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```TEACH RECURSE3                          Aaron Sloman November 1989

RECURSION AND THE TERMINATING VALUE

People learning about recursion often have considerable difficulty over
what to do with the terminating condition, especially when recursing
down a list. What result should be returned when the empty list [] is
found?

In particular, difficulties often arise if it is decided to make
-false- the result when the list is empty, and otherwise to return a
list of items.

This teach file explores some of the design issues that need to be
understood in dealing with such cases.

CONTENTS - (Use <ENTER> g to access required sections)

-- The problem
-- How can we think about taking this decision? Two issues
-- Addressing the first issue: Q1
-- SUMMARY of answer to Q1:
-- What if the answers to Q1 and Q2 are different ?
-- The complicated solution
-- The simpler (divide and conquer) solution
-- POP-11 allows nested procedure definitions
-- Final silly exercise

-- The problem --------------------------------------------------------

We are talking about a schema of the SECOND of the following two forms,
representing a procedure P that takes a list, and possibly other
arguments (e.g. a predicate), and recurses down the list, looking at
every element for some purpose, and returns some result.

Schema (A)
The procedure produces no result (e.g. because it is doing something
with some or all elements of the list such as printing them on the
terminal, or putting them into the database or writing them to a file).

define P(list, .....);

.....

enddefine;

Ignore Schema (A) for now. It is mentioned simply to point out that the
following discussion does not apply to it, as we are considering only
recursive procedures that produce a result (on the stack).

Schema (B)
A result is produced.

define P(list, ....) -> result;

......

enddefine;

The problem is: what should the result for the stopping condition
be?
define P(list, ....) -> result;

if list == [] then
???? -> result
else(if)
......
......

endif

enddefine;

-- How can we think about taking this decision? Two issues ------------

In order to answer this question you have to separate two different
questions and think VERY clearly about them.

Q1. What should the procedure P return when given an empty list?

Q2. What does the result of P have to be for the recursive terminating
case in order that the result can be used sensibly for
the other (non-terminating) cases?

Let's ignore Q2 for now, and concentrate on Q1.

-- Addressing the first issue: Q1 -------------------------------------

Q1 Has to be answered on the basis of how the problem is defined
originally, and THAT may depend on what you want to use the procedure P
for, e.g. which other procedures will call it.

CASE 1 The procedure is to be used in CONDITIONs like these:

if P(.........) then .....
until P(.........) do .....
while P(.........) do .....

then the answer to Q1 should be that the procedure produces a result
that is essentially BOOLEAN, i.e. sometimes false, sometimes not.
(Remember in Pop-11 anything other than false counts as true, in a
condition.) If P never returns false, in these contexts, then the
condition will always be true, and there is no point using a
condition.

CASE 2 The result of the procedure is to be used as one argument for
an operation or procedure that REQUIRES a list as input, as in:

....  ::  P(.....)              (Second arg of "::" MUST be a list)

P(.....) <> .......             (Both args of "<>" MUST be lists)

[^^(P(......)  ....]            ("^^" in a list REQUIRES a list)

rev(P(.....))                   (Arg of "rev" MUST be a list)

Then for these uses P MUST always return a list (which could be [], as
that is a list).

CASE 3 The result of P is used as one argument for an arithmetic
operation, e.g. "+" or "*", as in

x + P(x - 1)                    (Arguments for "+" MUST be numbers)

Then the result of P MUST be a number (e.g. it could be 0).

CASE 4 The result of P is used as input to some other procedure
that requires  a particular type of result (e.g. a string, or a
vector, etc), e.g.

subscrs(num, P(.....))
subscrv(num, P(.....))

then the result of P MUST be an object of the required type.

CASE 5 the result of P is used only as input for procedures that will
operate on ANY kind of object, e.g. the first argument of -member- can
be any kind of object.

In that case you don't get any programming constraints on the result of
P, though there may be other constraints, that depend on what the
procedure is supposed to be doing.

-- SUMMARY of answer to Q1: -------------------------------------------

The result produced by P depends on the kind of use to which the result
is to be put. So for some procedures the result should be boolean, for
others a (possibly empty) list, for others a (possibly zero) number,
etc.

-- How to answer Q2 ---------------------------------------------------

As you may already have noticed, in a recursive procedure P, the result
of a recursive call will itself be used in some way, and THAT use can
determine what sort of result the recursive call should produce,
including the terminating case.

E.g. if P is making a list of objects, and the result of the recursive
call is used in creating that list, i.e. the recursive call is used
something like this:

hd(list)  ::  P(tl(list), ....) -> result

P(tl(list), ....) <> .......    -> result

[^^(P(tl(list), ....) ....]     -> result

then the recursive call (on tl(list)) MUST always return a list,
possibly []. It must NOT return FALSE, or the program will produce

;;; MISHAP - LIST NEEDED
;;; INVOLVING:  <false>
;;; DOING    :  :: P(*3)  .....

where P(*3) means it was three levels deep in recursion in P when the
error occurred, and "::" in the DOING list tells you that it is
complaining about <false>  being given as argument to "::"

If P is traced the last bit of tracing before the error would be
something like this
!!!> P []
!!!< P <false>

i.e. it recursed on [], returned false, and then the PREVIOUS call of P
waiting for the result of P([]) complained.

