This is my first post on StackOverflow. I have been working on Exercise 1.11 from SICP and feel I have a viable solution. In transferring from paper to Emacs I seem to have some syntax error that I am unaware of. I tried my best to double and triple check the parenthesis and solve it but the terminal is still giving me an 'object #t is not applicable' message. Could someone please point me in the right direction of how to fix the code so I can test its output properly?"
Exercise 1.11: A function f
is defined by the rule that:
f(n)=n
if n<3
, and
f(n)=f(n−1)+2f(n−2)+3f(n−3)
if n>=3
Write a procedure that computes f by means of a recursive process.
Write a procedure that computes f by means of an iterative process.
(define (f-recur n)
(if ((< n 3) n)
(+ (f(- n 1))
(* 2 (f(n-2)))
(* 3 (f(n-3)))))
(define (f-iter n)
(define (counter n)
(if (<= n 3) 0)
(- n 3))
(define (d n) (+ n (* 2 n) (* 3 n)))
(define (c n) (+ d (* 2 n) (* 3 n)))
(define (b n) (+ c (* 2 d) (* 3 n)))
(define (a n) (+ b (* 2 c) (* 3 d)))
(define (f a b c d counter)
(if ((> (+ counter 3) n) a)
(f (+ b (* 2 c) (* 3 d)) a b c (+ counter 1)))))
(cond ((= counter 0) d)
((= counter 1) c)
((= counter 2) b)
((= counter 3) a)
(else (f a b c d counter))))
I'm pretty sure SICP is looking for a solution in the manner of this iterative fibonnacci:
(define (fib n)
(define (helper n a b)
(if (zero? n)
a
(helper (- n 1) b (+ a b))))
(helper n 0 1))
Fibonacci is f(n)=f(n−1)+f(n−2) so I guess f(n)=f(n−1)+2f(n−2)+3f(n−3) can be made exactly the same way with one extra variable. Your iterative solution looks more like Fortran than Scheme. Try avoiding set!.
Related
(define (pow b n)
"YOUR-DOC-HERE"
(cond ((= n 0) 1)
((even? n) (pow (pow b (/ n 2)) 2))
((odd? n) (* b (pow (pow b (/ (- n 1) 2)) 2)))))
(define (pow b n)
"YOUR-DOC-HERE"
(cond ((= n 0) 1)
((even? n) (* (pow b (/ n 2)) (pow b (/ n 2))))
((odd? n) (* b (pow b (/ (- n 1) 2)) (pow b (/ (- n 1) 2))))))
Here are two versions of my code for a power function with logarithmic efficiency. However, the first function would have a maximum recursion depth exceeded error and the second, though works, doesn't seem to function at a required efficiency. I am new to Scheme and I wonder what's wrong with these implementations?
Your 1st version uses itself to square every value, which creates an infinite loop in the even? clause.
Your 2nd version calls pow twice in each clause which reverses any gain from the logarithmic algorithm.
Your can fix it using let like this:
(define (pow b n)
"Recursive power in logarithmic depth."
(let ((square (lambda (x) (* x x))))
(cond ((= n 0) 1)
((even? n) (square (pow b (/ n 2))))
((odd? n) (* b (square (pow b (/ (- n 1) 2))))))))
or like this:
(define (pow b n)
"Recursive power in logarithmic depth."
(cond ((= n 0) 1)
((even? n)
(let ((x (pow b (/ n 2))))
(* x x)))
((odd? n)
(let ((x (square (pow b (/ (- n 1) 2)))))
(* b x x)))))
I am working through SICP. In exercise 1.28 about the Miller-Rabin test. I had this code, that I know is wrong because it does not follow the instrcuccions of the exercise.
