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Engel coding

What most people call Huffman coding is technically prefix coding. It predates Huffman's algorithm and often doesn't employ Huffman's algorithm. Catchy names tend to be stickier than implementation details.

Fast huffman coders tend to use length-limited codes. For example, Huff0 uses a 12-bit limit. As a result, it can use a 4096-entry table, which nicely fits into L1 cache and (on 64bit) can decode 4 symbols before refilling the shift register. If it allowed longer symbols, the decode would be slower.

But once you enforce a length limit, you can no longer use Huffman's algorithm as-is. Basically you are left with three options.

  1. Use the Package Merge algorithm. It is optimal, but somewhat hard to understand and there is currently no open-source implementation anywhere.
  2. Use Huffman's algorithm, then shorten long symbols to not exceed your limit (violating the Kraft invariant), then heuristically adjust other symbols to compensate and return to K=1. Huff0 does this.
  3. Use a heuristic without Huffman's algorithm as a starting point. Using Huffman to get an optimal intermediate step, then throwing away the optimality with a heuristic is a bit pointless, after all. Polar coding is an example of this.

For my own Huffman coder, I picked option 3. My encoder is faster than others, at least in my own benchmarks, so the heuristic I picked appears to have some merit. It also appears to be something new. Charles Bloom hasn't found a prior description and he coined the term Engel coding for it.


Step 1: Sort symbols by histogram.
Step 2: Create initial bitlen for each symbol.
Step 3: Calculate credit/debt.
Step 4: Use up credit/repay debt.

Sorting symbols has known solutions. Yann's version in Huff0 is pretty good. Initial bitlen is round(-log2(p(sym))), so a sym with p=.5 gets bitlen=1, p=.25 gets bitlen=2, etc. Boundary between 1-bit and 2-bit symbols has 1.5 = -log2(p) or p = 2^(-1.5), about .353.

Floating point math is slow, logarithms are slow, so let's do everything with integer math. Because symbols are already sorted, we can decide that all symbols receive 1 bit until we hit a symbols with p <≈ .353. The next symbols receive 2 bits until we reach p<≈ .176, etc. We don't even need to calculate p if we pick a boundary based on histogram count.

static const uint64_t sqrt_2_32 = 1518500250;
int bit_boundary = (slen * sqrt_2_32) >> 32;

1518500250 is 2^(-1.5) << 32 or 2^(30.5). Large enough for good precision, small enough to avoid integer overflows. After multiplication and shift we have the boundary between 1-bit and 2-bit based on histogram and can do a simple comparison for each symbol.

You have to be careful with the rare symbols, so make sure nothing gets a bitlen larger than 12 (or whatever). Give symbols with p=0 a special value, I use MAX_BITS+1 in my code.

Calculating credit/debt is the same as calculating the Kraft number. But I find it nicer use table slots as units, i.e. with MAX_BITS=12 and 4096 table slots you multiply K with 4096. It nicely avoids floating point math and rounding errors.

Repaying debt

This is the heuristic part where you fix one of two problems. Either you use more slots than you have in your table (debt) or fewer (credit). Credit is simply inefficient, debt breaks the decoder.

To repay debt, you make a symbol longer. That reduces the slots by half, up to MAX_BITS, which uses a single slot. So you only want to consider symbols with 1..MAX_BITS-1 bits. And you only want to consider the cheapest symbols, i.e. the symbols where lengthening would do the least harm in terms of compression ratio.

Since your symbols are still sorted, the cheapest symbols necessarily are right at the boundary between N bits and N+1 bits. So you only have to consider MAX_BITS-1 symbols and pick the one that costs you the least per slot. Cost is calculated as

sym.cost = sym.hgram << sym.bitlen;

Finally, you don't want to repay too much. If you need to free 3 slots, lengthening a 1-bit symbols is a bit excessive. But you might want to lengthen a MAX_BITS-3 symbol and free 4 slots. Now you have repaid too much, but absolute debt (credit is just negative debt) has been reduced.

Dealing with credit is pretty much the same, except that nearly everything is off-by-one. You consider symbols 2..MAX_BITS, consider the symbol on the other side of the bit-boundary, etc.

Once your debt reaches 0, you are done.

Dealing with rare cases

You can run into cases where the algorithm above cannot terminate. If you have debt of 1, you need to lengthen a symbol with MAX_BITS-1. But you might not have such a symbol. You could lengthen two symbols with MAX_BITS, but that would be illegal. So really you have to go back and lengthen a symbol with MAX_BITS-2, MAX_BITS-3 or so.

As a result, your absolute debt may have to grow. If you don't allow that, you are stuck. If you do allow it, you will next pick the least-cost symbol to shorten, which is exactly the symbol you just lengthened, and have an infinite loop.

My solution is to fall back to a naïve heuristic, described by Charles here.

End result should be faster than using Huffman's algorithm and give a similar result, sometimes a little better, sometimes a little worse. Definitely beats Charles' heuristic, which afaics is used by Huff0 as well. Package-merge would be optimal, but I expect it to be noticeably slower.

UPDATE: Here is the source code. If you find any bugs or improvements, I would appreciate a comment. License should probably be 2-clause BSD, but I haven't talked to legal folks yet.