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Define some primitives:

(define (square x) (* x x)) (define (compose f g) (lambda (x) (f (g x))))

Define the procedure:

```
(define (repeat f n)
(if (< n 1)
(lambda (x) x)
(compose f (repeat f (- n 1)))))
```

Test with:

```
((repeat square 2) 5)
```

Output:

625

Another solution using the linear iterative way.

```
(define (repeat f n)
(define (iter n result)
(if (< n 1)
result
(iter (- n 1) (compose f result))))
(iter n identity))
```

Note: This is not linearly iterative as described in the book as a chain of *deferred operations* is still being built.

The above answer does not follow the book's instructions. The book instructs "Write a procedure that takes as inputs a procedure that computes f and a positive integer n and returns the procedure that computes the nth repeated application of f." A correct answer is as follows:

```
(define (repeated f n)
(lambda (x) (cond ((= n 0) x)
(else
((compose (repeated f (- n 1)) f) x)))))
```

I think the following solution is more elegant. When we only need to apply the function once, we can just return the function (I'm not doing error-checking here to see of n is smaller than 1).

```
(define (repeated f n)
(if (= n 1)
f
(compose f (repeated f (- n 1)))))
```

An extremely succinct solution uses the accumulate procedure defined in 1.32:

```
(define (repeated f n)
(accumulate compose identity (lambda (i) f) 1 inc n))
```

An solution with O(log n) complexity using compose:

```
(define (repeated f n)
(cond ((= n 0) identity)
((even? n) (repeated (compose f f) (/ n 2)))
(else (compose f (repeated f (- n 1))))))
```

A logarithmic iterative solution.

(define (lcompose f g) (lambda (x) (g (f x)))) (define (identity x) x) (define (repeated f n) (define (iter g h m) (cond ((= m 0) g) ((odd? m) (iter (lcompose g h) h (- m 1))) (else (iter g (lcompose h h) (/ m 2))))) (iter identity f n))

I personally would rather like to avoid using compose (as the only actual use I see in it would be in the logarithmic solution given above).

```
(define (repeated f n)
(define (repeated-step counter input)
(if (> counter n)
input
(f (repeated-step (+ counter 1) input))
)
)
(lambda (x) (repeated-step 1 x))
)
```

What do you guys think about this way of doing it? I believe it generates a linear iterative process:

```
(define (repeated f n)
(lambda (x)
(define (iter n result)
(cond ((= n 0) result)
((odd? n) (iter (- n 1) (f result)))
(else
(iter (- n 2) ((compose f f) result)))))
(iter n x)))
```

Is this another correct way of solving? I get the right answer when testing the given example from the book: ((repeated square 2) 5) = 625. I know I'm missing something. New to Scheme and this course, so feedback/criticisms welcomed. Could someone provide a good explanation of this?

```
(define (repeated f n)
(lambda (x) (f (f x))))
```

^That's not quite right, that just applies f twice. The goal is to apply f *n* times, as below:

(repeated f 1) ;; f(x) (repeated f 2) ;; f(f(x)) (repeated f 3) ;; f(f(f(x))) ...

As a hint, notice how your solution takes *n* as input and yet doesn't do anything with it in the body - that's a good indicator that something's off.

iterative solution using double with log(n) time complexity

```
(define (repeated2-iter f n)
(define (iter n current)
(cond ((= 1 n) current)
((even? n) (double (iter (/ n 2) current)))
(else (compose f (iter (- n 1) (compose f current))))
)
)
(iter n f)
)
```

My own solution, that does not use compose. Should be considered silly.

(define (repeated f n) (if (= n 1) (lambda (x) (f x)) (lambda (x) (f ((repeated f (- n 1)) x))))) ((repeated square 2) 5)