Summary: We've learned a number of basic techniques for writing recursive functions over lists, including the a pattern of recursion. But these techniques aren't always the clearest or most elegant for every case. Here, we extend your recursion toolbox to include a few more techniques, particularly the identification of special base cases and ways to write recursive predicates.

Sometimes the problem that we need an algorithm for doesn't apply to the empty list, even in a vacuous or trivial way, and the base case for a direct recursion instead involves singleton lists -- that is, lists with only one element. For instance, suppose that we want an algorithm that finds the leftmost element of a given non-empty list of drawings. (The list must be non-empty because there is no leftmost element of an empty list.) > (drawings-leftmost (list (drawing-rectangle 10 0 20 20) (drawing-rectangle 3 5 20 20) (drawing-rectangle 6 2 20 20) (drawing-rectangle 1 50 20 20) (drawing-rectangle 8 5 20 20))) (drawing rectangle 0 "" 1 50 20 20) The assumption that the list is not empty is a precondition for the meaningful use of this procedure, just as a call to Scheme's built-in quotient procedure requires that the second argument, the divisor, be non-zero. A precondition is a requirement that must be met in order for your procedure to work correctly. You should form the habit of figuring out when such preconditions are appropriate. With the Six-P technique for documenting procedures, you should also make it a habit to document such preconditions as you write the initial comment for a procedure: ;;; Procedure: ;;; drawings-leftmost ;;; Parameters: ;;; drawings, a list of drawings. ;;; Purpose: ;;; Find the leftmost drawing in drawings. ;;; Produces: ;;; leftmost, a drawing. ;;; Preconditions: ;;; drawings is not empty. ;;; All the values in drawings are drawings. ;;; Postconditions: ;;; leftmost is an element of drawings (and is, by implication, a drawing). ;;; For each drawing, d, in drawings, leftmost either starts in the same ;;; column as di, or to the left of d. Alternately, we can make the structure of the list part of the specification of the parameters (and therefore an implicit precondition). ;;; Parameters: ;;; drawings, a nonempty list of drawings. Whether you specify the precondition in the parameters or the preconditions section is often a matter of personal taste. What is most important is that you specify it somewhere. Now that we've documented the procedure, let's think about how to implement it. If a list of drawings contains only one element, the answer is trivial -- its only element is its leftmost. Otherwise, we can take the list apart into its car and its cdr, invoke the procedure recursively to find the leftmost of the cdr, and then try to figure out which comes first. How do we figure whether or not one drawing is to the left of another? We compare their left edge, which we can do with drawing-left. Let's use a helper procedure to compare the two drawings and return the leftmost. (This is not a recursive helper procedure. Rather, like rgb-darker, it is a relatively straightforward procedure that simplifies our recursive definitions.) We can test whether the given list has a single element by checking whether its cdr is an empty list. The value of the expression (null? (cdr drawings)) is #t if drawings has a single element and #f if drawings has two or more elements. (It gives an error if drawings has zero elements.) Here, then, is the procedure definition: If someone who uses this procedure happens to violate its precondition, applying the procedure to the empty list, the Scheme interpreter notices the error and prints out a diagnostic message: > (drawings-leftmost null) cdr: expects argument of type <pair>; given ()

If you think back to the tail-recursive version of difference, you may note another time that we had a special singleton case. When we compute v1 - v2 - v3 - vk, the base case is not we have nothing to subtract, but rather we have nothing to subtract from v1. We didn't note the need for a singleton base case until we tried to write a tail-recursive version, but the need was there. Of course, that means that we might consider rewriting the non-tail-recursive version, but that version gave us the wrong answer, anyway.

