Why Linear Screen-Space Interpolation Fails

After projection and perspective division, triangle vertices live in normalized device coordinates. Attributes like color, texture coordinates, and normals are not linear in screen space under perspective projection. Linear interpolation stretches values near the camera and compresses them far away.

For example, a textured wall viewed at a steep angle shows visible warping if you interpolate UVs linearly across the triangle. PlayStation 1 titles skipped perspective correction to save cycles, and the swimming textures were a signature artifact.

The 1/w Trick

The fix: interpolate A/w and 1/w in screen space, then divide:

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Here u, v, w are screen-space barycentric coordinates and w0, w1, w2 are homogeneous w values from the projection matrix. Any attribute divided by w becomes linear in screen space; the final division corrects perspective distortion.

Linear interpolation: fast but wrong under perspective
Perspective-correct: standard in GPU hardware since 3dfx Voodoo
Software rasterizers: must implement this by hand

For This Exercise

You will compare linear and perspective-correct color interpolation at a test point with known barycentric coordinates and w values. This task asks you to print both results side by side so the difference is visible. Anyone writing a WebGL fallback or software renderer must implement this 1/w division themselves.

Your Task

Texture a quad drawn in perspective, such as a floor receding into the distance, with a checkerboard, twice:

once with affine texture coordinates (screen-space barycentrics, like the original PlayStation);
once perspective-correct (interpolating u/w, v/w and 1/w).

Print the two images side by side and count the pixels where they disagree. Everything is computed with exact integer and rational arithmetic, so the images are reproducible bit for bit.

Input

# starts a comment line.

c
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The quad is drawn as triangles (0, 1, 2) and (0, 2, 3).

Rules

Coverage: exactly as in the rasterizer tasks. Use pixel centres on a doubled grid (vertex (x, y) → (2x, 2y), centre (2px+1, 2py+1)) and integer edge functions E(a→b) = (bx-ax)(Py-ay) - (by-ay)(Px-ax). A triangle with negative area has vertices 1 and 2 swapped, and zero area is skipped. Zero edge values count only on top (ay == by && bx > ax) or left (by < ay) edges. A pixel belongs to the first triangle that covers it.
Weights: l0 = E(b→c), l1 = E(c→a), l2 = E(a→b) at the pixel centre.
Affine: u = Σ lᵢuᵢ / Σ lᵢ, and the same for v.
Perspective-correct: with L = lcm(w0, w1, w2) and kᵢ = lᵢ × (L / wᵢ), u = Σ kᵢuᵢ / Σ kᵢ, and the same for v. This is exactly Σ(lᵢuᵢ/wᵢ) / Σ(lᵢ/wᵢ).
Checker: cu = floor(u × N / 1000) and cv = floor(v × N / 1000), computed exactly from the fractions. The texel is # if cu + cv is odd, else ..

Output

c
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Header: affine padded to the canvas width, then | perspective-correct.
Rows: the affine row (uncovered pixels are spaces), then |, then the perspective row. Trailing spaces don't matter.
Summary: D of C covered pixel(s) differ.
Errors: canvas: W (1-30) H (1-30), checks: N (1-64), quad: 4 x (X Y W U V), W 1-20, U V 0-1000, render: define a quad first, unknown command X.

Example

Input:

c
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Output:

c
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Requirements

Keep u and v as exact fractions until the checker lookup. Floating-point rounding would flip texels on square boundaries.
When all four w are equal, the two methods must agree on every pixel. That is a good self-test.

Hidden tests cover equal w (no difference), strong foreshortening with fine checks, a degenerate quad that covers nothing, and argument errors.

Your Task

Texture a quad drawn in perspective, such as a floor receding into the distance, with a checkerboard, twice:

once with affine texture coordinates (screen-space barycentrics, like the original PlayStation);

once perspective-correct (interpolating u/w, v/w and 1/w).

Print the two images side by side and count the pixels where they disagree. Everything is computed with exact integer and rational arithmetic, so the images are reproducible bit for bit.

Rules

Coverage: exactly as in the rasterizer tasks. Use pixel centres on a doubled grid (vertex (x, y) → (2x, 2y), centre (2px+1, 2py+1)) and integer edge functions E(a→b) = (bx-ax)(Py-ay) - (by-ay)(Px-ax). A triangle with negative area has vertices 1 and 2 swapped, and zero area is skipped. Zero edge values count only on top (ay == by && bx > ax) or left (by < ay) edges. A pixel belongs to the first triangle that covers it.

Weights: l0 = E(b→c), l1 = E(c→a), l2 = E(a→b) at the pixel centre.

Affine: u = Σ lᵢuᵢ / Σ lᵢ, and the same for v.

Perspective-correct: with L = lcm(w0, w1, w2) and kᵢ = lᵢ × (L / wᵢ), u = Σ kᵢuᵢ / Σ kᵢ, and the same for v. This is exactly Σ(lᵢuᵢ/wᵢ) / Σ(lᵢ/wᵢ).

Checker: cu = floor(u × N / 1000) and cv = floor(v × N / 1000), computed exactly from the fractions. The texel is # if cu + cv is odd, else ..

Output

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Header: affine padded to the canvas width, then | perspective-correct.

Rows: the affine row (uncovered pixels are spaces), then |, then the perspective row. Trailing spaces don't matter.

Summary: D of C covered pixel(s) differ.

Errors: canvas: W (1-30) H (1-30), checks: N (1-64), quad: 4 x (X Y W U V), W 1-20, U V 0-1000, render: define a quad first, unknown command X.

Requirements

Keep u and v as exact fractions until the checker lookup. Floating-point rounding would flip texels on square boundaries.

When all four w are equal, the two methods must agree on every pixel. That is a good self-test.

Hidden tests cover equal w (no difference), strong foreshortening with fine checks, a degenerate quad that covers nothing, and argument errors.

Perspective-Correct Interpolation

Topic Overview

Why Linear Screen-Space Interpolation Fails

The 1/w Trick

For This Exercise

Implementation Task

Your Task

Input

Rules

Output

Example

Requirements

Solve it with a plan

Progressive Hints

Resources

Learn by building

Why Linear Screen-Space Interpolation Fails

The 1/w Trick

For This Exercise

Your Task

Input

Rules

Output

Example

Requirements