Task 4 of 5

Which Answers to Believe

Task 3 returned a number for every pixel it could solve, and some of those numbers are worthless. A useful tracker does not just answer — it says how much it should be believed, and the material for that is already sitting in the window matrix M = [[Sxx, Sxy], [Sxy, Syy]].

M's two eigenvalues measure how much intensity change the window sees in its two principal directions. Three cases, and they are the whole story:

        λmax     λmin    verdict
flat    ≈ 0      ≈ 0     nothing to track
edge    large    ≈ 0     one direction
corner  large    large   trustworthy

The middle row is task 2's aperture problem wearing a matrix. So the smaller eigenvalue is the confidence score: it is large only when the window is pinned down in both directions. That measure has a name — it is the Shi–Tomasi score, and thresholding it is literally what "good features to track" means. A 2×2 symmetric matrix has a closed-form spectrum, so this is one line of arithmetic, not an eigensolver:

trace = Sxx + Syy
det   = Sxx·Syy − Sxy²
λmin  = (trace − √(trace² − 4·det)) / 2

flat says nothing, an edge says half of it, only a corner says both — Three windows side by side. A flat patch with no gradients scores zero on both eigenvalues; a straight edge scores high on the larger and zero on the smaller; a corner scores high on both and is the only one marked trustworthy.

Goal: produce a 64×64 confidence map — the smaller eigenvalue of each pixel's 5×5 window matrix.

Requirements

Reuse the 5×5 window sums from task 3, but only the three you need: Sxx, Sxy, Syy
Output is plain 2D — output: [64, 64], one number per pixel
Return the smaller root: (trace - Math.sqrt(disc)) / 2, where disc = trace * trace - 4 * det
Clamp the discriminant with Math.max(0, …) before the square root — it is mathematically non-negative, but float32 rounding can nudge it below zero and hand you a NaN

Hint 1 — three sums, not five

It plays no part here. Confidence is a property of the window, not of the motion — you can decide a pixel is untrackable before you look at the second frame at all.

Hint 2 — the closed form

const trace = sxx + syy;
const det = sxx * syy - sxy * sxy;
const disc = Math.max(0, trace * trace - 4 * det);
return (trace - Math.sqrt(disc)) / 2;

Both roots share everything but a sign; + gives the larger eigenvalue, − the smaller. Take the smaller one — an edge scores well on the larger and is exactly what you are trying to reject.

Same idea elsewhere

Every corner detector is this matrix with a different scalar squeezed out of it: Shi–Tomasi takes λmin, Harris takes det − k·trace² to dodge the square root, FAST skips the matrix entirely and tests a pixel ring instead. OpenCV's goodFeaturesToTrack, the corner pass in ARKit and ARCore, and every visual-odometry front end run this per-pixel score and then keep the local maxima — which is a reduction, then a compaction, over a map you just computed in one launch.

All tasks in Optical Flow

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