Graphics & simulation

The rightmost matrix acts first. A normal must remain perpendicular to every transformed tangent.

T=[[1,0,3],[0,1,1],[0,0,1]] and R=[[0,−1,0],[1,0,0],[0,0,1]]. Rp=(0,2,1), so TRp=(3,3,1). Tp=(5,1,1), so RTp=(−1,5,1). Their difference demonstrates that transform order changes the result. For the direction, Td=d because its homogeneous coordinate is zero; both TRd and RTd equal (0,2,0).

If nᵀt=0 and a nonsingular linear transform sends tangent t to Mt, choose n′=M⁻ᵀn. Then n′ᵀMt=nᵀM⁻¹Mt=nᵀt=0. Applying M directly to both vectors does not generally preserve perpendicularity under nonuniform scaling. Renormalize the transformed normal when unit length is required. Singular transforms have no inverse and need separate treatment; the formula cannot be applied blindly to a scale that collapses an axis.

Check your reasoning

• State and follow the column-vector convention.
• Distinguish points from directions.
• Derive normal transformation from orthogonality.

Solve A+t(B−A) at z=1. For perspective correction interpolate u/w and 1/w, then divide.

The segment has z=0.5+1.5t, so t=1/3. The intersection is (2/3,0,1), which projects to (2/3,0). Clipping avoids treating geometry behind or too close to the camera as ordinary projected points; dividing near zero can produce unbounded coordinates and incorrect visibility.

Naive u is 1/4. Perspective-correct u is [Σλᵢuᵢ/wᵢ]/[Σλᵢ/wᵢ]=(1/8)/(1/2+1/8+1/16)=(1/8)/(11/16)=2/11≈0.1818. The screen-space weights are not the original surface's affine weights after perspective division. Depth interpolation follows the pipeline's defined representation and must not be confused with arbitrary attribute interpolation.

Two adjacent triangles should assign a sample on their shared edge consistently, usually to exactly one primitive under a documented fill rule. If both exclude it, a crack appears; if both include it, blending or stencil effects may double-count it. Numerical precision, winding, and edge rules must agree across the implementation.

Check your reasoning

• Clip in the specified camera space before division.
• Compute numerator and denominator separately.
• Explain shared-edge ownership and numerical consistency.

Black and white correspond to linear values zero and one. A pixel estimates an area or filtered integral, not necessarily the color at its center.

Direct encoded averaging gives 0.5. Linear-light averaging gives L=0.5, then E(0.5)≈0.73536. An encoded value of 0.5 decodes to roughly 0.21404 linear light, substantially darker than half the light. The correct operation depends on what is being combined, but physical light contributions and ordinary radiometric filtering should be handled in an appropriate linear representation before display encoding.

A high-frequency checkerboard can place every pixel-center sample on the same color or produce a low-frequency pattern unrelated to the true area average. Different continuous images can generate identical point samples, so a later contrast adjustment cannot recover the discarded distinctions. Use suitable prefiltering or multiple well-distributed samples to estimate the pixel footprint, with reconstruction choices matched to the signal and display. More samples reduce some errors but do not remove every bias or guarantee a sharp alias-free result at arbitrary resolution. Test phase shifts of the pattern, not just one conveniently aligned image.

Check your reasoning

• Compute the linear-light midpoint and encoded result.
• Explain aliasing as information loss, not only blur.
• Test sampling phase and state filtering assumptions.

Study the multiplier's magnitude and sign separately. The continuous solution remains positive even when a numerical sequence alternates.

Magnitude decays when |1−2h|<1, giving 0<h<1. Nonnegative monotone decay occurs for 0<h≤0.5; at h=0.5 the method reaches zero in one step, which is stable but not the exact positive solution. For 0.5<h<1 it alternates sign while shrinking. At h=1 the magnitude stays constant and the sign alternates, so it does not reproduce physical decay. For h>1 the magnitude grows.

At h=0.25 the sequence after each update is 0.5,0.25,0.125,0.0625, at times 0.25,0.5,0.75,1. The exact value at time one is exp(−2)≈0.13534. At h=1.1 the multiplier is −1.2 and the first four values are −1.2,1.44,−1.728,2.0736, while the exact solution rapidly decays. Local truncation error describes a single step started from the exact state; global behavior also depends on propagation of errors and the method's stability region. Stiff systems can require impractically small explicit steps even when the desired solution changes slowly.

Check your reasoning

• Separate magnitude stability from sign behavior.
• Compare at equal physical times.
• Explain error propagation and the step-size restriction.