Helper Functions

In class we discussed in addition to the stereotypical recursive patterns in the book, the ``helper function with something extra'' pattern. This is a page of notes about that.


Let's say we want to define a function rev that takes a sentence and returns that sentence with the words in reverse order. For example:
(rev '(aye bee sea))   ; (sea bee aye)
(rev '(ex))            ; (ex)
(rev '())              ; ()
  1. Earlier we learned a cute but tricky way to do this using higher order functions:
    (define (rev sen)
      (accumulate (lambda (x y) (sentence y x))
    but let's say we want to do it recursively.

  2. A straightforward thought is to take the reverse of the sentence with it's first element removed, and then put that first element on at the end.
    (define (rev sen)
       (if (empty? sen)
           (sentence (rev (butfirst sen)) (first sen))))
    Let's look at a call graph for (rev '(a b c d)), to make sure we understand how it works.

    That's nice, but maybe inefficient, because it is expensive to tack things onto the end of a sentence but cheap to tack them onto the beginning.

  3. We can use a ``helper function with something extra'' instead. The helper function will take two arguments: a sentence to go at the end of its result, and a sentence to be reversed and put at the beginning. For example:
    (rev-aux '(aye bee sea) '(one two three))  ; (three two one aye bee sea)
    So let's define it:
    (define (rev sen)
       (rev-aux '() sen))
    (define (rev-aux end-of-result to-be-reversed)
       (if (empty? to-be-reversed)
           (rev-aux (sentence (first to-be-reversed)
                    (butfirst to-be-reversed))))
    This should be efficient because it only moneys around with the beginning of sentences.

    Let's look at a call graph of (rev '(a b c d)), to see if we can understand this.

Parallel Prefix

Consider the function pp+, which is supposed to compute the ``parallel prefix'' of a sentence of numbers. That is, a sentence of the running totals.

For example:

(pp+ '(1 1 1 1 1))        ; (1 2 3 4 5)
(pp+ '(1000 200 30 4))    ; (1000 1200 1230 1234)
(pp+ '(14))               ; (14)
(pp+ '())                 ; ()
As with most problems, there are a bunch of different strategies we might use here.
  1. Simple recursion: cut off the first elt, figure out the parallel prefix of the rest, and modify it to give the right answer.
    (define (pp+ sen)
       (if (empty? sen)
           (let ((fir (first sen)))
              (sentence fir
                        (every (lambda (x) (+ x fir))
                               (pp+ (butfirst sen)))))))
    This looks nice, except one might worry about efficiency, in that shared partial sums are not being used, so it might take too much time to compute.

  2. Instead we can try to do it from the other direction, recursing by removing the last element:
    (define (pp+ sen)
       (if (empty? sen)
           (let ((partial-pre (pp+ (butlast sen))))
              (sentence partial-pre
                        (+ (last sen)
                           (if (empty? partial-pre)
                               (last partial-pre)))))))
    This seems elegant, in that it uses the already-computed total for the first part of the sentence in calculating the remaining numbers, so it will not do too many additions. But it might be inefficient for a different reason: first is fast, while last takes time of order the length of the sentence, as does tacking a new element onto the end of a sentence.

  3. In an attempt to have our cake and eat it too, we can define pp+ in term of a recursive helper function, which takes an extra argument: the running total so far.
    (define (pp+ sen) (pp+-aux 0 sen))
    (define (pp+-aux running-total sen)
       (if (empty? sen)
           (let ((new-running-total (+ running-total (first sen))))
              (sentence new-running-total
                        (pp+-aux new-running-total
                                 (butfirst sen))))))
    Which is efficient both in its manipulation of the sentence and in taking advantage of the left-to-right structure of the addition.

    Let's look at it's call graph for (pp+ '(1000 200 30 4))

The Fibonacci Sequence

Another problem we can use helper functions for is the Fibonacci function discussed in the text. (We all know where the Fibonacci sequence comes from: it's the number of female rabbits you have each year, if you buy a female rabbit the first year, each litter contains one female rabbit, there is one litter a year, rabbits live for two years, and there is no shortage of male rabbits. The Fibonacci sequence also comes up in spiral growth patterns, like the whorls in a sunflower.)

Each element in the sequence is the sum of the previous two elements: 1, 1, 2, 3, 5, 8, 13, 21, 34, etc.

We want to define our function such that

(fibb 0)     ; 1
(fibb 1)     ; 1
(fibb 2)     ; 2
(fibb 3)     ; 3
(fibb 4)     ; 5
(fibb 5)     ; 8
(fibb 6)     ; 13
(fibb 7)     ; 21
(fibb 8)     ; 34
The obvious recursive definition is:
(define (fibb n)
   (cond ((= n 0) 1)
         ((= n 1) 1)
         (else (+ (fibb (- n 1)) (fibb (- n 2))))))
This will be very inefficient for large arguments: the total number of recursive calls is going up exponentially with the argument! We can see that by sketching the call graph of (fibb 8)
If you think about it, you'll see that the function either recurses with two calls to itself or returns 1. So the total number of calls to fibb when you go (fibb n) must be twice the returned result minus one. This grows pretty quickly.

Instead we would like to sum things up as we would if we were doing this by hand. Doing it by hand, we would keep track of the previous two Fibonacci numbers to calculate the next one, and repeat that until we get up to where we want to be.

This can be done with a helper function that takes three arguments: the last two Fibonacci numbers, and the number of steps that remain to be taken.

(define (fibb n)
   (fibb-aux 1 1 (- n 1)))

(define (fibb-aux fibb-nm1 fibb-n steps-remaining)
   (cond ((= steps-remaining 0) fibb-n)
         ((= steps-remaining -1) fibb-nm1)
         (else (fibb-aux fibb-n
	                 (+ fibb-nm1 fibb-n)
			 (- steps-remaining 1)))))
This gives a much faster algorithm, as seen from the following call graph for (fibb 8)

Barak Pearlmutter <>