|Problem Set 3: L-System Fractals||
Out: 5 February 2007
Due: Monday, 12 February 2007
Collaboration Policy - Read Carefully
For this problem set, you are required to work with an assigned partner. You will receive an email before midnight Monday containing the partner assignments. You and your partner should turn in one assignment with both of your names on it and both people must participate fully in all of the work and completely understand everything you submit. You should read the whole problem set yourself and think about the questions before beginning to work on them with your partner.
You may discuss this assignment with other students in the class and ask and provide help in useful ways. You may consult any outside resources you wish including books, papers, web sites and people except for materials from previous cs150 courses.
If you use resources other than the class materials, indicate what you used along with your answer.
In this problem set, you will explore a method of creating fractals known as the Lindenmayer system (or L-system). Aristid Lindemayer, a theoretical biologist at the University of Utrecht, developed the L-system in 1968 as a mathematical theory of plant development. In the late 1980s, he collaborated with Przemyslaw Prusinkiewicz, a computer scientist at the University of Regina, to explore computational properties of the L-system and developed many of the ideas on which this problem set is based.
The idea behind L-system fractals is that we can describe a curve as a list of lines and turns, and create new curves by rewriting old curves. Everything in an L-system curve is either a forward line (denoted by F), or a right turn (denoted by Ra where a is an angle in degrees clockwise). We can denote left turns by using negative angles.
We create fractals by recursively replacing all forward lines in a curve list with the original curve list. Lindemayer found that many objects in nature could be described using regularly repeating patterns. For example, the way some tree branches sprout from a trunck can be described using the pattern: F O(R30 F) F O(R-60 F) F. This is interpreted as: the trunk goes up one unit distance, a branch sprouts at an angle 30 degrees to the trunk and grows for one unit. The O means an offshoot — we draw the curve in the following parentheses, and then return to where we started before the offshoot. The trunk grows another unit and now another branch, this time at -60 degrees relative to the trunk grows for one units. Finally the trunk grows for one more unit. The branches continue to sprout in this manner as they get smaller and smaller, and eventually we reach the leaves.
We can describe this process using replacement rules:
Start: (F)Here are the commands this produces after two iterations:
Rule: F ::= (F O(R30 F) F O(R-60 F) F)
Iteration 0: (F)
Iteration 1: (F O(R30 F) F O(R-60 F) F)
Iteration 2: (F O(R30 F) F O(R-60 F) F O(R30 F O(R30 F) F O(R-60 F) F) F O(R30 F) F O(R-60 F) F O(R-60 F O(R30 F) F O(R-60 F) F) F O(R30 F) F O(R-60 F) F)
Here's what that looks like:
Iteration 5 (with color)
The Great Lambda Tree of
Infinite Knowledge and Ultimate Power
Note that L-system command rewriting is similar to the replacement rules in a BNF grammar. The important difference is that with L-system rewriting, each iteration replaces all instances of F in the initial string instead of just picking one to replace.
We can divide the problem of producing an L-system fractal into two main parts:
Here is a BNF grammar for L-system commands:
- CommandSequence ::= ( CommandList )
- CommandList ::= Command CommandList
- CommandList ::=
- Command ::= F
- Command ::= RAngle
- Command ::= OCommandSequence
- Angle ::= Number
We need to find a way to turn strings in this grammar into objects we can manipulate in a Scheme program. We can do this by looking at the BNF grammar, and converting the non-terminals into Scheme objects.
;;; CommandSequence ::= ( CommandList ) (define make-lsystem-command list) ;;; We represent the different commands as pairs where the first item in the ;;; pair is a tag that indicates the type of command: 'f for forward, 'r for ;;; rotate and 'o for offshoot. We use quoted letters to make tags, which ;;; evaluate to the quoted letter. The tag 'f is short for (quote f). ;;; Command ::= F (define (make-forward-command) (cons 'f #f)) ;; No value, just use false. ;;; Command ::= RAngle (define (make-rotate-command angle) (cons 'r angle)) ;;; Command ::= OCommandSequence (define (make-offshoot-command commandsequence) (cons 'o commandsequence))
You will find the following procedures useful:
You should be able to make up similar test cases yourself to make sure the other procedures you defined work.