So if the result of the recursive call is used as input to a
list-processing procedure, then the result for the terminating condition
MUST be a list. This will usually be [], but it may be some other list
in more complex cases.

Similarly, if the result of the recursive call is used as input for
an arithmetic operation, then the result of the terminating condition
MUST be a number, e.g. 0 or 1, as in

define P(list) -> result
if list == [] then
0 -> result
else
hd(list) + P(tl(list)) -> result

endif
enddefine;

-- What if the answers to Q1 and Q2 are different ? -------------------

Then you have a problem. There are two different solutions. The
complicated one is to try to make one procedure definition for P handle
the situation. The simpler solution is to use two procedures, one that
does the work, and one that calls it and decides whether to return
false.

-- The complicated solution -------------------------------------------

Here for example is a version of findall that takes a
list and a predicate and returns false if NO items in the list satisfy
the predicate, otherwise a list of all the items that do.

define findall(list, pred) -> out;

if list == [] then
false -> out
else
;;; first get the result for the tail, and then decide what
findall(tl(list), pred) -> out;

if out then
;;; out is not false, so it must be a list. Use it
if pred(hd(list)) then
[^(hd(list)) ^^out] -> out
else
;;; do nothing - return previous value of out
endif
else
;;; decide what to by testing hd(list)
if pred(hd(list)) then
;;; ignore previous value of out, because it is false
[^(hd(list))] -> out
else
;;; do nothing - return previous (false) value of out
endif
endif
endif
enddefine;

NOTE: you can remove the empty "else" clauses.

We can test this to make sure it works.

findall([], isword) =>
** <false>
findall([cat],isword) =>
** [cat]
findall([1 cat on 2 mats],isword) =>
** [cat on mats]

-- The simpler (divide and conquer) solution --------------------------

The above definition is fairly complex, even if you ignore the empty
"else" clauses. But you can do it more simply by defining a recursive
sub-procedure called sub_find, that returns a LIST under all conditions,
and then define findall to see if the list is empty.

define findall(list, pred) -> out;

sub_find(list, pred) -> out;
;;; decide whether to replace the value of out with false
if out == [] then
false -> out
endif;

enddefine;

define sub_find(list, pred) -> out;

if list == [] then
[] -> out
elseif pred(hd(list)) then
[^(hd(list)) ^^(sub_find(tl(list), pred))] -> out
else
sub_find(tl(list), pred) -> out
endif
enddefine;

findall([], isword) =>
** <false>

findall([cat],isword) =>
** [cat]

findall([1 cat on 2 mats],isword) =>
** [cat on mats]

NOTE
You might have been tempted to write

[^^(sub_find(tl(list), pred))] -> out

Can you see why that is unnecessary complexity? (The square brackets say
make a list. "^^(...)" says put in all the elements of the list you
started with!).

-- POP-11 allows nested procedure definitions -------------------------

If the procedure -sub_find- is used nowhere but in findall, then we can
hide it from everything else, by actually putting its definition INSIDE
that of findall. If you do that, make sure that you give the nested
definition extra indentation to show very clearly that it is nested.

<ENTER> tidy (or <ENTER> jcp if the cursor is at the beginning of the
outer definition) will do this automatically. (See HELP * MARK )

This is how the nested procedures look:

define findall(list, pred) -> out;

define sub_find(list, pred) -> out;

if list == [] then
[] -> out
elseif pred(hd(list)) then
[^(hd(list)) ^^(sub_find(tl(list), pred))] -> out
else
sub_find(tl(list), pred) -> out
endif
enddefine;

sub_find(list, pred) -> out;

;;; decide whether to replace the value of out with false
if out == [] then
false -> out
endif;

enddefine;

You can make your nested procedure slightly more efficient if you
replace the nested "define" with "define lconstant", as in:

define findall(list, pred) -> out;

define lconstant sub_find(list, pred) -> out;

if list == [] then
[] -> out
elseif pred(hd(list)) then
[^(hd(list)) ^^(sub_find(tl(list), pred))] -> out
else
sub_find(tl(list), pred) -> out
endif
enddefine;

sub_find(list, pred) -> out;

;;; decide whether to replace the value of out with false
if out == [] then
false -> out
endif;

enddefine;

-- Final silly exercise -----------------------------------------------

As a test for whether you have understood all that, try to define
a recursive procedure called silly that takes a list of numbers and
returns the sum of all the numbers if the list is non-empty, otherwise,
for the empty list, returns the result FALSE, not 0!

silly([]) =>
** false

silly([3 4 5]) =>
** 12