet size, which has no more impact on memory allocation.
It results into 4K * 2 = 8 KB of memory, which is much more manageable, suitable for an L1 cache.

We now have all sub-range maps stored into a single common table. We just need to find the segment corresponding to the current symbol to be encoded. This is what symbolTT.deltaFindState does : it provides the offset to find the correct segment into the table.
Hence :
state = stateTable[subrangeID + symbolTT.deltaFindState];
This trick is very significant. It was a decisive factor in publishing an open source FSE implementation, as the initial naive version was unpractical, too memory hungry and too slow.

Tuesday, February 25, 2014

FSE encoding, Part 2

 In previous article, we learned how to determine the number of bits to write, and which bits to write, when encoding a symbol using FSE compression. We are therefore left with the conversion exercise, from newState to oldState (remember that we encode in reverse direction). This is the most important step of the compression process.

Remember that we said we could simplify the determination of nbBits by a simple threshold comparison, simply by looking at sub-ranges map, such as this one :

Well, in fact, thanks to this map, we know quite a bit more : we know in which exact sub-range fit the current newState.
Remember that we have numbered these sub-ranges, starting with the larger ones :

What we need to know, is the ordered position of symbols into the state table. Since we have 5 sub-ranges, it simply means we also have 5 symbols, directly associated.

And that's it, we know the sub-range, we know the destination state of the encoding process, oldState.

Encoding is therefore just a repeat of this process :
- get Symbol to encode
- look at current state value
- determine nbBits, flush them
- determine sub-Range Id
- look for Symbol position of same Id : you get your next state

If you look into FSE source code, you'll find that the following lines of code perform this job :

nbBitsOut = (state + symbolTT.deltaNbBits) >> 16; flushBits(bitC, state, nbBitsOut); state = stateTable[(state>>nbBitsOut) + symbolTT.deltaFindState];

If you feel that the correlation between the text and the code is not obvious, it's because it's not.
To reach this compact expression, a few non-trivial tricks have been used, reducing the amount of computation, and quite importantly, reducing the amount of memory required to describe the sub-ranges map.

These will be explained in a later post.