> (is-forward? (make-forward-command))
> (is-forward? (make-rotate-command 90))
> (get-angle (make-rotate-command 90))
> (get-angle (make-forward-command))
Yikes! Attempt to get-angle for a command that is not an angle command
Expression ::= BeginExpressionThe evaluation rule for begin is:
BeginExpression ::= (begin MoreExpressions Expression)
Evaluation Rule 6: Begin. To evaluate (begin Expression1 Expression2 ... Expressionk), evaluate each sub-expression in order from left to right. The value of the begin expression is the value of Expressionk.The begin special form is useful when we are evaluating expressions that have side-effects. This means the expression is important not for the value it produces (since the begin expression ignores the values of all expressions except the last one), but for some change to the state of the machine it causes.
The special define syntax for procedures includes a hidden begin expression. The syntax,
(define (Name Parameters) MoreExpressions Expression)is an abbreviation for:
(define name (lambda (Parameters) (begin MoreExpressions Expression)))
Expression ::= LetExpressionThe evaluation rule for a let expression is:
LetExpression ::= (let (Bindings) Body)
Body ::= MoreExpressions Expression
Bindings ::= Binding Bindings
Binding ::= (Name Expression)
Evaluation Rule 7: Let. To evaluate a let expression, evaluate each binding in order. To evaluate each binding, evaluate the binding expression and bind the name to the value of that expression. Then, evaluate the body expressions in order with the names in the expression that match binding names subsituted with their bound values. The value of the let expression is the value of the last body expression.A let expression can be transformed into an equivalent application expression. The let expression
(let ((Name1 Expression1) (Name2 Expression2) ... (Namek Expressionk)) MoreExpressions Expression)is equivalent to the application expression:
((lambda (Name1 Name2 ... Namek) (begin MoreExpressions Expression)) Expression1 Expression2 ... Expressionk)The advantage of the let expression syntax is it puts the expressions next to the names to which they are bound. For example, the let expression:
(let ((a 2) (b (* 3 3))) (+ a b))is easier to understand than the corresponding application expression:
((lambda (a b) (+ a b)) 2 (* 3 3))
Expression ::= CondExpressionThe evaluation rule is:
CondExpression ::= (cond ClauseList)
ClauseList ::= Clause ClauseList
Clause ::= (ExpressionTest ExpressionAction)
Clause ::= (else ExpressionAction)
Evaluation Rule 8: Conditionals. To evaluate a CondExpression, evaluate each clause's test expression in order until one is found that evaluates to a true value. Then, evaluate the action expression of that clause. The value of the CondExpression is the value of the action expression. If none of the test expressions evaluate to a true value, if the CondExpression includes an else clause, the value of the CondExpression is the value of the action expression associated with the else clause. If none of the test expressions evaluate to a true value, and the CondExpression has no else clause, the CondExpression has no value.Note that a conditional expression could straightforwardly be translated into an equivalent if expression:
(cond (Test1 Action1) (Test2 Action2) ... (Testk Actionk) (else Actionelse))is equivalent to:
(if Test1 Action1 (if Test2 Action2 ... (if Testk Actionk actionelse)...))
Start: (F)To produce levels of the tree fractal, we need a procedure that takes a list of L-system commands and replaces each F command with the list of L-system commands given by the rule.
Rule: F ::= (F O(R30 F) F O(R-60 F) F)
So, for every command in the list:
For example, consider a simple L-System rewriting:
Start: (F)We want to get:
Rule: F ::= (F R30 F)
Iteration1: (F R30 F)but if we just replace F's with (F R30 F) lists, we would get:
Iteration2: (F R30 F R30 F R30 F)
Iteration1: ((F R30 F))The easiest way to fix this problem is to flatten the result. The code should look similar to many recursive list procedures you have seen (this code is provided in lsystem.scm):
Iteration2: ((F R30 F) R30 (F R30 F))
(define (flatten-commands ll) (if (null? ll) ll (if (is-lsystem-command? (car ll)) (cons (car ll) (flatten-commands (cdr ll))) (flat-append (car ll) (flatten-commands (cdr ll)))))) (define (flat-append lst ll) (if (null? lst) ll (cons (car lst) (flat-append (cdr lst) ll))))
Here's the easy part:
Complete the definition of rewrite-lcommands.(define (rewrite-lcommands lcommands replacement) (flatten-commands (map ; Procedure to apply to each command lcommands)))
To make interesting L-system curves, we will need to apply rewrite-lcommands many times. We will leave that until the last question. Next, we will work on turning sequences of L-system commands into curves we can draw.
