Count Fibonacci "cost" in scheme - scheme

Question

I have to make a function that finds the "cost" of a Fibonacci number. My Fibonacci code is
(define fib (lambda (n) (cond
((< n 0) 'Error)
((= n 0) 0)
((= n 1) 1)
(else (+ (fib (- n 1)) (fib (- n 2)))))))
Each + or - that is used to evaluate a fib number is worth $1. Each < or > is worth $0.01. For example, 1 is worth $0.01, 2 is worth $3.03, etc. I don't know how to count the number of +, -, <, and >. Do I need the fib code in my fibCost code?

Answer 1

I'm not sure whether or not you wanted the solution to include the original code or not. There are direct ways of computing the cost, but I think it's interesting to look at ways that are similar to instrumenting the existing code. That is, what can we change so that something very much like the original code will compute what we want?
First, we can replace the arithmetic operators with a bit of indirection. That is, instead of calling (+ x y), you can call (op + 100 x y), which increments the total-cost variable.
(define (fib n)
(let* ((total-cost 0)
(op (lambda (fn cost . args)
(set! total-cost (+ total-cost cost))
(apply fn args))))
(let fib ((n n))
(cond
((op < 1 n 0) 'error)
((= n 0) 1)
((= n 1) 1)
(else (op + 100
(fib (op - 100 n 1))
(fib (op - 100 n 2))))))
total-cost))
That doesn't let us keep the original code, though. We can do better by defining local versions of the arithmetic operators, and then using the original code:
(define (fib n)
(let* ((total-cost 0)
(op (lambda (fn cost)
(lambda args
(set! total-cost (+ total-cost cost))
(apply fn args))))
(< (op < 1))
(+ (op + 100))
(- (op - 100)))
(let fib ((n n))
(cond
((< n 0) 'error)
((= n 0) 1)
((= n 1) 1)
(else (+ (fib (- n 1))
(fib (- n 2))))))
total-cost))
> (fib 1)
1
> (fib 2)
303
> (fib 3)
605
> (fib 4)
1209
What's nice about this approach is that if you start using macros to do some source code manipulation, you could actually use this as a sort of poor-man's profiler, or tracing system. (I'd suggest sticking with the more robust tools provided by the implementation, of course, but there are times when a technique like this can be useful.)
Additionally, this doesn't even have to compute the Fibonnaci number anymore. It's still computed because we do (apply fn args), but if we remove that, then we never even call the original arithmetic operation.

Answer 2

The quick and dirty solution would be to define a counter variable each time the cost procedure is started, and update it with the corresponding value at each branch of the recursion. For example:
(define (fib-cost n)
(let ((counter 0)) ; counter initialized with 0 at the beginning
(let fib ((n n)) ; inner fibonacci procedure
; update counter with the corresponding cost
(set! counter (+ counter 0.01))
(when (> n 1)
(set! counter (+ counter 3)))
(cond ((< n 0) 'Error)
((= n 0) 0)
((= n 1) 1)
(else (+ (fib (- n 1)) (fib (- n 2))))))
counter)) ; return the counter at the end
Answering your second question - no, we don't need the whole fib code; given that we're not interested in the actual value of fibonacci, the above can be further simplified to just make the required calls and ignore the returned values:
(define (fib-cost n)
(let ((counter 0)) ; counter initialized with 0 at the beginning
(let fib ((n n)) ; inner fibonacci procedure
; update counter with the corresponding cost
(set! counter (+ counter 0.01))
(when (> n 1)
(fib (- n 1))
(fib (- n 2))
(set! counter (+ counter 3))))
counter)) ; return the counter at the end

Answer 3

You have +/- just anytime you call the code recursively, in the else Part. So, easily anytime you enter the else part, you should count 3 of them. One for f(n-1), one for f(n-2) and one for f(n-1) + f(n-2).

Answer 4

Just for fun, a solution using syntactic extensions (aka "macros").
Let's define the following:
(define-syntax-rule (define-cost newf oldf thiscost totalcost)
(define (newf . parms)
(set! totalcost (+ totalcost thiscost))
(apply oldf parms)))
Now we create procedures based on the original procedures you want to have a cost:
(define-cost +$ + 100 cost)
(define-cost -$ - 100 cost)
(define-cost <$ < 1 cost)
so using +$ will do an addition and increase a cost counter by 1, and so on.
Now we adapt your inititial procedure to use the newly defined ones:
(define fib
(lambda (n)
(cond
((<$ n 0) 'Error)
((= n 0) 0)
((= n 1) 1)
(else
(+$ (fib (-$ n 1)) (fib (-$ n 2)))))))
For convenience, we create a macro to return both the result of a procedure and its cost:
(define-syntax-rule (howmuch f . args)
(begin
(set! cost 0)
(cons (apply f 'args) cost)))
then a cost variable
(define cost #f)
and off we go
> (howmuch fib 1)
'(1 . 1)
> (howmuch fib 2)
'(1 . 303)
> (howmuch fib 10)
'(55 . 26577)
> (howmuch fib 1)
'(1 . 1)