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Quote: Original post by SamLowry
Why the infinite computation?
(define (square x) (* x x))(define (sum-of-squares x y) (+ (square x) (square y)))(define (larger-sum-of-sqrs x y z) (cond ((and (> x z) (> y z)) (sum-of-squares x y)) ((and (> x y) (> z y)) (sum-of-squares x z)) ((and (> y x) (> z x)) (sum-of-squares y z))))(larger-sum-of-sqrs (insert test numbers x y z))(define (cont-frac n d k) (if (= k 0) 0 (/ (n k) (+ (d k) (cont-frac n d (- k 1))))))(define (cont-frac-iter n d k) (define (iter k result) (if (= 0 k) result (iter (- k 1) (/ (n k) (+ result (d k)))))) (iter k 0))(define (calculate-e k) (define (d i) (let ((shifted-index (- i 2))) (cond ((= i 2) 2.0) ((= 0 (remainder shifted-index 3)) (+ 2.0 (* 2.0 (/ shifted-index 3.0)))) (else 1.0)))) (+ 2.0 (cont-frac (lambda (i) 1.0) d k)))> (calculate-e 10000)2.5000374981248825(define (cont-frac n d k) (define (inner-cont-frac i) (if (= i k) (/ (n i) (+ (d i))) (/ (n i) (+ (d i) (inner-cont-frac (+ i 1)))))) (inner-cont-frac 1))(define (double f) (lambda (x) (f (f x))))(((double (double double)) inc) 5)((double (double (double inc))) 5)Quote: Original post by kevtimc
Just finished exercise 1.3 (I'm going slowly I know):
*** Source Snippet Removed ***
The assignment didn't say anything about arguments of the same number. Did any of you guys include a test for equal numbers?
(define (square x) (* x x))(define (sum-of-squares x y) (+ (square x) (square y)))(define (sum-of-squares-of-largest-two a b c) (cond ((and (> a c) (> b c)) (sum-of-squares a b)) ((and (> a b) (> c b)) (sum-of-squares a c)) (else (sum-of-squares b c))))Quote: Original post by kevtimcYes.
On the topic of exercise 1.5,
Is (define (p) (p)),
equivilant in Python to:
def badrecursion():
badrecursion()
?
Quote: Original post by Muhammad Haggag
So, my question rephrased would be: How do you go about deciphering such combinations, or visualize them in your head or on paper?
Quote: Original post by Rebooted
Muhammad Haggag: I'll try to come up with a full answer for you, but maybe carefully following the single-step substitution rules and giving names to intermediate values would help?
(((double (double double)) inc) 5)(((double ((lambda (x) (f (f x))) double)) inc) 5)(((double (lambda (x) (double (double x)))) inc) 5) <-- I was getting stuck after this, when expanding the outer doubleLet g = ((lambda (x) (double (double x))))(((double g) inc) 5)(((lambda (x) (g (g x))) inc) 5)((g (g inc)) 5)((g (double (double inc))) 5)((g (double inc2)) 5)((g inc4) 5)((double (double inc4)) 5)((double inc8) 5)(inc16 5)21Quote: In order to better understand such constructions, I often invent my own notations and work with these; such syntactic abstractions often help me better understand complicated things.
I used the mathematical notation fn(x) to designate multiple application, e.g. f3(x) = f(f(f(3))). Thus, (double f) returns f2.
...
I don't know if this helps you, but I was able to decipher it quite easily this way.
Quote: Also, I consider this example a fairly complex one; most other uses of higher-order functions are far more simple than this IMHO, so understanding other uses won't be as hard.
