## Solution to SICP Exercise 2.12 One solution to Exercise 2.12:

```(define (make-center-percent c p) (make-center-width c (* c (/ p 100))))```

``` ```

```(define (percent i) (* (/ (width i) (center i)) 100)) ```

## Solution to SICP Exercise 2.11 One solution to Exercise 2.11:

``````(define (mul-interval x y)
(let* ((lx (lower-bound x))
(ux (upper-bound x))
(ly (lower-bound y))
(uy (upper-bound y))
(pos-lx? (positive? lx))
(pos-ux? (positive? ux))
(pos-ly? (positive? ly))
(pos-uy? (positive? uy)))
(cond
; lx ux ly uy example
; ----------------------------------
; + - + + invalid interval
; + - + - invalid interval
; + - - + invalid interval
; + - - - invalid interval
((and pos-lx? (not pos-ux?))
(error "invalid interval" x))

; + + + - invalid interval
; - + + - invalid interval
; - - + - invalid interval
((and pos-ly? (not pos-uy?))
(error "invalid interval" y))

; + + + + (1.2)(2.3) = (2.6)
((and pos-lx? pos-ux? pos-ly? pos-uy?)
(make-interval (* lx ly) (* ux uy)))

; + + - + (1.2)(-2.3) = (-4.6)
((and pos-lx? pos-ux? (not pos-ly?) pos-uy?)
(make-interval (* ux ly) (* ux uy)))

; + + - - (1.2)(-2.-1) = (-4.-1)
((and pos-lx? pos-ux? (not pos-ly?) (not pos-uy?))
(make-interval (* ux ly) (* lx uy)))

; - + + + (-1.2)(2.3) = (-3.6)
((and (not pos-lx?) pos-ux? pos-ly? pos-uy?)
(make-interval (* lx uy) (* ux uy)))

; - + - + (-1.2)(-2.3) = (-4.6) *
((and (not pos-lx?) pos-ux? (not pos-ly?) pos-uy?)
(make-interval (min (* lx uy) (* ux ly))
(* ux uy)))

; - + - - (-1.2)(-2.-1) = (-4.2)
((and (not pos-lx?) pos-ux? (not pos-ly?) (not pos-uy?))
(make-interval (* ux ly) (* lx ly)))

; - - + + (-2.-1)(2.3) = (-6.-2)
((and (not pos-lx?) (not pos-ux?) pos-ly? pos-uy?)
(make-interval (* lx uy) (* ux ly)))

; - - - + (-2.-1)(-2.3) = (-6, 4)
((and (not pos-lx?) (not pos-ux?) (not pos-ly?) pos-uy?)
(make-interval (* lx uy) (* lx ly)))

; - - - - (-2.-1)(-2.-1) = (1.4)
((and (not pos-lx?) (not pos-ux?) (not pos-ly?) (not pos-uy?))
(make-interval (* ux uy) (* lx ly))))))``````

## Solution to SICP Exercise 2.10 One solution to Exercise 2.10:

```(define (spans-zero? x) (not (or (positive? (lower-bound x)) (negative? (upper-bound x)))))```

``` ```

```(define (div-interval x y) (if (spans-zero? y) (error "Divisor spans zero" y) (mul-interval x (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y)))))) ```

## Solution to SICP Exercise 2.9 A solution to Exercise 2.9:

Some preliminary definitions:

```(upper-bound (make-interval a b)) = a  (lower-bound (make-interval a b)) = b ```

``` (width i) = (/ (- (upper-bound i) (lower-bound i)) 2)  ```

```(add i1 i2) = (make-interval (+ (upper-bound i1) (upper-bound i2)) (+ (lower-bound i1) (lower-bound i2)))  ```

Rearranging :

```(* 2 (width i)) = (- (upper-bound i) (lower-bound i)) (+ (* 2 (width i)) (lower-bound i)) = (upper-bound i)  ```

From :

```(width (add i1 i2)) = (/ (- (upper-bound (add i1 i2)) (lower-bound (add i1 i2))) 2) ```

Simplifying with ,  and :

```(width (add i1 i2)) = (/ (- (+ (upper-bound i1) (upper-bound i2)) (+ (lower-bound i1) (lower-bound i2))) 2) ```

Substituting in :

``` = (/ (- (+ (+ (* 2 (width i1)) (lower-bound i1)) (+ (* 2 (width i2)) (lower-bound i2))) (+ (lower-bound i1) (lower-bound i2))) 2)```

``` ```

``` = (/ (- (+ (* 2 (width i1)) (* 2 (width i2)) (lower-bound i1) (lower-bound i2)) (+ (lower-bound i1) (lower-bound i2))) 2) ```

All the (lower-bound x)s cancel out, leaving:

``` = (/ (+ (* 2 (width i1)) (* 2 (width i2))) 2) ```

So do the 2s:

`(width (add i1 i2)) = (+ (width i1) (width i2))`

Clearly the width of the sum is a function only of the width of the operands.

## Solution to Exercise SICP 2.8 A solution to Exercise 2.8:

The maximum the difference could be is difference between the upper bound of the first interval and the lower bound of the second. The minimum difference is the difference between the lower bound of the first and the upper bound of the second. This holds true even if the second interval is greater than the first or the intervals overlap.

```(define (sub-interval x y) (make-interval (- (lower-bound x) (upper-bound y)) (- (upper-bound x) (lower-bound y)))) ```

## Solution to Exercise SICP 2.6 One solution to Exercise 2.6:

`(define zero (lambda (f) (lambda (x) x)))`

``` (define (add-1 n) (lambda (f) (lambda (x) (f ((n f) x))))) ; (add-1 zero) ; (lambda (f) (lambda (x) (f ((zero f) x)))) ; (lambda (f) (lambda (x) (f (((lambda (g) (lambda (y) y)) f) x)))) ; (lambda (f) (lambda (x) (f ((lambda (y) y) x)))) ; (lambda (f) (lambda (x) (f x))) (define one (lambda (f) (lambda (x) (f x)))) ; (add-1 one) ;(lambda (f) (lambda (x) (f ((one f) x)))) ;(lambda (f) (lambda (x) (f (((lambda (g) (lambda (y) (g (y)))) f) x)))) ;(lambda (f) (lambda (x) (f ((lambda (y) (f (y))) x)))) ;(lambda (f) (lambda (x) (f (f (x))))) (define two (lambda (f) (lambda (x) (f (f x))))) (define (add a b) (lambda (f) (lambda (x) ((a f) ((b f) x))))) ```

```; transform Church numerals to integers (for testing) (define (to-integer n) (define (inc x) (+ 1 x)) ((n inc) 0)) ```