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Brilliant! Hadn't seen this technique before.
latency optimization is a skill. I liked how you went till CSE pass. I myself wrote several passes to go to lowest latency possible
This is really surprising! I've never considered the possibility that using an equality test to skip a write that would be a no-op could break a dependency and thus lead to higher perf overall if the "equal" outcome occurs often enough. This might be applicable in many situations where you "edit" some data in-place, but most of the time there are few or no changes.
I think this call for something similar to "__builtin_expect" or linux' likely()/unlikely().

Not very clean, but better than inserting obscure optimisations in the source.

That would be great, if only it worked as intended! From the perspective of an optimizing compiler, `a == b ? a : b` is worse than `b` regardless of the probability you assign to `a == b`.
I was wondering the same thing. Also, could profile-guided optimisation help here?
Interesting. I thought modern CPU optimisation required avoiding branches, but here adding the branch allows the branch pediction to parallelise what it otherwise couldn't.
It does and the key here is that adding the if is akin to avoiding a branch, since getting data then doing something with it is a hidden branch if you already have the data. All this code does is formalise the hidden branch so that it can be avoided when possible.
That's pretty cool. Is there something obcluding the compiler from noticing this parallelization opportunity without the new `if` ?

My understanding is the assignment and the evaluation are somehow coupled in this case based on the essay, but I could use an explanation.

The optimization in the post is only advantageous if `next_j[i][j] == j` holds often enough. Without prior knowledge, the compiler can't know if it's going to improve performance, and the worst losses are greater than the best wins (branch misprediction is very expensive), so it decides not to interfere.
> since getting data then doing something with it is a hidden branch if you already have the data

You mean that the naive, simpler code, despite not having an if, has a "branch" on the microarchitectural state? (which is like.. if we have this already in cache, do something. if not, do something else)

    // Find the optimal encoding for each symbol.
    // Chunk boundaries are located where encodings change.
    uint8_t encoding[n_symbols];
    uint8_t j = 0; // always start with encoding 0 for simplicity
    for (int i = 0; i < n_symbols; i++) {
        j = next_j[i][j];
        encoding[i] = j;
    }
my js brain keeps thinking encoding[i] = next_j[i][j];
wow, what an interesting optimization! how would you even figure out that that if is what's needed to make it faster?
I knew the loop was latency-bound and I couldn't easily decrease the latency, so I knew I had to somehow avoid the dependency chain at all. I remembered that CPUs predict some properties of memory accesses (e.g. they might predict that a store and then a load from different addresses likely don't intersect), but not addresses, so I thought about another way to force it to predict `j` well. Branch prediction turned out to be the simplest way to do so.

Actually, since then I've found out that I could reduce latency by replacing a load on the critical chain with a vector shuffle instruction (`pshufb`, takes just 1 cycle on x86). Ironically, if I realized that sooner, I probably wouldn't have tried to use branch prediction at all!

You can utilize MLP (memory-level parallelism) and reduce memory latency by reading whole 8 bytes of next_j at every iteration. Just do something like

`uint64_t packed_next_j = (uint64_t*)next_j[i]`

and then just shift right to find proper j. This way you remove dependency on previous iteration. Did you tried that?

The right-shift is the problem. You need to shift right by `j * 8`, which itself requires a shift to compute (`j << 3`), so you have two shifts on the critical path, resulting in a latency of 2 cycles. It's better than a load, but it's still noticeable.
2 cycles vs ~5 cycles is still nice improvement plus its more deterministic (but not that fast in optimistic case; better in pessimistic case) than `if` appraoch and probably easier to "invent" and reason about. Definitely not that cool as your `if` solution though :)
Don't you need the previous iteration to find out how much you need to shift right?
Yes you need, but the problem here isnt shifting, but loading from L1 into registers.