Before starting to work with a partner, you should go through questions 1 and 2 yourself on paper. When you meet with your partner (if any), check if you made the same predictions. If there are any discrepancies, try to decide which is correct before using DrScheme to evaluate the expressions.
If you have a partner, you and your partner should work together on the rest of the assignment. You should read the whole problem set yourself and think about the questions before beginning to work on them with your partner. Both partners need to understand everything you submit.
Remember to follow the pledge you read and signed at the beginning of the semester. For this assignment, you may consult any outside resources, including books, papers, web sites and people, you wish except for materials from previous cs1120, cs150, and cs200 courses. You may consult an outside person (e.g., another friend who is a CS major but is not in this class) who is not a member of the course staff, but that person cannot type anything in for you and all work must remain your own. That is, you can ask general questions such as "can you explain recursion to me?" or "how do lists work in Scheme?", but outside sources should never give you specific answers to problem set questions. If you use resources other than the class materials, lectures and course staff, explain what you used in your turnin.
You are strongly encouraged to take advantage of the scheduled help hours and office hours for this course.
In this problem set, you will explore a method of creating fractals known as the Lindenmayer system (or Lsystem). Aristid Lindemayer, a theoretical biologist at the University of Utrecht, developed the Lsystem 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 Lsystem and developed many of the ideas on which this problem set is based.
The idea behind Lsystem 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 Lsystem 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 trunk can be described using the pattern: F O(R30 F) F O(R60 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(R60 F) F)
Iteration 0: (F)
Iteration 1: (F O(R30 F) F O(R60 F) F)
Iteration 2: (F O(R30 F) F O(R60 F) F O(R30 F O(R30 F) F O(R60 F) F) F O(R30 F) F O(R60 F) F O(R60 F O(R30 F) F O(R60 F) F) F O(R30 F) F O(R60 F) F)

Iteration 5 (with color) The Great Lambda Tree of Infinite Knowledge and Ultimate Power 
Note that Lsystem command rewriting is similar to the replacement rules in a BNF grammar. The important difference is that with Lsystem 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 Lsystem fractal into two main parts:
Here is a BNF grammar for Lsystem 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 nonterminals into Scheme objects.
;;; CommandSequence ::= ( CommandList ) (define makelsystemcommand 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 (makeforwardcommand) (cons 'f #f)) ;; No value, just use false. ;;; Command ::= RAngle (define (makerotatecommand angle) (cons 'r angle)) ;;; Command ::= OCommandSequence (define (makeoffshootcommand 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.> (isforward? (makeforwardcommand))
#t
> (isforward? (makerotatecommand 90))
#f
> (getangle (makerotatecommand 90))
90
> (getangle (makeforwardcommand))
Yikes! Attempt to getangle 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 Expression_{1} Expression_{2} ... Expression_{k}), evaluate each subexpression in order from left to right. The value of the begin expression is the value of Expression_{k}.The begin special form is useful when we are evaluating expressions that have sideeffects. 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
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 substituted 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 ((Name_{1} Expression_{1}) (Name_{2} Expression_{2}) ... (Name_{k} Expression_{k})) MoreExpressions Expression)is equivalent to the application expression:
((lambda (Name_{1} Name_{2} ... Name_{k}) (begin MoreExpressions Expression)) Expression_{1} Expression_{2} ... Expression_{k})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 ::=
ClauseList ::= Clause ClauseList
Clause ::= (Expression_{Test} Expression_{Action})
Clause ::= (else Expression_{Action})
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 (Test_{1} Action_{1}) (Test_{2} Action_{2}) ... (Test_{k} Action_{k}) (else Action_{else}))is equivalent to:
(if Test_{1} Action_{1} (if Test_{2} Action_{2} ... (if Test_{k} Action_{k} action_{else})...))
Start: (F)To produce levels of the tree fractal, we need a procedure that takes a list of Lsystem commands and replaces each F command with the list of Lsystem commands given by the rule.
