Section 30

The Loss of Knowledge

When we design recursive functions, we don't think about the context of their use. Whether they are applied for the first time or whether they are called for the hundredth time in a recursive manner doesn't matter. They are to work according to their purpose statement, and that's all we need to know as we design the bodies of the functions.

Altough this principle of context-independence greatly facilitates the development of functions, it also causes occasional problems. In this section, we illustrate the most important problem with two examples. Both concern the loss of knowledge that occurs during a recursive evaluation. The first subsection shows how this loss makes a structurally recursive function more complicated and less efficient than necessary; the second one shows how the loss of knowledge causes a fatal flaw in an algorithm.

30.1  A Problem with Structural Processing

Suppose we are given the relative distances between a series of points, starting at the origin, and suppose we are to compute the absolute distances from the origin. For example, we might be given a line such as this:

relative distance

Each number specifies the distance between two dots. What we need is the following picture, where each dot is annotated with the distance to the left-most dot:

absolute distance

;; relative-2-absolute : (listof number)  ->  (listof number)
;; to convert a list of relative distances to a list of absolute distances
;; the first item on the list represents the distance to the origin
(define (relative-2-absolute alon)
  (cond
    [(empty? alon) empty]
    [else (cons (first alon)
	        (add-to-each (first alon) (relative-2-absolute (rest alon))))]))

;; add-to-each : number (listof number)  ->  (listof number)
;; to add n to each number on alon
(define (add-to-each n alon)
  (cond
    [(empty? alon) empty]
    [else (cons (+ (first alon) n) (add-to-each n (rest alon)))]))

Figure 83:  Converting relative distances to absolute distances

Developing a program that performs this calculation is at this point an exercise in structural function design. Figure 83 contains the complete Scheme program. When the given list is not empty, the natural recursion computes the absolute distance of the remainder of the dots to the first item on (rest alon). Because the first item is not the actual origin and has a distance of (first alon) to the origin, we must add (first alon) to each and every item on the result of the recursive application. This second step, adding a number to each item on a list of numbers, requires an auxiliary function.

While the development of the program is straightforward, using it on larger and larger lists reveals a problem. Consider the evaluation of the following definition:64

(define x (relative-2-absolute (list 0 ... N)))

As we increase N, the time needed grows even faster:65

N time of evaluation
100 220
200 880
300 2050
400 5090
500 7410
600 10420
700 14070
800 18530
Instead of doubling as we go from 100 to 200 items, the time quadruples. This is also the approximate relationship for going from 200 to 400, 300 to 600, and so on.

Exercise 30.1.1.   Reformulate add-to-each using map and lambda.    Solution

Exercise 30.1.2.   Determine the abstract running time of relative-2-absolute.

Hint: Evaluate the expression

(relative-2-absolute (list 0 ... N))

by hand. Start by replacing N with 1, 2, and 3. How many natural recursions of relative-2-absolute and add-to-each are required each time?    Solution

Considering the simplicity of the problem, the amount of ``work'' that the two functions perform is surprising. If we were to convert the same list by hand, we would tally up the total distance and just add it to the relative distances as we take another step along the line.

Let's attempt to design a second version of the function that is closer to our hand method. The new function is still a list-processing function, so we start from the appropriate template:

(define (rel-2-abs alon)
  (cond
    [(empty? alon) ...]
    [else ... (first alon) ... (rel-2-abs (rest alon)) ...]))

Now imagine an ``evaluation'' of (rel-2-abs (list 3 2 7)):

  (rel-2-abs (list 3 2 7))

= (cons ... 3 ...
    (convert (list 2 7)))

= (cons ... 3 ...
    (cons ... 2 ...
      (convert (list 7))))

= (cons ... 3 ...
    (cons ... 2 ...
      (cons ... 7 ...
	(convert empty))))

The first item of the result list should obviously be 3, and it is easy to construct this list. But, the second one should be (+ 3 2), yet the second instance of rel-2-abs has no way of ``knowing'' that the first item of the original list is 3. The ``knowledge'' is lost.

Put differently, the problem is that recursive functions are independent of their context. A function processes the list L in (cons N L) in the exact same manner as L in (cons K L). Indeed, it would also process L in that manner if it were given L by itself. While this property makes structurally recursive functions easy to design, it also means that solutions are, on occasion, more complicated than necessary, and this complication may affect the performance of the function.

To make up for the loss of ``knowledge,'' we equip the function with an additional parameter: accu-dist. The new parameter represents the accumulated distance, which is the tally that we keep when we convert a list of relative distances to a list of absolute distances. Its initial value must be 0. As the function processes the numbers on the list, it must add them to the tally.

Here is the revised definition:

(define (rel-2-abs alon accu-dist)
  (cond
    [(empty? alon) empty]
    [else (cons (+ (first alon) accu-dist)
	        (rel-2-abs (rest alon) (+ (first alon) accu-dist)))]))

The recursive application consumes the rest of the list and the new absolute distance of the current point to the origin. Although this means that two arguments are changing simultaneously, the change in the second one strictly depends on the first argument. The function is still a plain list-processing procedure.

