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submitted 2 days ago by solo@slrpnk.net to c/technology@beehaw.org
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[-] wicked@programming.dev 3 points 1 day ago

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).
x
x = (-(1.0-1e-15))^2 = (1.0-1e-15)^2 = 1.0 - 2e-15 + 1e-30.
yy = (-5e-8)^2 = 2.5e-15.
x
x + 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.

this post was submitted on 10 May 2025
125 points (100.0% liked)

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