We will use a coordinate system from (0, 0) to (1, 1):
|(0.0, 1.0)||(1.0, 1.0)|
|(0.0, 0.0)||(1.0, 0.0)|
Points have x and y coordinates. To represent points we would like to define procedures make-point, x-of-point and y-of-point. Our pictures will be more interesting if points can have color too. So, we represent a colored point using a list of three values: x, y and color:
Note that we have defined points so we can have both colored points and colorless points that appear black.(define (make-point x y) (list x y)) (define (make-colored-point x y c) (list x y c)) (define (is-colored-point? (= (length point) 3) (define (x-of-point point) (car point)) (define (y-of-point point) (cadr point)) ;; (cadr x) = (car (cdr x)) ;;; Regular points are black. Colored points have a color. (define (color-of-point point) (if (is-colored-point? point) (caddr point) ;; == (car (cdr (cdr point))) (make-color 0 0 0)))
We have provided some procedures for drawing on the window in graphics.scm:
(define (mid-line t) (make-point t 0.5))defines a curve that is a horizontal line across the middle of the window. If we apply mid-line to a value x, we get the point (x, 0.5). Hence, if we apply mid-line to all values between 0.0 and 1.0, we get a horizontal line.
Predict what (x-of-point (mid-line 0.7)) and (y-of-point (mid-line 0.7)) should evaluate to. Try them in your Interactions window.
Of course, there are infinitely many values between 0.0 and 1.0, so we can't apply the curve function to all of them. Instead, we select enough values to show the curve well. To draw a curve, we need to apply the curve procedure to many values in the range from 0.0 to 1.0 and draw each point it evaluates to. Here's a procedure that does that:
(define (draw-curve-points curve n) (define (draw-curve-worker curve t step) (if (<= t 1.0) (begin (window-draw-point (curve t)) (draw-curve-worker curve (+ t step) step)))) (draw-curve-worker curve 0.0 (/ 1 n)))The procedure draw-curve-points takes a procedure representing a curve, and n, the number of points to draw. It calls the draw-curve-worker procedure. The draw-curve-worker procedure takes three parameters: a curve, the current time step values, and the difference between time step values. Hence, to start drawing the curve, draw-curve-points evaluates draw-curve-worked with parameters curve (to pass the same curve that was passed to draw-curve-points), 0.0 (to start at the first t value), and (/ 1 n) (to divide the time values into n steps).
The draw-curve-worker procedure is defined recursively: if t is less than or equal to 1.0, we draw the current point using (window-draw-point (curve t)) and draw the rest of the points by evaluating (draw-curve-worker curve (+ t step) step)).
We stop once t is greater than 1.0, since we defined the curve over the interval [0.0, 1.0].
The draw-curve-worker code uses a being expression. The first expression in the begin expression is (window-draw-point (curve t)). The value it evaluates to is not important, what matters is the process of evaluating this expression draws a point on the display.
For example, the procedure rotate-ccw takes a curve and rotates it 90 degrees counter-clockwise by swapping the x and y points:
(define (rotate-ccw curve) (lambda (t) (let ((ct (curve t))) (make-colored-point (- (y-of-point ct)) (x-of-point ct) (color-of-point ct)))))We use a let expression here to avoid needing to evaluate (curve t) more than once. It binds the value (curve t) evaluates to, to the name ct.
Note that (rotate-ccw c) evaluates to a curve. The function rotate-ccw is a procedure that takes a procedure (a curve) and returns a procedure that is a curve.