(define (fast-prime? n times)
(define (even? x)
(= (remainder x 2) 0))
(define (miller-rabin-test n)
(try-it (+ 1 (random (- n 1)))))
(define (try-it a)
(= (expmod a (- n 1) n) 1))
(define (expmod base exp m)
(cond ((= exp 0) 1)
((even? exp)
(if (and (not (= exp (- m 1))) (= (remainder (square exp) m) 1))
0
(remainder (square (expmod base (/ exp 2) m)) m)))
(else
(remainder (* base (expmod base (- exp 1) m)) m))))
(cond ((= times 0) true)
((miller-rabin-test n) (fast-prime? n (- times 1)))
(else false)))
In it I test if the square of the exponent is congruent to 1 mod n. Which according
to what I have read, and other correct implementations I have seen is wrong. I should test
the entire number as in:
...
(square
(trivial-test (expmod base (/ exp 2) m) m))
...
The thing is that I have tested this, with many prime numbers and large Carmicheal numbers,
and it seems to give the correct answer, though a bit slower. I don't understand why this
seems to work.
Your version of the function "works" only because you are lucky. Try this experiment: evaluate (fast-prime? 561 3) a hundred times. Depending on the random witnesses that your function chooses, sometimes it will return true and sometimes it will return false. When I did that I got 12 true and 88 false, but you may get different results, depending on your random number generator.
> (let loop ((k 0) (t 0) (f 0))
(if (= k 100) (values t f)
(if (fast-prime? 561 3)
(loop (+ k 1) (+ t 1) f)
(loop (+ k 1) t (+ f 1)))))
12
88
I don't have SICP in front of me -- my copy is at home -- but I can tell you the right way to perform a Miller-Rabin primality test.
Your expmod function is incorrect; there is no reason to square the exponent. Here is a proper function to perform modular exponentiation:
(define (expm b e m) ; modular exponentiation
(let loop ((b b) (e e) (x 1))
(if (zero? e) x
(loop (modulo (* b b) m) (quotient e 2)
(if (odd? e) (modulo (* b x) m) x)))))
Then Gary Miller's strong pseudoprime test, which is a strong version of your try-it test for which there is a witness a that proves the compositeness of every composite n, looks like this:
(define (strong-pseudoprime? n a) ; strong pseudoprime base a
(let loop ((r 0) (s (- n 1)))
(if (even? s) (loop (+ r 1) (/ s 2))
(if (= (expm a s n) 1) #t
(let loop ((r r) (s s))
(cond ((zero? r) #f)
((= (expm a s n) (- n 1)) #t)
(else (loop (- r 1) (* s 2)))))))))
Assuming the Extended Riemann Hypothesis, testing every a from 2 to n-1 will prove (an actual, deterministic proof, not just a probabilistic estimate of primality) the primality of a prime n, or identify at least one a that is a witness to the compositeness of a composite n. Michael Rabin proved that if n is composite, at least three-quarters of the a from 2 to n-1 are witnesses to that compositeness, so testing k random bases demonstrates, but does not prove, the primality of a prime n to a probability of 4−k. Thus, this implementation of the Miller-Rabin primality test:
(define (prime? n k)
(let loop ((k k))
(cond ((zero? k) #t)
((not (strong-pseudoprime? n (random (+ 2 (- n 3))))) #f)
(else (loop (- k 1))))))
That always works properly:
> (let loop ((k 0) (t 0) (f 0))
(if (= k 100) (values t f)
(if (prime? 561 3)
(loop (+ k 1) (+ t 1) f)
(loop (+ k 1) t (+ f 1)))))
0
100
I know your purpose is to study SICP rather than to program primality tests, but if you're interested in programming with prime numbers, I modestly recommend this essay at my blog, which discusses the Miller-Rabin test, among other topics. You should also know there are better (faster, less likely to report erroneous result) primality tests available than randomized Miller-Rabin.
It seems to me, you still got correct answer, because in each iteration of expmod you check conditions for previous iteration. You could try to debug exp value using display function inside expmod. Really, your code is not very different from this one.