Rule: F ::= (F O(R30 F) F O(R60 F) F)
So, for every command in the list:
For example, consider a simple LSystem 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.ss):
Iteration2: ((F R30 F) R30 (F R30 F))
(define (flattencommands ll) (if (null? ll) ll (if (islsystemcommand? (car ll)) (cons (car ll) (flattencommands (cdr ll))) (flatappend (car ll) (flattencommands (cdr ll)))))) (define (flatappend lst ll) (if (null? lst) ll (cons (car lst) (flatappend (cdr lst) ll))))
Here's the easy part:
Complete the definition of rewritelcommands.(define (rewritelcommands lcommands replacement) (flattencommands (map ; Procedure to apply to each command lcommands)))
We have predefined some simple lsystem commands (e.g., f, fr30f, ffr30) for easy testing. If you define these procedures correctly, you should produce these evaluations:
> (rewritelcommands f fr30f)
((f . #f) (r . 30) (f . #f))
> (rewritelcommands fr30f ffr30)
((f . #f) (f . #f) (r . 30) (r . 30) (f . #f) (f . #f) (r . 30))
To make interesting Lsystem curves, we will need to apply rewritelcommands many times. We will leave that until the last question. Next, we will work on turning sequences of Lsystem 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 makepoint, xofpoint and yofpoint. 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 (makepoint x y) (list x y)) (define (makecoloredpoint x y c) (list x y c)) (define (iscoloredpoint? (= (length point) 3) (define (xofpoint point) (car point)) (define (yofpoint point) (cadr point)) ;; (cadr x) = (car (cdr x)) ;;; Regular points are black. Colored points have a color. (define (colorofpoint point) (if (iscoloredpoint? point) (caddr point) ;; == (car (cdr (cdr point))) (makecolor 0 0 0)))
We have provided some procedures for drawing on the window in graphics.ss:
Read through graphics.ss and look for other handy functions.
(define (midline t) (makepoint t 0.5))defines a curve that is a horizontal line across the middle of the window. If we apply midline to a value x, we get the point (x, 0.5). Hence, if we apply midline to all values between 0.0 and 1.0, we get a horizontal line.
Predict what (xofpoint (midline 0.7)) and (yofpoint (midline 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 (drawcurvepoints curve n) (define (drawcurveworker curve t step) (if (<= t 1.0) (begin (windowdrawpoint (curve t)) (drawcurveworker curve (+ t step) step)))) (drawcurveworker curve 0.0 (/ 1 n)))The procedure drawcurvepoints takes a procedure representing a curve, and n, the number of points to draw. It calls the drawcurveworker procedure. The drawcurveworker procedure takes three parameters: a curve, the current time step values, and the difference between time step values. Hence, to start drawing the curve, drawcurvepoints evaluates drawcurveworked with parameters curve (to pass the same curve that was passed to drawcurvepoints), 0.0 (to start at the first t value), and (/ 1 n) (to divide the time values into n steps).
The drawcurveworker procedure is defined recursively: if t is less than or equal to 1.0, we draw the current point using (windowdrawpoint (curve t)) and draw the rest of the points by evaluating (drawcurveworker 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 drawcurveworker code uses a being expression. The first expression in the begin expression is (windowdrawpoint (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 rotateccw takes a curve and rotates it 90 degrees counterclockwise by swapping the x and y points:
(define (rotateccw curve) (lambda (t) (let ((ct (curve t))) (makecoloredpoint ( (yofpoint ct)) (xofpoint ct) (colorofpoint 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 (rotateccw c) evaluates to a curve. The function rotateccw is a procedure that takes a procedure (a curve) and returns a procedure that is a curve.
Predict what (drawcurvepoints (rotateccw midline) 1000) and (drawcurvepoints (rotateccw (rotateccw midline)) 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))) (makecoloredpoint (* scale (xofpoint ct)) (* scale (yofpoint ct)) (colorofpoint ct)))))Predict what (drawcurvepoints (shrink midline .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))) (makecoloredpoint (+ x (xofpoint ct)) (+ y (yofpoint ct)) (colorofpoint ct)))))Now we have translate, it makes more sense to define midline this way:
(define (horizline t) (makepoint t 0)) (define midline (translate horizline 0 0.5))
When you are done, (drawcurvepoints halfline 1000) should produce a horizontal line that starts in the middle of the window and extends to the right boundary.
Hint: If you do not see anything when you are drawing a curve, it may be that you haven't yet applied translate and the points are being drawn along the bottom edge of the screen.
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 (firsthalf 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
curves including:
It is also useful to have curve transforms where curves may be combined. An example is (connectrigidly 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 connectrigidly is defined:
(define (connectrigidly curve1 curve2) (lambda (t) (if (< t (/ 1 2)) (curve1 (* 2 t)) (curve2 ( (* 2 t) 1)))))Predict what (drawcurvepoints (connectrigidly verticalmidline midline) 1000) will do. Is there any difference between that and (drawcurvepoints (connectrigidly midline verticalmidline) 1000)? Check your predictions in the Interactions window.