If you consider the examples you've studied over the past few days, you will see that there is a common form for most of the procedures. The form goes something like this: (define recursive-proc (lambda (val) (if (base-case-test? val) (base-case-computation val) (combine (partof val) (recursive-proc (simplify val)))))) For example, for the drawings-leftmost procedure, The recursive-proc is drawings-leftmost. The val is drawings, our list of numbers. The base-case-test is (null? (cdr drawings)), which checks whether drawings has only one element. The base-case-computation is car, which extracts the one drawing left in drawings. The partof procedure is also car, which extracts the first drawing in drawings. The simplify procedure is cdr, which drops the first element, thereby giving us a simpler (well, smaller) list. Finally, the combine procedure is drawing-leftmost. Similarly, consider the first complete version of sum. In the sum procedure, The recursive-proc is sum. The val is again numbers, a list of numbers. The base-case-test is (null? numbers), which checks if we have no numbers. The base-case-computation is 0. (This computation does not quite match the form above, since we don't apply the 0 to numbers. As this example suggests, sometimes the base case does not involve the parameter.) The partof procedure is car, which extracts the first value in numbers. The simplify procedure is cdr, which drops the the first element. When you write your own recursive procedures, it's often useful to start with the general structure and then to fill in the pieces. When you are recursing over lists (as you have in our first explorations of recursion), partof is almost always car and simplify is almost always cdr. There's a bit more to the base-case-test, since we've used both (null? ___) and (null? (cdr? ___)). We may also find other techniques. However, when you work with other kinds of information (as you will do soon), you'll have different techniques for extracting a piece of information, for simplifying the argument, and for deciding when you're done. Note, also, that examples like filtering suggest a similar, but more complex structure for recursive procedures. (define recursive-proc (lambda (val) (cond ((one-base-case-test?) (one-base-case-computation val)) ((another-base-case-test?) (another-base-case-computation val)) ... ((special-recursive-case-test-1?) (combine-1 (partof-1 val) (recursive-proc (simplify-1 val)))) ((special-recursive-case-test-2?) (combine-2 (partof-2 val) (recursive-proc (simplify-2 val)))) ... (else (combine (partof val) (recursive-proc (simplify val))))))) However, in practice you will find that you rarely have more than two base-case tests (mostly one) and rarely more than two recursive cases. When we write tail-recursive procedures, we simply use the result of the recursive call, and don't combine it with anything. Here's a simple tail recursive pattern. (define procedure (lambda (val) (procedure-helper initial-result val))) (define procedure-helper (lambda (so-far remaining) (if (base-case-test? remaining) (final-computation so-far) (procedure-helper (combine (part-of remaining) so-far) (simplify remaining)))))

Of course, these common forms are not the only way to define recursive procedures. In particular, when we define a predicate that uses direct recursion on a given list, the definition is usually a little simpler if we use and- and or-expressions rather than if-expressions. For instance, consider a predicate rgb-all-dark? that takes a given list of colors and determines whether all of them are dark. As usual, we consider the cases of the empty list and non-empty lists separately: Since the empty list has no elements, it is (as mathematicians say) vacuously true that all of its elements are dark -- there is certainly no counterexample that one could use to refute the assertion. So rgb-all-dark? should return #t when given the empty list. For a non-empty list, we separate the car and the cdr. If the list is to count as rgb-all dark, the car must clearly be dark, and in addition the cdr must be an all-dark list. We can use a recursive call to determine whether the cdr is all dark, and we can combine the expressions that test the car and cdr conditions with and to make sure that they are both satisfied. Thus, rgb-all-dark? should return #t when the given list either is empty or has a dark first element and all dark elements after that. This yields the following definition: When colors is the empty list, rgb-all-dark? applies the first test in the or-expression, finds that it succeeds, and stops, returning #t. In any other case, the first test fails, so rgb-all-dark? proceeds to evaluate the first test in the and-expression. If the first element of colors is not dark, the test fails, so rgb-all-dark? stops, returning #f. However, if the first element of colors is dark, the test succeeds, so rgb-all-dark? goes on to the recursive procedure call, which checks whether all of the remaining elements are dark, and returns the result of this recursive call, however it turns out. Here's a template for that solution. (define all-____? (lambda (lst) (or (null? lst) (and (____? (car lst)) (all-____? (cdr lst)))))) We can use a similar technique to ask if any value in a list is dark. In this case, if there are no values in the list, we know that no values are dark. Otherwise, we check if the first value is dark. If it is, then some value must be dark. The complicated part is getting the base case right (particularly if we want to avoid using if). The standard technique is to require that the list not be null (using not and and). If the list is null, (not (null? lst)) returns #f. And, since (and #f ...) is #f, we get false back for the empty list, just as we wanted. And, once again, we can generalize. (define any-____? (lambda (lst) (and (not (null? lst)) (or (____? (car lst)) (any-____? (cdr lst))))))