Node:No deferment solution, Previous:No Deferment, Up:Recursion
The solution to the problem of deferred operations is to write in a manner that does not defer operations1. This requires writing to a different pattern, often one that involves writing two function definitions, an `initialization' function and a `helper' function.
The `initialization' function sets up the job; the `helper' function does the work.
Here are the two function definitions for adding up numbers. They are so simple, I find them hard to understand.
(defun triangle-initialization (number) "Return the sum of the numbers 1 through NUMBER inclusive. This is the `initialization' component of a two function duo that uses recursion." (triangle-recursive-helper 0 0 number))
(defun triangle-recursive-helper (sum counter number)
"Return SUM, using COUNTER, through NUMBER inclusive.
This is the `helper' component of a two function duo
that uses recursion."
(if (> counter number)
sum
(triangle-recursive-helper (+ sum counter) ; sum
(1+ counter) ; counter
number))) ; number
Install both function definitions by evaluating them, then call
triangle-initialization with 2 rows:
(triangle-initialization 2)
=> 3
The `initialization' function calls the first instance of the `helper' function with three arguments: zero, zero, and a number which is the number of rows in the triangle.
The first two arguments passed to the `helper' function are
initialization values. These values are changed when
triangle-recursive-helper invokes new instances.2
Let's see what happens when we have a triangle that has one row. (This triangle will have one pebble in it!)
triangle-initialization will call its helper with
the arguments 0 0 1. That function will run the conditional
test whether (> counter number):
(> 0 1)
and find that the result is false, so it will invoke
the then-part of the if clause:
(triangle-recursive-helper
(+ sum counter) ; sum plus counter => sum
(1+ counter) ; increment counter => counter
number) ; number stays the same
which will first compute:
(triangle-recursive-helper (+ 0 0) ; sum
(1+ 0) ; counter
1) ; number
which is:
(triangle-recursive-helper 0 1 1)
Again, (> counter number) will be false, so again, the Lisp
interpreter will evaluate triangle-recursive-helper, creating a
new instance with new arguments.
This new instance will be;
(triangle-recursive-helper
(+ sum counter) ; sum plus counter => sum
(1+ counter) ; increment counter => counter
number) ; number stays the same
which is:
(triangle-recursive-helper 1 2 1)
In this case, the (> counter number) test will be true! So the
instance will return the value of the sum, which will be 1, as
expected.
Now, let's pass triangle-initialization an argument
of 2, to find out how many pebbles there are in a triangle with two rows.
That function calls (triangle-recursive-helper 0 0 2).
In stages, the instances called will be:
sum counter number (triangle-recursive-helper 0 1 2) (triangle-recursive-helper 1 2 2) (triangle-recursive-helper 3 3 2)
When the last instance is called, the (> counter number) test
will be true, so the instance will return the value of sum,
which will be 3.
This kind of pattern helps when you are writing functions that can use many resources in a computer.
The phrase tail recursive is used to describe such a process, one that uses `constant space'.
The
jargon is mildly confusing: triangle-recursive-helper uses a
process that is iterative in a procedure that is recursive. The
process is called iterative because the computer need only record the
three values, sum, counter, and number; the
procedure is recursive because the function `calls itself'. On the
other hand, both the process and the procedure used by
triangle-recursively are called recursive. The word
`recursive' has different meanings in the two contexts.