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sicp-exercises.scm
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(define (average x y)
(/ (+ x y) 2))
(define (improve guess x)
(average guess (/ x guess)))
(define (square x)
(* x x))
(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))
(define (sqrt-iter guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))
(define (sqrt-iter-2 guess old-guess x)
(if (< (abs (- guess old-guess)) .0001)
guess
(sqrt-iter-2 (improve guess x) guess x)))
(define (sqrt-3 x)
(sqrt-iter-2 1.0 100.0 x))
(define (sqrt x)
(sqrt-iter 1.0 x))
(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))
(define (sqrt-iter2 guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))
(define (sqrt2 x)
(sqrt-iter2 1.0 x))
; exercise 1.23
; for 1.24 need a working runtime function
; exercise 1.26
; exercise 1.27
; exercise 1.28
; non-trivial expmod
(define (expmod-nt base exp m)
(cond ((= exp 0) 1)
((even? exp)
(if (and (not (= base 1))
(not (= base (- m 1)))
(= (remainder (square base) m) 1))
0
(remainder (square (expmod base (/ exp 2) m)) m)))
(else (remainder (* base (expmod base (- exp 1) m)) m))))
; doesn't seem to be working 561 should return #f
(define (miller-rabin-test n)
(define (try-it a)
(if (= (expmod-nt a n n) 0)
#f
(= (expmod-nt a n n) a)))
(try-it (+ 1 (random (- n 1)))))
; from the text:
(define (fermat-test n)
(define (try-it a)
(= (expmod a n n) a))
(try-it (+ 1 (random (- n 1)))))
; next section notes:
; in general lambda is used to create procedures in the same way as define, except that no name is specified for the procedure.
(lambda (<formal parameters>) <body>)
;like:
(lambda (x) (+ x 1))
(define (plus4 x) (+ x 4)) == (define plus4 (lambda (x) (+ 4 x)))
(define (search f neg-point pos-point)
(let ((midpoint (average neg-point pos-point)))
(if (close-enough? neg-point pos-point)
midpoint
(let ((test-value (f midpoint)))
(cond ((positive? test-value)
(search f neg-point midpoint))
((negative? test-value)
(search f midpoint pos-point))
(else midpoint))))))
(define (close-enough? x y) (< (abs (- x y)) 0.001))
(define (half-interval-method f a b)
(let ((a-value (f a))
(b-value (f b)))
(cond ((and (negative? a-value) (positive? b-value))
(search f a b))
((and (negative? b-value) (positive? a-value))
(search f b a))
(else
(error "Values are not of opposite sign" a b)))))
(half-interval-method sin 2.0 4.0)
(define (average x y) (/ (+ x y) 2))
(define tolerance 0.00001)
(define (fixed-point f first-guess)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try next))))
(try first-guess))
(define (sqrt x)
(fixed-point (lambda (y) (average y (/ x y))) 1.0))
; ^^ this technique is called average damping
; from other file:
(define (cube a) (* a a a))
(define (sum-cubes a b)
(if (> a b) 0 (+ (cube a) (sum-cubes (+ a 1) b))))
(sum-cubes 1 2)
(+ (cube 1) (sum-cubes (+ 1 1) 2))
(+ 1 (sum-cubes 2 2))
(+ 1 (+ (cube 2) (sum-cubes (+ 2 1) 2)))
(+ 1 (+ 8 0))
9
(define (inc n) (+ n 1))
(define (sum-cubes a b)
(sum cube a inc b))
(define (sum term a next b)
(if (> a b)
0
(+ (term a)
(sum term (next a) next b))))
(define (identity x) x)
(define (sum-ints a b)
(sum identity a inc b))
(define (integral f a b dx)
(define (add-dx x) (+ x dx))
(* (sum f (+ a (/ dx 2.0)) add-dx b) dx))
(integral cube 0 1 0.01)
(* (sum cube (+ 0 (/ 0.01 2.0)) add-dx 1) 0.01)
(* (+ (cube 0) (sum cube (add-dx 0) add-dx 1)) 0.01)
(* (+ (cube 0) (sum cube (+ 0 0.01) add-dx 1)) 0.01)
(* (+ (cube 0) (sum cube 0.01 add-dx 1)) 0.01)
(* (+ (cube 0) (+ (cube 0.01) (sum cube (add-dx 0.01) add-dx 1))) 0.01)
(* (+ 0 (+ 1.0e-6 (sum cube (add-dx 0.01) add-dx 1))) 0.01)
(* (+ 0 (+ 1.0e-6 (sum cube 0.02 add-dx 1))) 0.01)
(* (+ 0 (+ 1.0e-6 (+ 8.0e-6 (sum cube (add-dx 0.02) add-dx 1)))) 0.01)
(integral cube 0 1 0.01)
.24998750000000042
(integral cube 0 1 0.001)
.249999875000001
(The exact value of the integral of cube between 0 and 1 is 1/4.)
