tepples wrote:

Related question: Is there a sequence of 32767-step and 93-step clocks that ends up in a single defined state?

Hm. From a probabilistic point of view, exactly 1/4 each of all 32768 states clock in a 0, a 1, the tonal noise mode bit "M", or "not M".

So ... it's not obviously the case that such a path will exist.

edit: but the real question is whether there's any usefully different probabilities between the state at a given time, and some finite number of clocks later. Obviously the only correlations would be after some combination of 8, 13, and 13-8 clocks (e.g. starting from state 256, which becomes either 512 or 513 depending on "M". When 512 or 513 hit 8192 or 8208, the value of the "M" bit will matter again:

256 -nM→ 512 → 1024 → 2048 → 4096 → 8192

256 -M→ 513 → 1026 → 2052 → 4104 → 8208

8192 -nM → 16385 → 3 → 6 → 12 → 24 → 48 → 96 → 192 → 384

8192 -M→ 16384 → 1 → 2 → 4 → 8 → 16 → 32 → 64 → 128 → 256

8208 -nM→ 16417 → 67 → 134 → 268

8208 -M→ 16416 → 65 → 130 → 260

But I don't think I have enough knowledge to handle this.

rainwarrior wrote:

The noise LFSR is seeded with

all zeroes on reset, AFAIK.

To be clear: the noise LFSR just happens to have all 0s on initial power up. A warm reset (or even an "insufficiently" cold boot) won't reset its contents.

edit:

tepples wrote:

I assume you want to reset the noise LFSR without using the

"test mode" at $4018-$401A. But even that allows only a triangle phase reset, not noise.

But it

*does* let you read out the current noise LFSR state, and because LFSRs, reading 15 bits from it will let you know what the current contents are, and knowing the current state lets you calculate a path to any desired state.