Im want to make a function where rootcheck has a list L as input, L always is 3 atoms (a b c) where a is coefficient of x^2, b coef of x and c is the constant. it checks if the equation is quadratic, using discriminant (b^2 - 4ac) and should output this (num 'L) where num is the number of roots and L is a list that contains the roots themselves (using quadratic formula), L is empty in case of no roots. here is my code:
(define roots-2
(lambda (L)
(let ((d (- (* (cdr L) (cdr L)) (4 (car L) (caddr L))))))
(cond ((< d 0) (cons(0 null)))
((= d 0) (cons(1 null)))
(else((> d 0) (cons(2 null)))))
))
its giving me no expression in body error.
also I tried to code the quadratic function and even tried some that are online, one compiled fint but gave me an error when I inserted input this is the code for the quadratic function, NOT MINE!
(define quadratic-solutions
(lambda (a b c) (list (root1 a b c) (root2 a b c))))
(define root1
(lambda (a b c) (/ (+ (- b) (sqrt (discriminant a b c)))
(* 2 a))))
(define root2
(lambda (a b c) (/ (- (- b) (sqrt (discriminant a b c)))
(*2 a))))
(define discriminant
(lambda (a b c) (- (square b) (* 4 (* a c)))))
There are several mistakes in the code:
Some parentheses are incorrectly placed, use a good IDE to detect such problems. This is causing the error reported, the let doesn't have a body
You forgot to multiply in the 4ac part
You're incorrectly accessing the second element in the list
The else part must not have a condition
The output list is not correctly constructed
This should fix the errors, now replace null with the actual call to the function that calculates the roots for the second and third cases (the (< d 0) case is fine as it is):
(define roots-2
(lambda (L)
(let ((d (- (* (cadr L) (cadr L)) (* 4 (car L) (caddr L)))))
(cond ((< d 0) (list 0 null))
((= d 0) (list 1 null))
(else (list 2 null))))))
for the quadractic function part, I found a code online and tweaked it to provide both roots of a quadratic equation. returns a list of both roots
(define (solve-quadratic-equation a b c)
(define disc (sqrt (- (* b b)
(* 4.0 a c))))
(list (/ (+ (- b) disc) (* 2.0 a))
(/ (- (- b) disc) (* 2.0 a))
))
I am trying to determine the number of marbles that fall within a given circle (radius 1) given that they have random x and y coordinates.
My overall goal is to find an approximate value for pi by using monte carlo sampling by multiplying by 4 the (number of marbles within the circle)/(total number of marbles).
I intended for my function to count the number of marbles within the circle, but I am having trouble following why it does not work. Any help on following the function here would be appreciated.
Please comment if my above request for help is unclear.
(define(monte-carlo-sampling n)
(let ((x (- (* 2 (random)) 1))
(y (- (* 2 (random)) 1)))
(cond((= 0 n)
* 4 (/ monte-carlo-sampling(+ n 1) n)
((> 1 n)
(cond((< 1 (sqrt(+ (square x) (square y))) (+ 1 (monte-carlo-sampling(- n 1)))))
((> 1 (sqrt(+ (square x) (square y))) (monte-carlo-sampling(- n 1))))
)))))
Your parentheses are all messed up, and your argument order for < is wrong. Here's how the code should look like after it's corrected:
(define (monte-carlo-sampling n)
(let ((x (- (* 2 (random)) 1))
(y (- (* 2 (random)) 1)))
(cond ((= n 0)
0)
(else
(cond ((< (sqrt (+ (square x) (square y))) 1)
(+ 1 (monte-carlo-sampling (- n 1))))
(else
(monte-carlo-sampling (- n 1))))))))
This returns the number of hits. You'd have to convert the number of hits into a pi estimate using an outer function, such as:
(define (estimate-pi n)
(* 4 (/ (monte-carlo-sampling n) n)))
Here's how I'd write the whole thing, if it were up to me:
(define (estimate-pi n)
(let loop ((i 0)
(hits 0))
(cond ((>= i n)
(* 4 (/ hits n)))
((<= (hypot (sub1 (* 2 (random)))
(sub1 (* 2 (random)))) 1)
(loop (add1 i) (add1 hits)))
(else
(loop (add1 i) hits)))))
(Tested on Racket, using the definition of hypot I gave in my last answer. If you're not using Racket, you have to change add1 and sub1 to something appropriate.)