The way connectrigidly is defined above, we use all the tvalues below 0.5 on the first curve, and use the tvalues between 0.5 and 1.0 on the second curve. If the second curve is the result of connecting two other curves, like (connectrigidly c1 (connectrigidly c2 c3)) then 50% of the points will be used to draw c1, 25% to draw c2 and 25% to draw c3.
(connectrigidly c1 (connectrigidly c2 (connectrigidly curve3 (... cn))))The first argument to numpoints is the number of tvalues left. The second argument is the number of curves left.
Think about this yourself first, but look in ps3.ss 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 n^{th} curve.
Your numpoints procedure should produce results similar to:
This means if we connected just 20 curves using connectrigidly, and passed the result to drawcurvepoints with one million as the number of points, there would still be only one or two points drawn for the 20^{th} curve. If we are drawing thousands of curves, for most of them, not even a single point would be drawn!> (exact>inexact (numpoints 1000 10))
1.953125
> (exact>inexact (numpoints 1000 20))
0.0019073486328125
> (exact>inexact (numpoints 1000000 20))
1.9073486328125
To fix this, we need to distribute the tvalues between our curves more fairly. We have provided a procedure connectcurvesevenly in graphics.ss 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 tvalues 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 (connectcurvesevenly curvelist) (lambda (t) (let ((whichcurve (if (>= t 1.0) ( (length curvelist) 1) (inexact>exact (floor (* t (length curvelist))))))) ((getnth curvelist whichcurve) (* (length curvelist) ( t (* (/ 1 (length curvelist)) whichcurve)))))))
(define (constocurvelist curve curvelist) (let ((endpoint (curve 1.0))) ;; The last point in curve (cons curve (map (lambda (thiscurve) (translate thiscurve (xofpoint endpoint) (yofpoint endpoint))) curvelist))))
The convertlcommandstocurvelist procedure converts a list of LSystem commands into a curve. Here is the code for convertlcommandstocurvelist (with some missing parts that you will need to complete). It will be explained later, but try to understand it yourself first.
(define (convertlcommandstocurvelist lcommands) (cond ((null? lcommands) (list ;;; We make a leaf with just a single point of green: (lambda (t) (makecoloredpoint 0.0 0.0 (makecolor 0 255 0))) )) ((isforward? (car lcommands)) (constocurvelist verticalline (convertlcommandstocurvelist (cdr lcommands)))) ((isrotate? (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 ;; Lsystem turns are clockwise, so we need to use  angle ((rotateangle ( (getangle (car lcommands))))) (map (lambda (curve) ;;; Question 9: fill this in ) ;;; Question 9: fill this in ))) ((isoffshoot? (car lcommands)) (append ;;; Question 10: fill this in )) (#t (error "Bad lcommand!"))))We define convertlcommandstocurvelist 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 convertlcommandstocurvelist 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 constocurvelist. The recursive call to convertlcommandstocurve produces the curve list corresponding to the rest of the Lsystem 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 Lsystem commands that does not use offshoots. For example, evaluating
should produce a "V".(drawcurvepoints (positioncurve (translate (connectcurvesevenly (convertlcommandstocurvelist (makelsystemcommand (makerotatecommand 150) (makeforwardcommand) (makerotatecommand 120) (makeforwardcommand)))) 0.3 0.7) 0 .5) 10000)
Hint 1: See the next few paragraphs for help testing Question 10.
Hint 2: Evaluate treecommands and look at its definition in lsystem.ss. This may help you to visualize relevant car and cdr combinations.
We have provided the positioncurve procedure to make it easier to fit fractals into the graphics window:
(positioncurve 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 positioncurve is in curve.ss. 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 lsystem command list using positioncurve and the convertlcommandstocurvelist function you completed in Questions 9 and 10. Try drawing a few simple Lsystem command lists before moving on to the next part. For example, given this input:
Your output should look like this:(drawcurvepoints (positioncurve (connectcurvesevenly (convertlcommandstocurvelist treecommands)) 0.5 0.1) 50000)
Hint: You should use the rewritelcommands you defined in Question 5. You may also find it useful to use the ntimes function (which we may have described in lecture):
(define (ntimes proc n) (if (= n 1) proc (compose proc (ntimes proc ( n 1)))))
(define (maketreefractal level) (makelsystemfractal treecommands (makelsystemcommand (makeforwardcommand)) level))
(define (drawlsystemfractal lcommands) (drawcurvepoints (positioncurve (connectcurvesevenly (convertlcommandstocurvelist lcommands)) 0.5 0.1) 50000))
For example, (drawlsystemfractal (maketreefractal 3)) will create a tree fractal with 3 levels of branching.
