Raster graphics

Raster graphics describe a picture by an array of pixels on a grid. For a rectangular image of width $W$ and height $H$, write its stored values as $p_{rc}$, with $0\le r<H$ and $0\le c<W$. A grayscale pixel can have one shade value; a colour pixel can have several component values. The grid specifies where values belong, while the encoding specifies what those values mean. A raster image therefore needs dimensions and an interpretation of its samples, not just a string of numbers.

The appealing feature is that pictures become explicit data: selecting a rectangle selects entries of an array; comparing two images compares their pixel values; image compression can exploit repeated or related values. Raster sampling and reconstruction explains why the stored grid and the displayed picture are different objects. A square drawn around a sample is one display convention, not a claim that the original scene consisted of little squares.

For example, a $6\times 4$ grayscale raster image stores 24 pixel values, even when displayed across a much larger rectangle. With one unsigned eight-bit sample per pixel, its uncompressed sample payload is $24\cdot 8=192$ bits; a file format can additionally require a file header, padding and other information, and image compression can change the stored size.

Enlarging that image can copy each pixel into a block through nearest-neighbor interpolation, or infer intermediate values through interpolation. Either procedure supplies a reconstruction rule; neither determines which unrecorded detail was present in the original. Vector graphics use a different representation, specifying geometric objects to be rendered onto a chosen grid. Re-rendering such a description and enlarging an existing raster image are different operations.

Jack Bresenham's raster graphics algorithm selects a thin sequence of pixels to represent a line segment by exact incremental integer arithmetic. For a first-octant segment with $0\le \Delta y\le \Delta x$, a scaled midpoint line decision chooses between the next same-row and next higher-row pixels.

For the implicit equation $f(X,Y)=\Delta y(X-x_0)-\Delta x(Y-y_0)$, the sign of $d$ compares the exact line with the midpoint between candidate pixel centres $(x+1,y)$ and $(x+1,y+1)$. Multiplication by two makes the test integral. A consistent equality rule specifies which tied pixel is selected.

A pixel is one picture element at an indexed position in a raster image. It can contain a single grayscale value or several colour components. The word can also refer to a physical display element; the indexed element of the stored image is the meaning used here. One stored pixel need not occupy exactly one physical display element when the image is resized. Microsoft's bitmap documentation likewise distinguishes an image's grid and its numerical colour representation.^[ref]

Separate position from value: raster sampling and reconstruction concerns the spatial grid, while sample depth concerns how many bits encode each component. Four times as many pixels can describe a finer grid without improving the quantization of any value. Conversely, more bits per sample can distinguish more shades without locating an edge more precisely.

In the binary example under run-length encoding, one bit per pixel selects black or white. In the grayscale example under lossy block averaging, eight bits per pixel select one of 256 codes. Calling them codes keeps their numerical values distinct from the physical light that a display emits.

The sample depth $b$ is the number of bits allocated to one stored sample. An unsigned $b$-bit integer has $2^b$ codes, $0,\dots ,2^b-1$. This is per component, not automatically per pixel: three eight-bit components require 24 payload bits per pixel. A raster image with $WH$ pixels and $k$ such components requires $WHkb$ payload bits before data compression, padding or a file header.

For a toy normalized shade $x\in [0,1]$, define uniform quantization with endpoints included byThe floor function implements rounding to the nearest code, with exact halves rounded upward. The reconstruction step is ${\delta }_b=1/(2^b-1)$, and $|{\hat{x}}_b-x|\le {\delta }_b/2$. For $b=2$, the levels are $0,1/3,2/3,1$; for $b=8$ they are $0,1/255,\dots ,1$. Increasing sample depth refines this value grid without changing the spatial grid. These levels are uniform in $x$; if $x$ is encoded sRGB, they are not uniform in relative luminance.