I wrote a solution to this problem at my blog; the inner function is called sand because I was throwing grains of sand instead of marbles:
(define (pi n)
(define (sand?) (< (+ (square (rand)) (square (rand))) 1))
(do ((i 0 (+ i 1)) (p 0 (+ p (if (sand?) 1 0))))
((= i n) (exact->inexact (* 4 p (/ n))))))
This converges very slowly; after a hundred thousand iterations I had 3.14188. The blog entry also discusses a method for estimating pi developed by Archimedes over two hundred years before Christ that converges very quickly, with 27 iterations taking us to the bound of double-precision arithmetic.
Here's a general method of doing monte-carlo it accepts as arguments the number of iterations, and a thunk (procedure with no arguments) that should return #t or #f which is the experiment to be run each iteration
(define (monte-carlo trials experiment)
(define (iter trials-remaining trials-passed)
(cond ((= trials-remaining 0)
(/ trials-passed trials))
((experiment)
(iter (- trials-remaining 1) (+ trials-passed 1)))
(else
(iter (- trials-remaining 1) trials-passed))))
(iter trials 0))
Now it's just a mater of writing the specific experiment
You could write in your experiment where experiment is invoked in monte-carlo, but abstracting here gives you a much more flexible and comprehensible function. If you make a function do too many things at once it becomes hard to reason about and debug.
(define (marble-experiment)
(let ((x ...) ;;assuming you can come up with
(y ...)) ;;a way to get a random x between 0 and 1
;;with sufficient granularity for your estimate)
(< (sqrt (+ (* x x) (* y y))) 1)))
(define pi-estimate
(* 4 (monte-carlo 1000 marble-experiment)))
I'm going through "Structure and Interpretation of Computer Programs" and I'm having a bit of trouble doing one of the exercises ( 2.1 ) . I'm coding in DrRacket in R5RS mode.
here's my code :
(define (make-rat n d)
(let (((c (gcd n d))
(neg (< (* n d) 0))
(n (/ (abs n) c))
(d (/ (abs d) c)))
(cons (if neg (- n) n) d))))
and here's the error message DrRacket is giving me:
let: bad syntax (not an identifier and expression for a binding) in: ((c (gcd n d)) (neg (< (* n d) 0)) (pn (/ (abs n) c)) (pd (/ (abs d) c)))
I think I've messed up let's syntax. but I'm not sure how to fix it.
I added an extra set of parentheses around the variable declarations, whoops.
Also, since I used c to define n and d, I had to change let into let* to make it work properly
my fixed code:
(define (make-rat n d)
(let* ((c (gcd n d))
(neg (< (* n d) 0))
(n (/ (abs n) c))
(d (/ (abs d) c)))
(cons (if neg (- n) n) d)))
As your edit indicates, you're using the c identifier prematurely. (Which is why it isn't working after fixing the syntax issue of the extra parenthesis.) Identifiers in "let" don't see each other. You'd need to nest your second three lets under the first.
(let ((c (gcd ...)))
(let ((...))
exps ...))
I don't recall when/if SICP introduces other let forms, but if you are stuck using a lot of nested lets, you can use let* in which each subsequent identifier is in the scope of all the previous. That is, the following two definitions are equivalent:
(define foo
(let* ((a 1)
(b (+ 1 a))
(c (+ 1 b)))
(+ 1 c)))
(define foo
(let ((a 1))
(let ((b (+ 1 a)))
(let ((c (+ 1 b)))
(+ 1 c)))))
The scoping rules of the different let forms can be a bit much for a beginner, unfortunately.
Try this:
(define (make-rat n d)
(let ([c (gcd n d)]
[neg (< (* n d) 0)]
[n (/ (abs n) c)]
[d (/ (abs d) c)])
(cons (if neg
(- n)
n)
d)))