h = (b - a) / n ; for some even integer n
ysub k = f (a + kh)
(define (simpson f a b n)
(let (h (/ (- b a) n))
(define (next-a k)
(* (+ 1 (remainder k 2)) (+ a (* k h))))
(* (/ h 3) (sum f a next-a b))))
; again!
(define (sum term a next b)
(if (> a b)
0
(+ (term a)
(sum term (next a) next b))))
(define (simpsons f a b n)
(define h (/ (- b a) n))
(define (inc x) (+ x 1))
(define (y k)
(f (+ a (* k h))))
(define (term k)
(* (cond ((odd? k) 4)
((or (= k 0) (= k n)) 1)
((even? k) 2))
(y k)))
(/ (* h (sum term 0 inc n)) 3))
(define (cube x) (* x x x))
(simpsons cube 0 1 4)
(/ (* (/ (- 1 0) 4) (sum term 0 inc 3)) 3)
(/ (* .25 (sum term 0 inc 4)) 3)
(/ (* .25 (+ (term 0) (sum term (inc 0) inc 4))) 3)
(/ (* .25 (+ (* 1 (y 0)) (sum term (inc 0) inc 4))) 3)
(/ (* .25 (+ (* 1 (f (+ 0 (* 0 .25))))) (sum term (inc 0) inc 4))) 3)
(define (reverse l)
(if (null? (cdr l))
l
(append (reverse (cdr l)) (list (car l)))))
; accumulate form:
(define (reverse l)
(define (helper li ac)
(if (null? (cdr li))
(cons (car li) ac)
(helper (cdr li) (cons (car li) ac))))
(helper (cdr l) (cons (car l) ())))
; TODO this returns incorrectly for (list 1 2 3 4)
(define (deep-reverse l)
(if (not (list? l))
l
(if (null? (cdr l))
(deep-reverse (car l))
;; probably just need to use append instead of cons here
(cons (deep-reverse (cdr l)) (list (deep-reverse (car l)))))))
(define x (list (list 1 2) (list 3 4)))
(reverse x)
; ((3 4) (1 2))
(deep-reverse x)
; ((4 3) (2 1))
; exercise 2.30
(define (square-tree tree)
(cond ((null? tree) ())
((not (pair? tree)) (* tree tree))
(else (cons (square-tree (car tree))
(square-tree (cdr tree))))))
(define (square-tree tree)
(map (lambda (sub-tree)
(if (pair? sub-tree)
(square-tree sub-tree)
(* sub-tree sub-tree)))
tree))
(square-tree
(list 1
(list 2 (list 3 4) 5)
(list 6 7)))
; exercise 2.31
(define (tree-map fn tree)
(map (lambda (sub-tree)
(if (pair? sub-tree)
(square-tree sub-tree)
(fn sub-tree)))
tree))
(define (square-tree tree) (tree-map square tree))
; exercise 2.32
(define (subsets s)
(if (null? s)
(list ())
(let ((rest (subsets (cdr s))))
(append rest (map (lambda (x)
(cons (car s) x)) rest)))))
(subsets (list 1 2 3))
(() (3) (2) (2 3) (1) (1 3) (1 2) (1 2 3))
rest will be a list of lists, so when you cons the car of the current s
with each of rest's members the result is a list of lists with the first element
of the passed list (s) added to it. The reason you don't end up with any lists containing
a distinct element and nil (for example (2 ()) ) is due to the base case returning a
list containing nil, which gets decomposed by map (which returns a list) and
then getting appended to the recursive result of the (subsets) of the cdr of the current set.