Draw some fractals by playing with the Lsystem commands. Try changing the rewrite rule, the starting commands, level and leaf curve (in convertlcommandstocurvelist) 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 saveimage procedure (defined in lsystem.ss). It can save images as .png files. For example, to save your fractal in yggdrasil.png evaluate
(saveimage "yggdrasij.png")Especially ambitious students may find the Viewport Graphics documentation useful for enhancing your pictures.
Cognitive psychologists, such as Roger Shepard and Jacqueline Metzler, have been interested in studying how humans manipulate mental representations of objects. To study this, they conducted a scientific experiment with a falsifiable hypothesis. In essence, their main hypothesis was that the time required to recognize that two perspective drawings portray objects of the same threedimensional shape is found to be a linearly increasing function of the angular difference in the portrayed orientations of the two objects. To test this hypothesis, they presented subjects with a series of image pairs. Each pair of images was either one drawing and its rotation (as in Figure 2), or one drawing and another drawing that cannot be obtained from it by rotation (as in Figure 1). They then measured the time it took for human subjects to determine whether or not the two shapes were equivalent modulo rotation.
It turns out that the number of seconds it takes humans to determine if the images are rotations or not is a linear function of the angle of rotation. For example, if the angle is 40 degrees, it takes 2 seconds. If the angle is 60 degrees, it takes 3 seconds. If the angle is 80 degrees, it takes 4 seconds. In this example, it takes one second to reason about a 20 degree rotation.
There are many tasks in perception that take constant time. For example, detecting a red X against a random pattern of blue Os takes constant time, regardless of how many Os are present. Some aspects of perception are thus handled in parallel. The "linear function" result for mental rotation suggested that humans actually form a mental model of the object and rotate that model internally to see if it matches up with the other image. Subsequent experimentation with nuclear magnetic resonance and functional magnetic resonance imaging showed that there are particular areas of the brain associated with such mental rotation. Moreover, when a subject is performing a mental rotation tasks, mappings of which neurons are firing for shapes that themselves rotate over time.
This extra credit assignment involves reproducing parts of the original mental rotation experiment. The first part is to read Shepard and Metzler's Mental Rotation of ThreeDimensional Objects from Science volume 701, February 1971. It's only three pages long.
Eventually we will have datapoints from human subjects: rotational angles and the time taken to perform the task. We will want to analyze those datapoints to determine the slop of the line (as in Figure 2 of Shepard and Metzler's article). Write a function calculateslope that takes a single argument: a list of datapoints. Each datapoint is a cons cell containing a rotation angle measurement and a time value measurement. Your function should return the average slope (rotation/time) for all of the datapoints.
For example:
In that dataset, it takes one second to reason about a 30.16 degree rotation.(define dataset (list (cons 32 1) (cons 54 2) (cons 95 3)))
(calculateslope dataset)
30 ^{1}/_{6}
We'll also need some way of getting timed user input. We'll assume a function gettimedinput that asks the user for a string and returns a pair containing the time taken and the string. You need not be able to write such a function, but you should be able to understand it:
The basic idea of the experiment is to present the user with a series of image pairs and record the times and answers. We'll use the following image:(define (gettimedinput) (let ((starttime (currentmilliseconds)) (input (readline (currentinputport))) (endtime (currentmilliseconds))) (cons ( endtime starttime) input)))(gettimedinput)
y
(498 . "y")(gettimedinput)
n
(651 . "n")(gettimedinput)
y
(322 . "y")
Define a procedure runmrexperiment that accepts a single parameter. That parameter is a list of cons cells. Each cons cell holds an angle in degrees and a boolean — the boolean is #t if the image should be flipped before being displayed. For each item in the list, you should clear the screen, display (mrcurve 0) on the left side of the screen, and display a rotatedandpossiblyflipped version of mrcurve on the right side of the screen based on the current list element. The user should enter "y" if it is a rotation and "n" if it is not (i.e., if a flip was involved). You should gather all of the times for which the user was correct as a dataset  a list suitable for passing to (calculateslope).(define (mrcurve degrees) (connectcurvesevenly (convertlcommandstocurvelist (makelsystemcommand (makerotatecommand degrees) (makeforwardcommand) (makeforwardcommand) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makeforwardcommand) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makeforwardcommand) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) (makeforwardcommand) (makerotatecommand 90) (makeforwardcommand) ))))
Hint: you may need to write your own version of positioncurve to make this work.