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

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

Recall that a list is a data structure defined recursively as follows: The empty list is a list. Cons of a value and a list is a list. In the reading on pairs and pair structures, the section entitled Recursion with Pairs includes a procedure that works on number trees, nested structures built with the pair procedure. Write a recursive definition for number trees, trees built from only numbers and cons cells, similar to that for lists.

Using your recursive definition of number trees from the previous problem, write a procedure, (number-tree? val) that returns true if val is a number tree and false otherwise.

Consider again the sum-of-number-tree procedure from the reading, reproduced here. a. Verify that sum-of-number-tree works as advertised on a single number. b. Verify that sum-of-number-tree works as advertised on a pair of numbers. c. Verify that it works as advertised on the first example. > (sum-of-number-tree (cons (cons (cons 0 1) (cons 2 3)) (cons (cons 4 5) (cons 6 7)))) d. What do you expect sum-of-number-tree to return when given (cons 10 11) as a parameter? Verify your answer experimentally. e. What do you expect sum-of-number-tree to return when given the empty list as a parameter? Verify your answer experimentally. f. What do you expect sum-of-number-tree to return when given (list 1 2 3 4 5) as a parameter? Verify your answer experimentally.

a. What preconditions should sum-of-number-tree have? b. Use the number-tree? predicate from earlier to rewrite sum-of-number-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. Some programmers consider it inappropriate to scan a tree twice, once to make sure that it's valid and once to compute a value based on the tree. Rewrite sum-of-number-tree so that it checks for and reports errors only when it is at one of the non-pair values.

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.

Recall the render-color-tree 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 color-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 color-black color-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 color-green color-yellow) color-orange) canvas 0 0 200 200) h. Check your answer experimentally. i. Create a color tree that you might be able to render in some interesting fashion. j. Rewrite render-color-tree so that it splits the image vertically rather than horizontally. j. Write a procedure that, given a list of colors and a maximum depth, generates an interesting color tree. Render your trees to see what they look like!

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)))))