Summary: In this laboratory, you will further explore issues of recursion over trees introduced in the reading on pairs and pair structures and continued in the reading on trees.

a. Make sure that you have the reading on pairs and pair structures and the reading on trees open in separate tabs or windows. b. Make sure that you have a piece of paper and writing instrument handy.

As you may recall from the reading, we have defined color trees as follows. Any color is a color tree. If t1 and t2 are color trees, then so is (cons t1 t2). Nothing else is a color tree. Using this recursive definition, write a procedure, (color-tree? val) that returns true (#t) if val is a color tree and false (#f) otherwise.

Recall the render-color-tree! procedure from the reading. a. Add render-color-tree! to your definitions window. b. Create a new 200x200 image named canvas. c. What effect do you expect the following instruction to have? > (render-color-tree! "blue" canvas 0 0 200 200) d. Check your answer experimentally. e. What effect do you expect the following instruction to have? > (render-color-tree! (cons "black" "red") canvas 0 0 200 200) f. Check your answer experimentally. g. What effect do you expect the following instruction to have? > (render-color-tree! (cons (cons "green" "yellow") "orange") canvas 0 0 200 200) h. Check your answer experimentally. i. What effect do you expect the following instruction to have? > (render-color-tree! (cons "red" (cons (cons "blue" "purple") "black")) canvas 0 0 200 200) j. Check your answer experimentally. k. Create a color tree that you might be able to render in some interesting fashion.

Rewrite render-color-tree! so that it splits the image vertically rather than horizontally.

a. What preconditions should render-color-tree! have? b. Use the color-tree? predicate from an earlier exercise to rewrite render-color-tree! so that it reports an appropriate error if its preconditions are not met. Use husk-and-kernel style, checking the preconditions in the husk. c. In your implementation from (b), the tree is traversed twice: Once in color-tree? to check that the tree is valid, and once in the kernel to do the work of rendering the tree. Some programmers see this as inefficient. Write another version of render-color-tree! that does not use color-tree? to verify preconditions. Instead, the kernel should verify as it goes along that each leaf is a color. That is, the kernel should issue an error message if it finds any tree value that is neither a pair nor a color.

Let's explore trees more generally. As you've noted, trees are built from cons cells (a.k.a. pairs). Sometimes it is useful to find out how many cons cells are in a tree. a. Define and test a procedure named cons-cell-count that takes any Scheme value and determines how many boxes would appear in its box-and-pointer diagram. (The data structure that is represented by such a box, or the region of a computer's memory in which such a structure is stored is called a cons cell. Every time the cons procedure is used, explicitly or implicitly, in the construction of a Scheme value, a new cons cell is allocated, to store information about the car and the cdr. Thus cons-cell-count also tallies the number of times cons was invoked during the construction of its argument.) For example, the structure in the following box-and-pointer diagram contains seven cons-cells, so when you apply cons-cell-count to that structure, it should return 7. On the other hand, the string "sample" contains no cons-cells, so the value of (cons-cell-count "sample") is 0. In answering this question, you should consider whether each value, in turn, is a pair using the pair? predicate. b. Use cons-cell-count to find out how many cons cells are needed to construct the list (0 (1 (2 (3 (4))))) See the notes at the end of the lab if you have trouble creating that list. c. Draw a box-and-pointer diagram of this list to check the answer.

If you have extra time, you should try some of these explorations or some of the for those with extra time problems below.

One potentially interesting way to experiment with color trees is to have a procedure build those trees. How? We might provide the procedure with an intended size of the tree. If the number of colors is 1, we return a randomly-selected color. If the number of colors is 2, we cons together two trees of size 1. Otherwise, we pick a random size for the left subtree (the size should be at least 1 and strictly less than the intended size of the whole tree). We then recursively build a tree of that size and another tree of an appropriate size and then cons them together. a. Using this technique, write a procedure, (random-bw-tree size), that randomly builds a black and white tree of the appropriate size. b. Using this technique, write a procedure, (random-color-tree size colors), that builds a random color tree of the desired size, randomly selecting the color from the list colors.

Create a new version of render-color-tree! that has an extra parameter, vsplit?, a Boolean value. If the Boolean is true, when render-color-tree! should split the area vertically (as it currently does). If the Boolean is false, render-color-tree! should split the area horizontally. In both cases, the recursive calls should negate that Boolean value so that the splits alternate between vertical and horizontal.

If you find that you have extra time, you might want to attempt one or more of the following problems.

As you may recall, a tree is either (a) a non-pair value or (b) the cons of two trees. In the reading, you saw a procedure that counted the number of values in a tree. In this lab, you wrote a procedure that counted the number of cons cells (pairs) in a tree. What is the relationship between the numbers returned by those two procedures?

a. Write a procedure, (color-tree-contains? ctree color), that determines whether color appears anywhere in ctree. (We may have written a similar procedure during the class discussion. If so, you can start with that procedure.) b. Rewrite the procedure to determine if a color nearby to color appears somewhere in ctree. (You might say that two colors are nearby if their red, green, and blue components all differ by less than sixteen.) c. Write (tree-contains tree value), a generalized version of color-tree-contains that works with a heterogeneous tree (that is, a tree that contains multiple kinds of values).

If, for some reason, you are having trouble creating the list (0 (1 (2 (3 (4))))) try (list 0 (list 1 (list 2 (list 3 (list 4))))) Return to the problem.