Formula & theorem reference

For an image I and a kernel H of size (2a+1)\times(2b+1), the 2D discrete correlation is (I \star H)(x,y) = \sum_{i=-a}^{a}\sum_{j=-b}^{b} I(x+i,\, y+j)\, H(i,j), and the 2D discrete convolution is (I * H)(x,y) = \sum_{i=-a}^{a}\sum_{j=-b}^{b} I(x-i,\, y-j)\, H(i,j). The two differ by exactly one sign inside the image's index — a 180° rotation of the kernel — and coincide whenever the kernel is symmetric under that rotation. Both are a weighted sum over a window of input pixels centered at (x,y), with H supplying the weights.

The Prewitt operator computes the horizontal and vertical gradient components with H_x^P = \begin{bmatrix}-1&0&1\\-1&0&1\\-1&0&1\end{bmatrix}, \qquad H_y^P = \begin{bmatrix}-1&-1&-1\\0&0&0\\1&1&1\end{bmatrix}. Each factors as a central-difference kernel in one direction and a uniform box-average in the other — differentiate one way, smooth the other, in a single kernel. The Sobel operator uses the identical structure with a weighted average instead of a uniform one, H_x^S = \begin{bmatrix}-1&0&1\\-2&0&2\\-1&0&1\end{bmatrix}, weighting the center row (or column, for H_y^S) twice as heavily — a discrete approximation closer to Gaussian-weighted smoothing, giving Sobel a slightly better noise response than Prewitt at identical cost.

For an H\times W image f(x,y), the 2D discrete Fourier transform is F(u,v) = \sum_{x=0}^{H-1}\sum_{y=0}^{W-1} f(x,y)\, e^{-j2\pi\left(\frac{ux}{H}+\frac{vy}{W}\right)}, \qquad u=0,\dots,H-1,\ v=0,\dots,W-1. Each F(u,v) is one complex number; a full H\times W image produces a full H\times W grid of them, no information lost or gained. The inverse DFT recovers the image exactly: f(x,y) = \frac{1}{HW}\sum_{u=0}^{H-1}\sum_{v=0}^{W-1} F(u,v)\, e^{+j2\pi\left(\frac{ux}{H}+\frac{vy}{W}\right)}. Because the exponential kernel factors as a product of a purely-x term and a purely-y term, the 2D sum separates into a 1D DFT along rows composed with a 1D DFT along columns.

Otsu's method picks the threshold T that best separates a histogram into two classes, by maximizing the between-class variance \sigma_B^2(T) = w_0(T)\,w_1(T)\,\big(\mu_0(T)-\mu_1(T)\big)^2, where w_0,w_1 are the two classes' own population fractions (v\le T and v>T respectively) and \mu_0,\mu_1 their own mean intensities. Evaluating this at every candidate split and keeping the largest is exhaustive but cheap (at most 255 candidates for an 8-bit image), and requires nothing but the image's own histogram — no assumption about where the "right" cutoff should sit. Otsu's method still produces a single global threshold; it removes the guesswork from choosing T, not the uneven-lighting weakness of using one T for the whole image.

For a symmetric matrix A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}, the eigenvalues are the roots of the characteristic equation \det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad - b^2) = 0, namely \lambda = \frac{(a+d) \pm \sqrt{(a-d)^2 + 4b^2}}{2}. Because the discriminant (a-d)^2 + 4b^2 is a sum of squares, it is never negative — a symmetric matrix always has two real eigenvalues (counted with multiplicity), never a complex pair.

For a symmetric matrix A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}, the eigenvalues are the roots of the characteristic equation \det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad - b^2) = 0, namely \lambda = \frac{(a+d) \pm \sqrt{(a-d)^2 + 4b^2}}{2}. Because the discriminant (a-d)^2 + 4b^2 is a sum of squares, it is never negative — a symmetric matrix always has two real eigenvalues (counted with multiplicity), never a complex pair.

Around a candidate pixel, the structure tensor M=\sum_{(x,y)\in W}\begin{bmatrix}I_x^2 & I_xI_y\\ I_xI_y & I_y^2\end{bmatrix} is built from a window's own gradient products. Computing M's eigenvalues by hand takes more work than needed. The Harris and Stephens response function avoids it: R = \det(M) - k\,\operatorname{trace}(M)^2. Here, \det(M)=\lambda_1\lambda_2 and \operatorname{trace}(M)=\lambda_1+\lambda_2 combine M's two eigenvalues, and the constant k is chosen by hand, usually 0.04–0.06. This number, R, alone sorts the window into one of three classes. A positive R means both eigenvalues are large and close in size: a corner. A negative R means one eigenvalue is near zero: an edge. A small trace, regardless of R, means flat.

Suppose a dataset's true inlier fraction is w. That means a point picked at random is an inlier with probability w, independent of the others. RANSAC fits a model from a randomly sampled minimal set of n points, and repeats this for k independent iterations. The probability that at least one of those k samples is entirely inliers -- which means it produces a correct model fit -- is

This is only a probabilistic guarantee. For any finite k, P<1. RANSAC never certifies that its returned fit is correct. It only guarantees that a correct fit was found with probability P, under the stated assumptions.

Append a 1 to every point, (x,y)\to(x,y,1). That one extra coordinate lets any affine map p'=Ap+t (a 2\times 2 linear part A, plus a translation vector t) become a single 3\times 3 matrix multiply: \begin{bmatrix}x'\\y'\\1\end{bmatrix} = \begin{bmatrix}a&b&t_x\\c&d&t_y\\0&0&1\end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix}. Translation, scaling, rotation, reflection, and shear are each one specific choice of a,b,c,d,t_x,t_y in this same shape. The bottom row is always [0,0,1] for an affine map.