Predict what (draw-curve-points (rotate-ccw mid-line) 1000) and (draw-curve-points (rotate-ccw (rotate-ccw mid-line)) 1000) will do. Confirm your predictions by trying them in your Interactions window.
Here's another example:
(define (shrink curve scale) (lambda (t) (let ((ct (curve t))) (make-colored-point (* scale (x-of-point ct)) (* scale (y-of-point ct)) (color-of-point ct)))))Predict what (draw-curve-points (shrink mid-line .5) 1000) will do, and then try it in your Interactions window.
The shrink procedure doesn't produce quite what we want because in addition to changing the size of the curve, it moves it around. Why does this happen? Try shrinking a few different curves to make sure you understand why the curve moves.
One way to fix this problem is to center our curves around (0,0) and then translate them to the middle of the screen. We can do this by adding or subtracting constants to the points they produce:
(define (translate curve x y) (lambda (t) (let ((ct (curve t))) (make-colored-point (+ x (x-of-point ct)) (+ y (y-of-point ct)) (color-of-point ct)))))Now we have translate, it makes more sense to define mid-line this way:
(define (horiz-line t) (make-point t 0)) (define mid-line (translate horiz-line 0 0.5))
In addition to altering the points a curve produces, we can alter a curve by changing the t values it will see. For example,
(define (first-half curve) (lambda (t) (curve (/ t 2))))is a function that takes a curve and produces a new curve that is just the first half of the passed curve.
Predict what each of these expressions will do:
The provided code includes several other functions that transform
It is also useful to have curve transforms where curves may be combined. An example is (connect-rigidly curve1 curve2) which evaluates to a curve that consists of curve1 followed by curve2. The starting point of the new curve is the starting point of curve1 and the end point of curve2 is the ending point of the new curve. Here's how connect-rigidly is defined:
(define (connect-rigidly curve1 curve2) (lambda (t) (if (< t (/ 1 2)) (curve1 (* 2 t)) (curve2 (- (* 2 t) 1)))))Predict what (draw-curve-points (connect-rigidly vertical-mid-line mid-line) 1000) will do. Is there any difference between that and (draw-curve-points (connect-rigidly mid-line vertical-mid-line) 1000)? Check your predictions in the Interactions window.
The way connect-rigidly is defined above, we use all the t-values below 0.5 on the first curve, and use the t-values between 0.5 and 1.0 on the second curve. If the second curve is the result of connecting two other curves, like (connect-rigidly c1 (connect-rigidly c2 c3)) then 50% of the points will be used to draw c1, 25% to draw c2 and 25% to draw c3.
(connect-rigidly c1 (connect-rigidly c2 (connect-rigidly curve3 (... cn))))Think about this yourself first, but look in ps3.scm for a hint if you are stuck. There are mathematical ways to calculate this efficiently, but the simplest way to calculate it is to define a procedure that keeps halving the number of points n times to find out how many are left for the nth curve.
Your num-points procedure should produce results similar to:
This means if we connected just 20 curves using connect-rigidly, and passed the result to draw-curve-points with one million as the number of points, there would still be only one or two points drawn for the 20th curve. If we are drawing thousands of curves, for most of them, not even a single point would be drawn!
> (exact->inexact (num-points 1000 10))
> (exact->inexact (num-points 1000 20))
> (exact->inexact (num-points 1000000 20))
To fix this, we need to distribute the t-values between our curves more fairly. We have provided a procedure connect-curves-evenly in graphics.scm that connects a list of curves in a way that distributes the range of t values evenly between the curves.
The definition is a bit complicated, so don't worry if you don't understand it completely. You should, however, be able to figure out the basic idea for how it distributed the t-values evenly between every curve in a list of curves.