Or explained via the example given:
in the example given the recursive subset calls result in ((2 3) (2) (3) ()) so when cons'ing with
1 we get ((1 2 3) (1 2) (1 3) (1)) - which gets appended to the result
this repeats for (2 3) where we get ((3) ())
when cons'ed with 2 we get ((2 3) (2)) - which gets appended to the result
this repeats for (3) where we get ()
when cons'ed with 3 we get ((3) ()) - which gets appended to the result
; exercise 2.34
(define (horner-eval x coefficient-sequence)
(accumulate (lambda (this-coeff higher-terms)
(if (pair? higher-terms)
(+ this-coeff (* x (horner-eval x higher-terms)))
(+ this-coeff (* x higher-terms))))
0
coefficient-sequence))
(horner-eval 2 (list 1 3 0 5 0 1)); 79
; exercise 2.35
; from the text:
(define (count-leaves x)
(cond ((null? x) 0)
((not (pair? x)) 1)
(else (+ (count-leaves (car x))
(count-leaves (cdr x))))))
; from the text:
(define (enumerate-tree tree)
(cond ((null? tree) ())
((not (pair? tree)) (list tree))
(else (append (enumerate-tree (car tree))
(enumerate-tree (cdr tree))))))
(enumerate-tree (list 1 (list 2 (list 3 4)) 5))
; (1 2 3 4 5)
(define (count-leaves t)
(accumulate + 0 (map (lambda (r) 1) (enumerate-tree t))))
(define y (list 1 (list 2 (list 3 4)) 5))
(define x (list (list 1 2) (list 3 4)))
(length x); 3
(count-leaves x); 4
(count-leaves y); 5
; exercise 2.36
(define (accumulate-n op init seqs)
(if (null? (car seqs))
()
(cons (accumulate op init (map (lambda (x) (car x)) seqs))
(accumulate-n op init (map (lambda (x) (cdr x)) seqs)))))
(define y (list (list 1 2 3) (list 4 5 6) (list 7 8 9) (list 10 11 12)))
(accumulate-n + 0 y)
; exercise 2.37
(define mat (list (list 1 2 3 4) (list 4 5 6 6) (list 6 7 8 9)))
(define vec (list 2 2 2 2))
(define (dot-product v w)
(accumulate + 0 (map * v w)))
(define (matrix-*-vector m v)
(map (lambda (x) (accumulate-n * 1 (list x v))) m))
(matrix-*-vector mat vec)
;Value 35: ((2 4 6 8) (8 10 12 12) (12 14 16 18))
(define (transpose mat)
(accumulate-n cons () mat))
(transpose mat)
; ((1 4 6) (2 5 7) (3 6 8) (4 6 9))
; mat:
((1 2 3 4)
(4 5 6 6)
(6 7 8 9))
; transpose of mat:
((1 4 6)
(2 5 7)
(3 6 8)
(4 6 9))
(define (matrix-*-matrix m n)
(let ((cols (transpose n)))
(map (lambda (y)
(map (lambda (x)
(accumulate +
0
(accumulate-n *
1
(list y x))))
mat))
mat)))
(define b (transpose mat))
(matrix-*-matrix mat b)
;((30 56 80) (56 113 161) (80 161 230))
; exercise 2.38
(define (fold-left op initial sequence)
(define (iter result rest)
(if (null? rest)
result
(iter (op result (car rest))
(cdr rest))))
(iter initial sequence))
(define fold-right accumulate)
(fold-right / 1 (list 1 2 3)); 3/2
(fold-left / 1 (list 1 2 3)); 1/6
(fold-right list () (list 1 2 3)); (1 (2 (3 ())))
(fold-left list () (list 1 2 3)); (((() 1) 2) 3)
The operation must produce the same value regardless of the
input order. For example addition and multiplication produce the same output:
(fold-right * 1 (list 1 2 3)); 6
(fold-left * 1 (list 1 2 3)); 6
(fold-right + 1 (list 1 2 3)); 7
(fold-left + 1 (list 1 2 3)); 7
; exercise 2.39
(define (reverse sequence)
(fold-right (lambda (x y) (append y (list x))) () sequence))
(reverse (list 1 2 3))
(define (reverse sequence)
(fold-left (lambda (x y) (cons y x)) () sequence))
(reverse (list 1 2 3))