A homography acts on homogeneous coordinates. Its bottom row need not be [0,0,1]: \begin{bmatrix}x_h\\y_h\\w\end{bmatrix} = \begin{bmatrix}h_{00}&h_{01}&h_{02}\\h_{10}&h_{11}&h_{12}\\h_{20}&h_{21}&h_{22}\end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix}. Dividing out the third coordinate recovers the actual point: (x',y') = (x_h/w,\ y_h/w). For an affine map, w always equals 1, so the divide is invisible. A homography's one defining trait is that w genuinely varies with (x,y).

Take a point \mathbf{X} in space, seen by two calibrated cameras. "Calibrated" means known intrinsics, so image coordinates are already normalized to camera coordinates. Let \mathbf{x} and \mathbf{x}' be \mathbf{X}'s homogeneous projections in the first and second camera. Let the second camera's pose relative to the first be the rotation R and translation \mathbf{t}. Then

where E is the essential matrix. [\mathbf{t}]_\times is the skew-symmetric cross-product matrix of \mathbf{t}: it satisfies [\mathbf{t}]_\times \mathbf{v} = \mathbf{t}\times\mathbf{v} for every \mathbf{v}. (For uncalibrated cameras, the same constraint holds with the fundamental matrix F = K'^{-\mathsf T} E K^{-1} in place of E. It acts directly on pixel coordinates: \mathbf{x}'^{\mathsf T} F\, \mathbf{x} = 0.)

Take a point \mathbf{X} in space, seen by two calibrated cameras. "Calibrated" means known intrinsics, so image coordinates are already normalized to camera coordinates. Let \mathbf{x} and \mathbf{x}' be \mathbf{X}'s homogeneous projections in the first and second camera. Let the second camera's pose relative to the first be the rotation R and translation \mathbf{t}. Then

where E is the essential matrix. [\mathbf{t}]_\times is the skew-symmetric cross-product matrix of \mathbf{t}: it satisfies [\mathbf{t}]_\times \mathbf{v} = \mathbf{t}\times\mathbf{v} for every \mathbf{v}. (For uncalibrated cameras, the same constraint holds with the fundamental matrix F = K'^{-\mathsf T} E K^{-1} in place of E. It acts directly on pixel coordinates: \mathbf{x}'^{\mathsf T} F\, \mathbf{x} = 0.)

Take one axis of an input of size N, a kernel of size K, total padding P added along that axis (for example P=2 for one pixel of padding per side), and stride S. The output size along that axis is O = \left\lfloor \frac{N - K + P}{S} \right\rfloor + 1. Valid padding sets P=0. Same-size padding at stride 1 instead chooses P so that O=N, which requires P=K-1.

IoU and Dice are two different ways to score the same overlap question. IoU divides shared area by the union. Dice divides twice the shared area by the sum of the two areas. Because both formulas measure the same overlap, they never disagree about which region overlaps more. But they are not numerically interchangeable: Dice is always at least as large as IoU for the same pair of regions, since \text{Dice}=\dfrac{2\,\text{IoU}}{1+\text{IoU}}.

Precision and recall ask a genuinely different pair of questions. Precision is about the correctness of what was flagged; recall is about the completeness of what was found. The two trade off against each other as the confidence threshold moves. A detector's precision-recall curve is built by sweeping that single threshold and recomputing both precision and recall at every point. AP and mAP compress that whole curve into one number, so detectors can be ranked without picking one threshold by hand.

IoU and Dice are two different ways to score the same overlap question. IoU divides shared area by the union. Dice divides twice the shared area by the sum of the two areas. Because both formulas measure the same overlap, they never disagree about which region overlaps more. But they are not numerically interchangeable: Dice is always at least as large as IoU for the same pair of regions, since \text{Dice}=\dfrac{2\,\text{IoU}}{1+\text{IoU}}.

Precision and recall ask a genuinely different pair of questions. Precision is about the correctness of what was flagged; recall is about the completeness of what was found. The two trade off against each other as the confidence threshold moves. A detector's precision-recall curve is built by sweeping that single threshold and recomputing both precision and recall at every point. AP and mAP compress that whole curve into one number, so detectors can be ranked without picking one threshold by hand.

For nonzero vectors \mathbf{u}, \mathbf{v}, \theta = \cos^{-1}\!\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}\right). In particular, \mathbf{u}\cdot\mathbf{v} = 0 if and only if \mathbf{u} and \mathbf{v} are orthogonal.

IoU and Dice are two different ways to score the same overlap question. IoU divides shared area by the union. Dice divides twice the shared area by the sum of the two areas. Because both formulas measure the same overlap, they never disagree about which region overlaps more. But they are not numerically interchangeable: Dice is always at least as large as IoU for the same pair of regions, since \text{Dice}=\dfrac{2\,\text{IoU}}{1+\text{IoU}}.

Precision and recall ask a genuinely different pair of questions. Precision is about the correctness of what was flagged; recall is about the completeness of what was found. The two trade off against each other as the confidence threshold moves. A detector's precision-recall curve is built by sweeping that single threshold and recomputing both precision and recall at every point. AP and mAP compress that whole curve into one number, so detectors can be ranked without picking one threshold by hand.