Evaluating our running example with rel-2-abs shows how much the use of an accumulator simplifies the conversion process:

= (rel-2-abs (list 3 2 7) 0)
= (cons 3 (rel-2-abs (list 2 7) 3))
= (cons 3 (cons 5 (rel-2-abs (list 7) 5)))
= (cons 3 (cons 5 (cons 12 (rel-2-abs empty 12))))
= (cons 3 (cons 5 (cons 12 empty)))

Each item in the list is processed once. When rel-2-abs reaches the end of the argument list, the result is completely determined and no further work is needed. In general, the function performs on the order of N natural recursion steps for a list with N items.

One minor problem with the new definition is that the function consumes two arguments and is thus not equivalent to relative-2-absolute, a function of one argument. Worse, someone might accidentally misuse rel-2-abs by applying it to a list of numbers and a number that isn't 0. We can solve both problems with a function definition that contains rel-2-abs in a local definition: see figure 84. Now, relative-2-absolute and relative-2-absolute2 are indistinguishable.

;; relative-2-absolute2 : (listof number)  ->  (listof number)
;; to convert a list of relative distances to a list of absolute distances
;; the first item on the list represents the distance to the origin
(define (relative-2-absolute2 alon)
  (local ((define (rel-2-abs alon accu-dist)
	    (cond
	      [(empty? alon) empty]
	      [else (cons (+ (first alon) accu-dist)
		          (rel-2-abs (rest alon) (+ (first alon) accu-dist)))])))
    (rel-2-abs alon 0)))

Figure 84:  Converting relative distances with an accumulator

30.2  A Problem with Generative Recursion

Let us revisit the problem of finding a path in a graph from section 28. Recall that we are given a collection of nodes and connections between nodes, and that we need to determine whether there is a route from a node labeled orig to one called dest. Here we study the slightly simpler version of the problem of simple graphs where each node has exactly one (one-directional) connection to another node.

Consider the example in figure 85. There are six nodes: A through F, and six connections. To get from A to E, we go through B, C, and E. It is impossible, though, to reach F from A or from any other node (besides F itself).

  
(define SimpleG 
  '((A B)
    (B C)
    (C E)
    (D E)
    (E B)
    (F F)))    
 

Figure 85:  A simple graph

The right part of figure 85 contains a Scheme definition that represents the graph. Each node is represented by a list of two symbols. The first symbol is the label of the node; the second one is the reachable node. Here are the relevant data definitions:

A node is a symbol.

A pair is a list of two nodes:

  (cons S (cons T empty))  
where S, T are symbols.

A simple-graph is a list of pairs:

 (listof pair)

They are straightforward translations of our informal descriptions.

Finding a route in a graph is a problem of generative recursion.

We have data definitions, we have (informal) examples, and the header material is standard:

;; route-exists? : node node simple-graph  ->  boolean
;; to determine whether there is a route from orig to dest in sg
(define (route-exists? orig dest sg) ...)

What we need are answers to the four basic questions of the recipe for generative recursion:

What is a trivially solvable problem?
The problem is trivial if the nodes orig and dest are the same.

What is a corresponding solution?
Easy: true.

How do we generate new problems?
If orig is not the same as dest, there is only one thing we can do, namely, go to the node to which orig is connected and determine whether a route exists between it and dest.

How do we relate the solutions?
There is no need to do anything after we find the solution to the new problem. If orig's neighbor is connected to dest, then so is orig.

From here we just need to express these answers in Scheme, and we get an algorithm. Figure 86 contains the complete function, including a function for looking up the neighbor of a node in a simple graph.

;; route-exists? : node node simple-graph  ->  boolean
;; to determine whether there is a route from orig to dest in sg
(define (route-exists? orig dest sg)
  (cond
    [(symbol=? orig dest) true]
    [else (route-exists? (neighbor orig sg) dest sg)]))

;; neighbor : node simple-graph  ->  node
;; to determine the node that is connected to a-node in sg
(define (neighbor a-node sg)
  (cond
    [(empty? sg) (error "neighbor: impossible")]
    [else (cond
	    [(symbol=? (first (first sg)) a-node)
	     (second (first sg))]
	    [else (neighbor a-node (rest sg))])]))

Figure 86:  Finding a route in a simple graph (version 1)

Even a casual look at the function suggests that we have a problem. Although the function is supposed to produce false if there is no route from orig to dest, the function definition doesn't contain false anywhere. Conversely, we need to ask what the function actually does when there is no route between two nodes.