It will also be useful to connect curves so that the next curve begins where the first curve ends. We can do this by translating the second curve to begin where the first curve ends. To do this for a list of curves, we translate each curve in the list the same way using map:(define (connect-curves-evenly curvelist) (lambda (t) (let ((which-curve (if (>= t 1.0) (- (length curvelist) 1) (inexact->exact (floor (* t (length curvelist))))))) ((get-nth curvelist which-curve) (* (length curvelist) (- t (* (/ 1 (length curvelist)) which-curve)))))))
(define (cons-to-curvelist curve curvelist) (let ((endpoint (curve 1.0))) ;; The last point in curve (cons curve (map (lambda (thiscurve) (translate thiscurve (x-of-point endpoint) (y-of-point endpoint))) curvelist))))
The convert-lcommands-to-curvelist procedure converts a list of L-System commands into a curve. Here is the code for convert-lcommands-to-curvelist (with some missing parts that you will need to complete). It will be explained later, but try to understand it yourself first.
(define (convert-lcommands-to-curvelist lcommands) (cond ((null? lcommands) (list ;;; We make a leaf with just a single point of green: (lambda (t) (make-colored-point 0.0 0.0 (make-color 0 255 0))) )) ((is-forward? (car lcommands)) (cons-to-curvelist vertical-line (convert-lcommands-to-curvelist (cdr lcommands)))) ((is-rotate? (car lcommands)) ;;; If this command is a rotate, every curve in the rest ;;; of the list should should be rotated by the rotate angle (let ;; L-system turns are clockwise, so we need to use - angle ((rotate-angle (- (get-angle (car lcommands))))) (map (lambda (curve) ;;; Question 9: fill this in ) ;;; Question 9: fill this in ))) ((is-offshoot? (car lcommands)) (append ;;; Question 10: fill this in )) (#t (error "Bad lcommand!"))))We define convert-lcommands-to-curvelist recursively. The base case is when there are no more commands (the lcommands parameter is null). It evaluates to the leaf curve (for now, we just make a point of green — you may want to replace this with something more interesting to make a better fractal). Since convert-lcommands-to-curvelist evaluates to a list of curves, we need to make a list of curves containing only one curve.
Otherwise, we need to do something different depending on what the first command in the command list is. If it is a forward command we draw a vertical line. The rest of the fractal is connected to the end of the vertical line using cons-to-curvelist. The recursive call to convert-lcommands-to-curve produces the curve list corresponding to the rest of the L-system commands. Note how we pass (cdr lcommands) in the recursive call to get the rest of the command list.
You can test your code by drawing the curve that results from any list of L-system commands that does not use offshoots. For example, evaluating
should produce a "V".(draw-curve-points (position-curve (translate (connect-curves-evenly (convert-lcommands-to-curvelist (make-lsystem-command (make-rotate-command 150) (make-forward-command) (make-rotate-command -120) (make-forward-command)))) 0.3 0.7) 0 .5) 10000)
We have provided the position-curve procedure to make it easier to fit fractals onto the graphics window:
(position-curve curve startx starty) evaluates to a curve that translates curve to start at (startx, starty) and scales it to fit into the graphics window maintaining the aspect ratio (the x and y dimensions are both scaled the same amount)The code for position-curve is in curve.scm. You don't need to look at it, but should be able to understand it if you want to.
Now, you should be able to draw any l-system command list using position-curve and the convert-lcommands-to-curvelist function you completed in Questions 9 and 10. Try drawing a few simple L-system command lists before moving on to the next part.
Hint: You should use the rewrite-lcommands you defined in Question 5. You may also find it useful to use the n-times function from lecture.
(define (make-tree-fractal level) (make-lsystem-fractal tree-commands (make-lsystem-command (make-forward-command)) level))
(define (draw-lsystem-fractal lcommands) (draw-curve-points (position-curve (connect-curves-evenly (convert-lcommands-to-curvelist lcommands)) 0.5 0.1) 50000))
For example, (draw-lsystem-fractal (make-tree-fractal 3)) will create a tree fractal with 3 levels of branching.
Draw some fractals by playing with the L-system commands. Try changing the rewrite rule, the starting commands, level and leaf curve (in convert-lcommands-to-curvelist) to draw an interesting fractal. You might want to make the branches colorful also. Try an make a fractal picture that will make a better course logo than the current Great Lambda Tree Of Infinite Knowledge and Ultimate Power.
To save your fractal in an image file, use the save-image procedure (defined in lsystem.scm). It can save images as .png files. For example, to save your fractal in wondertree.png evaluate