PNG uses a bit-depth field per sample or palette index. Its grayscale type without alpha permits 1, 2, 4, 8 or 16 bits per sample; its truecolour type permits 8 or 16 bits per component. These are format-specific restrictions, not a rule for every raster image.^[ref]

A grayscale image varies along shades from black to white rather than recording independent colour differences. A numerical grayscale image assigns one shade sample to each pixel; an encoding determines how that sample maps to a displayed shade. A colour raster image can also look grayscale, for example when its three sRGB components are equal.

The convention in our lossy block averaging example is black at the integer code 0 and white at 255. Its arithmetic means act on those stored codes. This is a fully specified numerical experiment, but it does not claim that code 128 means half the physical light of code 255. PNG explicitly allows samples that are not proportional to light: its grayscale interpretation depends on the indicated transfer curve, and can be device-dependent when that information is absent.^[ref]

Two different questions therefore need different operations. To average stored codes, average the integers. To average relative luminance for neutral sRGB shades, decode the codes to linear light, average there, and encode the result again. Equal weights on black and white give encoded value $1/2$ by the first method, but approximately $0.73536$ by the second. Rounded to eight-bit codes these are 128 and 188. This example concerns ideal neutral shades and the specified sRGB conversion; it does not assert a universal grayscale conversion for every colour encoding.

The upper panels hold the spatial grid fixed and change sample depth: four levels produce broad constant bands, while 256 levels make smaller numerical steps. The lower panels hold the colour interpretation fixed and compare encoded sRGB values with relative luminance. All panels are calculated synthetic examples, not measured display data. Grayscale is useful when shade structure is the quantity of interest: a synthetic scalar image signal can be inspected without introducing separate colour components.

A point-sampling model assigns $p_{rc}=f(x_c,y_r)$ to a pixel on a grid, where $f$ is a continuous function representing an image signal. An area-sampling model instead assigns the average value of a function over a specified footprint. These are different models: a narrow feature between point samples can be missed, whereas its contribution to an area sample depends on its width and shade. Neither model says that the recorded value uniquely determines everything inside a grid cell.

The upper panels start with exactly the same $6\times 4$ array. Nearest-neighbor interpolation repeats its values into a $24\times 16$ display; successive linear interpolations in the two coordinate directions produce a smoother $24\times 16$ display. Original sample positions are marked. Both displays use 384 output pixels, but only the same 24 recorded values. The extra values follow the chosen interpolation rule. Between-grid appearance is inferred, not independently recorded.

Spatial resolution needs a stated meaning. A grid over a rectangle of width $L_x$ and height $L_y$ has nominal cell pitches $\Delta x=L_x/W$ and $\Delta y=L_y/H$. Over the same rectangle, doubling both dimensions halves these pitches and quadruples the pixel count. That is a finer grid, not a guarantee of more recoverable detail. Blurring before sampling, quantization and aliasing can make different originals share the same recorded array. Enlarging the rectangle without adding samples instead increases the displayed pitch.

A reproducible computational demonstration makes the ambiguity exact. Take the one-dimensional profiles $f(x)=0.5+0.4\sin (2\pi x)$ and $g(x)=0.5+0.4\sin (10\pi x)$, repeated identically along every row. Sample at $x_j=j/4$ for $j=0,\dots ,8$. Since $10\pi x_j=2\pi x_j+2\pi j$, the fact that the sine is a periodic function gives $f(x_j)=g(x_j)$. Both yield $0.5,0.9,0.5,0.1,0.5,0.9,0.5,0.1,0.5$, although at $x=1/8$ they give approximately $0.78284$ and $0.21716$. The lower panel draws both originals and their common samples: no operation on those samples alone can identify the original profile. This is aliasing in an ideal point-sampling model, without image noise or rounding.

The Nyquist–Shannon sampling theorem describes reconstruction under a bandwidth restriction and ideal sampling assumptions; it does not grant uniqueness for arbitrary image signals from a finite raster image.^[ref] The beautiful lesson is constructive: choose what detail the grid should preserve before sampling, and choose a reconstruction rule suited to the information actually retained.