sicp exercise 1.41

;  Exercise 1.41.  Define a procedure double that takes a procedure of one argument as argument and
;  returns a procedure that applies the original procedure twice. For example, if inc is a procedure
;  that adds 1 to its argument, then (double inc) should be a procedure that adds 2. What value is returned
;  by

; (((double (double double)) inc) 5)


(define (inc x) (+ x 1))

(define (double f) (lambda (x) (f (f x))))

(display (inc 10)) (newline)
(display ((double inc) 10)) (newline)

(display (((double (double double)) inc) 5)) (newline)

; (double double) means 4 times
; ((double (4 times)) means 4 times 4 = 16 times increment
; Ans: 21

(display (((double (double (double double))) inc) 10)) (newline)
; gives 16 times 16 = 256 increments