Hmm, there's something wrong with the [fexpr~] in my first post because if I trigger it on every sample (i.e. send [sig~ 1] to the [pd srNoise]), the result sounds and looks like it has a short period.
01-260606_0606.wav
Oddly, if I modify it to use the 13th and 14th taps, it sounds more like what I would expect from using the 14th and 15th taps. What am I misunderstanding/doing wrong?

Edit: oh wow, check this out. When I rewrite this patch to record the value of the feedback and shift register at each step, I see that the fb is getting shifted into the register one sample late, shown here on line 14 of each array's List View.

srNoiseFexpr~.pd
But when I step through the patch manually and snapshot~ those same values, the fb value is shifted into the register exactly when I expect, the equivalent of line 13. What the....?

Edit 2: Oh interesting: even though $y2's output expression is evaluated before $y1's, the value of $y2[-1] is not updated until all the expressions are evaluated. Which means that Edit 2 of my first post was right but for the wrong reason. Or I could still be completely wrong.

Last edit I promise because I'm just digging myself deeper and deeper into a false statement hole: my [fexpr~] can be fixed by substituting the 2nd expression into the first, i.e.

fexpr~ if($x1 == 1, ($y1 >> 1) | (((($y1 >> 1) & 1) ^ ($y1 & 1)) << 14), $y1);
(($y1 >> 1) & 1) ^ ($y1 & 1);
(($y1 & 1) * 2) - 1;

You can then delete the 2nd expression because it's only used for debugging. Unless someone corrects me, I still believe that output variable values like $y2 are not updated until all expressions are evaluated. I'm really not sure if it makes sense to talk about the order of expression evaluation, except for when values are referenced. So now I'll stop obsessing and will eagerly wait to be properly skooled.

Or this
control rate version.pd

@ice-ice I have to correct a false statement I made.
@jameslo said:

Edit 2: Hmm, there might be something wrong with the expression for the first outlet of fexpr~. See how it references the last value of the 2nd outlet? I don't think that value is calculated yet, and so it's using 1 sample old data. Maybe I should've xor'd the 2 bits that are 1 bit to the left?

Nope. When [fexpr~] has multiple expressions, they're evaluated bottom to top, i.e. the outlets are evaluated right to left. Therefore, $y2 is calculated before it is potentially used to calculate $y1.

@Ice-Ice Ha! I went down this same rabbit hole but I was thinking about the TR-909
https://electricdruid.net/tr-909-noise-generator/
and I didn't use fexpr~ or expr~. The algorithm definitely requires feedback.
TR909 LFSR.zip
It might not be the most efficient way, but I like how the patch looks like the circuit.

Edit: Thanks for the fun problem. I tried to make the example in the last link you posted using [fexpr~], see what you think.
srNoiseFexpr~.pd (use [metro~] from the zipfile I uploaded earlier)

Edit 2: Hmm, there might be something wrong with the expression for the first outlet of fexpr~. See how it references the last value of the 2nd outlet? I don't think that value is calculated yet, and so it's using 1 sample old data. Maybe I should've xor'd the 2 bits that are 1 bit to the left?

@xaverius Because Pd evaluates the control graph depth first, and you happened to connect the left inlet of [_pan] to the right spigot before you connected it to the left spigot. That means Pd will go as deep as it can down the right side (including the feedback on the outside of the abstraction) before it evaluates the left side. That depth-first plunge stops at [delay 1] because the whole point of [delay 1] is to schedule the evaluation of the rest of the graph for 1 ms later. With this understanding, print out what you're sending to the right inlet of [_pan] (both with the delay and without) and I think you'll see what's happening.

Personally, I wouldn't do this with feedback, just with a toggle and select, e.g.

@jorbecke I am not qualified to opine on your mental health, but if you are referring to the hurleur links going nowhere, then it might be a Google Chrome thing. Is that what you're using? If so, try another browser. Otherwise I'd have more modest expectations of a 13 year old thread

@y0g1 My friend who uses Max/MSP just explained to me that phasor~ resets when the right input signal changes to a non-zero value, which means that in my fexpr~ version the test should be "$x != 0 && $x != $x[-1]"

Edit: argh, I even misunderstood my friend. @lacuna is correct below--it's when the signal changes to a non-zero value from zero

@y0g1 said:

"phasor~"'s phase is reset when receiving a non zero signal.

If this was a test in college I'd be there during office hours arguing that the question was misleading

@y0g1 Assuming the sample frequency is 44.1khz, the phasor frequency is fixed at 1, and you don't need to support any other feature than what you mentioned above, try [fexpr~ if($x != 0, 0, ($y + 1) % 44100)] to replace both the phasor and following multiplication.

@playinmyblues The object named a12 should be an array, not a canvas. Replace it and things should start working.