A fern frond looks far too detailed to describe with a few numbers. Yet a point that jumps around the plane by one of four short coordinate rules, chosen at random at each step, leaves behind dots that fill in a convincing frond with a stem, a curling tip and leaflets on every scale. How can four rules do that? This page lists the four affine maps, shows what each one contributes, and separates what the maps decide from what the probabilities of choosing them decide.
Key terms
Barnsley fern
Definition Barnsley’s fern is the attractor of the iterated function system made of the four maps in the table. Each map sends a point to the point with the new coordinates shown.
| Map | New | New | Probability |
|---|---|---|---|
It is usually drawn with the chaos game. Start at the point . At each step choose one of the four maps at random, with the probabilities in the last column and independently of all earlier choices, apply it to the current point, and plot the result. All four maps are contractions, with ratios of about , , and , so by Hutchinson’s theorem exactly one nonempty, closed and bounded set satisfies
and that set is the fern. The fixed point of each map lies on the attractor, because applying that map again and again to a point of the attractor stays on the attractor and approaches the fixed point, which the closed attractor must then contain. The point is the fixed point of , so every plotted point lies on the fern. With probability the plotted points come arbitrarily close to every point of it. The probabilities only affect how quickly and how evenly its parts fill in. The fern is a geometric model that resembles a plant, not a description of how a living fern grows, which involves cells, light, water and genes that appear nowhere in these maps. Michael Barnsley presented it in his 1988 book Fractals Everywhere.
In plain words The map flattens the whole fern onto a short vertical segment, the stem at its base. The map shrinks the whole fern to about 85 percent of its size, turns it slightly clockwise and lifts it by , which gives the rest of the frond above the base stem and the lowest pair of leaflets. The maps and make small, squashed and turned copies of the whole fern, which become the lowest leaflet on the left and the lowest leaflet on the right. The leaflets higher up are copies of these two carried up the frond by repeated uses of , and every leaflet has smaller leaflets of its own for the same reason.
Example Start at and suppose the first three choices are , and . The point moves
The second step uses and . The two uses of climb the central stem, and then carries the point to the left, into the lowest left leaflet. The side map does not start a separate leaf. It moves the current point by the same fixed rule as always, and the next map acts on the new position in turn.
Affine transformation
Definition An affine transformation of the plane is a map that sends to the point with
for fixed numbers , , , , and . The numbers , , , form its linear part, which keeps the origin in place, and is the shift applied afterward. When the map sends straight lines to straight lines, parallel lines to parallel lines and squares to parallelograms, and it multiplies every area by . A negative value of means the map also flips the picture over, like a mirror. When it flattens the whole plane onto a line or a single point. The map is a similarity, which stretches or shrinks every segment by the same factor, exactly when and , or and .
In plain words An affine transformation first turns, stretches, squashes, shears or flips the plane around the origin, and then slides the result to a new place. Straight lines stay straight or collapse to points, so a square becomes a parallelogram, a segment or a point.
Example The map sends the corners , , and of the unit square to
the corners of a small parallelogram leaning to the left. Its area is . For the same number is , so its parallelogram has area and is a mirror image, which is why the right leaflet is a flipped copy of the fern. For the number is , and the square is flattened onto the segment from to .
Self-affine set
Definition A set is self-affine when it is the attractor of an iterated function system whose maps are affine transformations and contractions, so that it is made of finitely many affine copies of itself. It is self-similar when all of those maps are similarities, each shrinking every segment by the same factor and possibly turning or flipping. Every self-similar set is self-affine, but not every self-affine set is self-similar. Barnsley’s fern is self-affine, since it is the union of its four affine copies , , and . Its four maps are not all similarities, so the fern is not described here as made of exact scaled copies of itself. The simple dimension formula for sets built from similar copies, all shrunk by the same ratio, does not apply to it, and this page gives no value for the fern’s dimension.
In plain words Each leaflet is a squashed and turned copy of the whole fern rather than an exact miniature of it. The fern repeats itself at every scale, but with distortions.
Example An affine map stretches segments by a factor that can depend on their direction. For that factor lies between about and , and for between about and , so the two lowest leaflets are copies of the fern with changed proportions. The map is a similarity, since its coefficients satisfy and . It shrinks every segment by the factor , so is an exact copy of the whole fern, turned clockwise by about degrees. Each leaflet anywhere on the fern is the image of the whole fern under a chain of the four maps, applied one after another, and a chain of affine maps is again affine.
Map probabilities
Definition In the chaos game for an iterated function system , the map probabilities are numbers with every and
At each step the map is chosen with probability , independently of the earlier steps. For the fern they are , , and . Whichever positive values are used, with probability the plotted points come arbitrarily close to every point of the same attractor. The probabilities decide how often each map is used, since over a long run the fraction of steps that use approaches , and so they decide how densely the points cover each part of the attractor. If some is , that map is never used, and the points fall on the attractor of the remaining maps instead, which can be a different set.
In plain words The maps decide the shape, and the probabilities decide where the dots pile up while it is being drawn. A part that is rarely visited looks faint in a finite picture, but it is still part of the shape.
Example Give all four maps probability . Then one step in four uses , whose points lie on the vertical segment from to , so about a quarter of all dots pile up on the short base stem. At the other end, ten choices of in a row put the point into the set , the image of the fern under applied ten times. It is a copy of the fern shrunk by the factor that forms the tip of the frond. With the usual probabilities this happens on about percent of the steps, since . With equal probabilities it happens on only about one step in a million, since . So the upper frond stays faint for a long time, although the shape being drawn is exactly the same.
Check yourself
- Find the image of the unit square under . Show that it is again a square, and find its side and its area.
Show the answer
The corners , , and go to
The two sides leaving are the vectors and . Both have length , and their dot product is , so they are perpendicular. The image is a square with side about and area , which equals . Compared with the unit square it is turned clockwise by the angle whose tangent is , about degrees, and moved so that the corner lands at .
- Change the probabilities to for each map. Does the fern change? What happens instead if the stem map gets probability and the other three get , and ?
Show the answer
With all four probabilities positive the attractor does not change, because it is fixed by the maps alone. Only the density of the dots changes. About a quarter of them now land on the base stem and the upper frond fills in very slowly, so a picture with a limited number of points looks uneven.
With probability for the condition that every probability be positive fails. The stem map is never used, and the dots fall on the attractor of the three maps , and . This set lies inside the fern, because the fern is mapped into itself by those three maps, but it is a different set. Every point of lies higher than , while the fern contains , so the bottom of the stem is missing from the new picture. The new probabilities add up to , so the problem is not the sum but the zero.
- In , change the coefficient of in the new from to and keep everything else, including the probabilities. Is the new map still a contraction? Its fixed point is where the central stem ends, because the copies of the stem made by repeated uses of the map shrink toward it. Where is that point before and after the change?
Show the answer
Take two points whose difference is the vector . The new map sends them to two points whose difference is with , since the shift cancels. The length of is at most times the length of , so the image of every segment is at most times as long. The map is still a contraction, and Hutchinson’s theorem still applies. It is no longer a similarity, because while .
The fixed point solves and . The first equation gives , and the second then gives , so
The same steps with give , then and . The end of the central stem moves from about to about , so the frond leans much farther to the right. Here the probabilities did not change at all, yet the shape did, because a map changed.