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;

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.