Take another look at figure 85. In this simple graph there is no route from C to D. The connection that leaves C passes right by D and instead goes to E. So let's look at how route-exists? deals with the inputs 'C and 'D for SimpleG:

  (route-exists? 'C 'D '((A B) (B C) (C E) (D E) (E B) (F F)))
= (route-exists? 'E 'D '((A B) (B C) (C E) (D E) (E B) (F F)))
= (route-exists? 'B 'D '((A B) (B C) (C E) (D E) (E B) (F F)))
= (route-exists? 'C 'D '((A B) (B C) (C E) (D E) (E B) (F F)))

The hand-evaluation confirms that as the function recurs, it calls itself with the exact same arguments again and again. In other words, the evaluation never stops.

Our problem with route-exists? is again a loss of ``knowledge,'' similar to that of relative-2-absolute in the preceding section. Like relative-2-absolute, route-exists? was developed according to the recipe and is independent of its context. That is, it doesn't ``know'' whether some application is the first or the hundredth of a recursive chain. In the case of route-exists? this means, in particular, that the function doesn't ``know'' whether a previous application in the current chain of recursions received the exact same arguments.

The solution follows the pattern of the preceding section. We add a parameter, which we call accu-seen and which represents the accumulated list of origination nodes that the function has encountered, starting with the original application. Its initial value must be empty. As the function checks on a specific orig and moves to its neighbors, orig is added to accu-seen.

Here is a first revision of route-exists?, dubbed route-exists-accu?:

;; route-exists-accu? : node node simple-graph (listof node)  ->  boolean
;; to determine whether there is a route from orig to dest in sg, 
;; assuming the nodes in accu-seen have already been inspected 
;; and failed to deliver a solution 
(define (route-exists-accu? orig dest sg accu-seen)
  (cond
    [(symbol=? orig dest) true]
    [else (route-exists-accu? (neighbor orig sg) dest sg
	                      (cons orig accu-seen))]))

The addition of the new parameter alone does not solve our problem, but, as the following hand-evaluation shows, provides the foundation for one:

  (route-exists-accu? 'C 'D '((A B) (B C) (C E) (D E) (E B) (F F)) empty)
= (route-exists-accu? 'E 'D '((A B) (B C) (C E) (D E) (E B) (F F)) '(C))
= (route-exists-accu? 'B 'D '((A B) (B C) (C E) (D E) (E B) (F F)) '(E C))
= (route-exists-accu? 'C 'D '((A B) (B C) (C E) (D E) (E B) (F F))
                      '(B E C))

In contrast to the original function, the revised function no longer calls itself with the exact same arguments. While the three arguments proper are again the same for the third recursive application, the accumulator argument is different from that of the first application. Instead of empty, it is now '(B E C). The new value represents the fact that during the search of a route from 'C to 'D, the function has inspected 'B, 'E, and 'C as starting points.

All we need to do at this point is exploit the accumulated knowledge in the function definition. Specifically, we determine whether the given orig is already an item on accu-seen. If so, the problem is trivially solvable with false. Figure 87 contains the definition of route-exists2?, which is the revision of route-exists?. The definition refers to contains, our first recursive function (see part II), which determines whether a specific symbol is on a list of symbols.

;; route-exists2? : node node simple-graph  ->  boolean
;; to determine whether there is a route from orig to dest in sg
(define (route-exists2? orig dest sg)
  (local ((define (re-accu? orig dest sg accu-seen)
            (cond
              [(symbol=? orig dest) true]
              [(contains orig accu-seen) false]
              [else (re-accu? (neighbor orig sg) dest sg (cons orig accu-seen))]))) 
    (re-accu? orig dest sg empty)))

Figure 87:  Finding a route in a simple graph (version 2)

The definition of route-exists2? also eliminates the two minor problems with the first revision. By localizing the definition of the accumulating function, we can ensure that the first call to re-accu? always uses empty as the initial value for accu-seen. And, route-exists2? satisfies the exact same contract and purpose statement as route-exists?.

Still, there is a significant difference between route-exists2? and relative-to-absolute2. Whereas the latter was equivalent to the original function, route-exists2? is an improvement over the route-exists? function. After all, it corrects a fundamental flaw in route-exists?, which completely failed to find an answer for some inputs.

Exercise 30.2.1.   Complete the definition in figure 87 and test it with the running example. Use the strategy of section 17.8 to formulate the tests as boolean-valued expressions.

Check with a hand-evaluation that this function computes the proper result for 'A, 'C, and SimpleG.    Solution

Exercise 30.2.2.   Edit the function in figure 87 so that the locally defined function consumes only those arguments that change during an evaluation.    Solution

Exercise 30.2.3.   Develop a vector-based representation of simple graphs. Adapt the function in figure 87 so that it works on a vector-based representation of simple graphs.    Solution

Exercise 30.2.4.   Modify the definitions of find-route and find-route/list in figure 77 so that they produce false, even if they encounter the same starting point twice.    Solution


64 The most convenient way to construct this list is to evaluate (build-list (add1 N) identity).

65 The time of evaluation will differ from computer to computer. These measurements were conducted on a Pentium 166Mhz running Linux. Measuring timings is also difficult. At a minimum, each evaluation should be repeated several times, and the time reported should be the average of those measurements.