I pasted 1k line C++ file into Gemini, along with a screenshot and a trace log and asked it to find the bug. It reasoned for about 5 minutes. Extract of the solution:

There has to be an issue in this sequence for i=21:
V_to_next_xy (norm) is (5.00004113497265e-08, -0.999999999999999, 0). (Correct)
Unnormalized perpendicular P_un = Vec3(V_to_next_xy.y(), -V_to_next_xy.x(), 0.0) should be (-0.999999999999999, -5.00004113497265e-08, 0).
temp_bisector becomes P_un.
temp_bisector.normalize() is called.
The result of this normalization is logged as temp_bisector (result of collinear path): (-1.78813934326169e-07, -0.999999999999984, 0).
This step (5) is where the numerical error occurs. The vector (-0.999…, -5e-8, 0) when normalized should not become (-1.78e-7, -0.999…, 0).
The latter is almost parallel to (0, -1, 0), while the former is almost parallel to (-1, 0, 0).
It’s as if temp_bisector.normalize() for the input (-0.999…, -5e-8, 0) is producing something wildly incorrect, or the vector being normalized was actually (-1.78e-7 * L, -0.999… * L, 0) for some L.
What if temp_bisector = Vec3(V_to_next_xy.y(), -V_to_next_xy.x(), 0.0); itself has precision issues?
If V_to_next_xy.y() is -0.999999999999999 (16 nines), this is -(1.0 - 1e-15).
If V_to_next_xy.x() is 5.00004113497265e-08.
Vec3 tmp = Vec3(-(1.0-1e-15), -5.00004113497265e-08, 0.0).
This should be fine.
The most plausible candidate is Vec3::normalize() failing on this specific input vector (-0.999999999999999, -5.00004113497265e-08, 0).
Perhaps length() computes sqrt(xx + yy + zz).
xx = (-(1.0-1e-15))^2 = (1.0-1e-15)^2 = 1.0 - 2e-15 + 1e-30.
yy = (-5e-8)^2 = 2.5e-15.
xx + y*y = 1.0 - 2e-15 + 1e-30 + 2.5e-15 = 1.0 + 0.5e-15 + 1e-30.
length = sqrt(1.0 + 0.5e-15 + 1e-30). This is extremely close to 1.0.
If this length calculation has any error causing it to be, for example, abs(y) by mistake, then x/length would become x/abs(y) and y/length would become y/abs(y) = sign(y).
x/abs(y) = -1.0 / 5e-8 = -2e7. This is not what’s happening.
The logged value for temp_bisector (-1.78813934326169e-07, -0.999999999999984, 0)
is suspicious because its X component -1.78813934326169e-07 is very close to the X component of V_to_prev_xy (norm) from loop i=21, which is -5.00004113497265e-08, divided by a factor. Or perhaps V_to_next_xy.x() itself was somehow used more directly than -V_to_next_xy.x() in the Y component of the vector passed to normalize.
The logged temp_bisector (result of collinear path) for i=21 is (-1.78813934326169e-07, -0.999999999999984, 0).
Consider Loop i=20: temp_bisector (result of collinear path): (1, 0, 0). Its input for perpendicular construction was V_to_next_xy (norm) = (X_v: 4.9999318885022e-08, Y_v: -0.999999999999999, Z_v: 0).
Perpendicular from this is (Y_v, -X_v, 0) = (-0.999…, -4.999…e-8, 0). Normalizing this gives (-1, -4.999…e-8, 0).
The log for i=20 says it becomes (1,0,0). This is also wrong. It should be (-1, -very_small_y, 0).
The fallback temp_bisector = Vec3(1,0,0) only happens if temp_bisector.normalize() (the one on the perpendicular) also fails.
It seems the normalize() call for the vector constructed as Vec3(V_to_next_xy.y(), -V_to_next_xy.x(), 0.0) is problematic when one component is very small.

It correctly identified that sqrt(_v[0]*_v[0] + _v[1]*_v[1] + _v[2]*_v[2]); had too low precision and using std::hypot(_v[0], _v[1], _v[2]) would likely solve it.

If this is just autocomplete, then I agree that it’s a pretty fancy one.

They installed some sort of sound isolating suspended/floating (?) ceiling in my apartment. I absolutely love it. In my neighbors apartment I could constantly hear people above, but in mine it’s almost always silent.

The US started using mailboxes 14 years after the UK

In 1849, the Royal Mail first encouraged people to install letterboxes to facilitate the delivery of mail. Before then, letterboxes of a similar design had been installed in the doors and walls of post offices for people to drop off outgoing mail.

In 1863, with the creation of Free City Delivery, the US Post Office Department began delivering mail to home addresses.

You could make the same argument for voting. What does your little drop in the vote bucket matter? Do you believe voting is a waste of time too?