Racket streams slower than custom streams? - performance

I implemented a simple and not very efficent sieve of eratosthenes. Once for the built in racket-streams, and once for self defined streams. The only known difference to me is that the built in streams are evaluating the first item on call and not on construction. When evaluating the first 1000 primes on both implementations the self defined streams are running 10-20 times as fast. Is there any explanation?
(define (integers-starting-from-stream n)
(stream-cons n (integers-starting-from-stream (+ n 1))))
(define (stream-limit s limit)
(when (not (= limit 0)) (stream-limit (stream-rest s) (- limit 1))))
(define x (integers-starting-from-stream 2))
(define (divisible? x y)
(= (remainder x y) 0))
(define (sieve s)
(stream-cons
(stream-first s)
(sieve (stream-filter
(lambda (x)
(not (divisible? x (stream-first s))))
(stream-rest s)))))
(time (stream-limit (sieve x) 1000))
Or am i mising something that's affecting the performance?
(define-syntax-rule (s-delay exp)
(λ() exp))
(define (s-force delayedObject)
(delayedObject))
(define empty-s 'S-EMPTY-STREAM)
(define (s-empty? s)
(eq? s empty-s))
(define (s-first s)
(car s))
(define (s-rest s)
(s-force (cdr s)))
(define-syntax-rule (s-cons a b)
(cons a (s-delay b)))
(define (s-filter p s)
(cond ((s-empty? s) empty-s)
((p (s-first s))
(s-cons (s-first s)
(s-filter p (s-rest s))))
(else (s-filter p (s-rest s)))))
;;;;;;;;;;;;;;;;;;;;
(define (divisible? x y)
(= (remainder x y) 0))
(define (integers-starting-from n)
(s-cons n (integers-starting-from (+ n 1))))
(define (s-limit s limit)
(when (not (= limit 0)) (s-limit (s-rest s) (- limit 1))))
(define x (integers-starting-from 2))
(define (sieve s)
(s-cons (s-first s) (sieve (s-filter (lambda(x) (not (divisible? x (s-first s)))) (s-rest s)))))
(time (s-limit (sieve x) 1000))

Here is an observation:
Using a version of integers-starting-from-stream that prints the numbers
as they are generated:
(define (integers-starting-from-stream n)
(stream-cons n
(begin
(display (~a n " "))
(integers-starting-from-stream (+ n 1)))))
And similarly:
(define (integers-starting-from n)
(s-cons n
(begin (display (~a n " "))
(integers-starting-from (+ n 1)))))
We can test with:
(collect-garbage) (collect-garbage) (collect-garbage)
(time (stream-limit (sieve x) 10))
(collect-garbage) (collect-garbage) (collect-garbage)
(time (s-limit (s-sieve s-x) 10))
We observe that the stream-version prints the numbers from 2 to 51
where as the s-version prints the numbers from 2 to 30.
The list generated by the stream version is nearly of double size.
This is the first (and most important) reason the stream version is slower than the custom version.
The second reason the stream version is slower, is that the stream version caches the
result of stream-first. Caching will be in general be faster, when the computation of the element is slow.
(define (integers-starting-from-stream n)
(stream-cons (begin (sleep 1) n)
(integers-starting-from-stream (+ n 1))))
and
(define (integers-starting-from n)
(s-cons (begin (sleep 1) n)
(integers-starting-from (+ n 1))))
And then run:
(collect-garbage) (collect-garbage) (collect-garbage)
(define x (integers-starting-from-stream 2))
(time (stream-limit x 10))
(time (stream-limit x 10))
(collect-garbage) (collect-garbage) (collect-garbage)
(define s-x (integers-starting-from 2))
(time (s-limit s-x 10))
(time (s-limit s-x 10))

Related

Queues in Racket?

I used the following code to solve Sum by Factors:
#lang racket
(provide sum-of-divided)
(define (sum-of-divided lst)
(define (go ps n l)
(define ((exhaust d) x)
(define q (/ x d))
(if (integer? q)
((exhaust d) q)
(if (> x 1) `(,x) '())))
(if (null? l)
ps
(if
(for/or
([p ps])
#:break (< n (sqr p))
(= 0 (modulo n p)))
(go ps (+ n 1) l)
(go
(append ps `(,n))
(+ n 1)
(append-map (exhaust n) l)))))
(for*/list
([m (go '() 2 (map abs lst))]
[s `(,(for/fold
([a '(0 #f)])
([x lst])
(if (= 0 (modulo x m))
`(,(+ (car a) x) #t)
a)))]
#:when (cadr s))
`(,m ,(car s))))
To my surprise, it passed the tests, which have a time limit of 12 s, only after I changed sequence-append in L20 to append. The documentation for sequence-append says:
The new sequence is constructed lazily.
But, as it turns out, it apparently means that the subsequent sequences aren't concatenated unless needed. But when their elements are needed, i.e. the sequence resulting from sequence-append is consumed far enough, the time cost linear in the sum of lengths of all previous sequences is incurred. Right? Is that why it was slow?
If so, how to work around it? (In this case append was performant enough, but suppose I really needed a structure which is at least a FIFO queue with the usual complexities.) Is there a good alternative within the racket language, without requireing additional packages (which may be unavailable, as is the case on Codewars)? Difference lists maybe (quite easy to implement from scratch)?
I ended up using the obvious, hitherto purposely avoided: mutable lists:
#lang racket
(provide sum-of-divided)
(define (sum-of-divided lst)
(define ps (mcons 0 '()))
(define t ps)
(for*/list
([m
(let go ([n 2] [l (map abs lst)])
(if (null? l)
(mcdr ps)
(go
(+ n 1)
(if
(for/or
([p (mcdr ps)])
#:break (< n (sqr p))
(= 0 (modulo n p)))
l
(begin
(set-mcdr! t (mcons n '()))
(set! t (mcdr t))
(remq*
'(1)
(map
(λ (x)
(let exhaust ([s x])
(define q (/ s n))
(if (integer? q)
(exhaust q)
s)))
l)))))))]
[s `(,(for/fold
([a '(0 #f)])
([x lst])
(if (= 0 (modulo x m))
`(,(+ (car a) x) #t)
a)))]
#:when (cadr s))
`(,m ,(car s))))
I also tried a purely functional approach with streams:
#lang racket
(provide sum-of-divided)
(define primes
(letrec
([ps
(stream*
2
(for*/stream
([i (in-naturals 3)]
#:unless
(for/or
([p ps])
#:break (< i (sqr p))
(= 0 (modulo i p))))
i))])
ps))
(define (sum-of-divided lst)
(for/fold
([l lst]
[r '()]
#:result (reverse r))
([d primes])
#:break (null? l)
(values
(remq*
'(1)
(map
(λ (x)
(let exhaust ([s x])
(define q (/ s d))
(if (integer? q)
(exhaust q)
s)))
l))
`(,#(for/fold
([a 0]
[f #f]
#:result
(if f
`((,d ,a))
'()))
([n lst])
(if (= 0 (modulo n d))
(values (+ a n) #t)
(values a f)))
,#r))))
Surprisingly, it consistently times out, whereas the imperative one above never does. Having believed Racket implementors cared at least equally for performance with functional style, I'm disappointed.

Scheme:R5RS:: Make it ten-thousandth place value

(define m (expt 2 32))
(define a 22695477)
(define c 1.0)
(define (integers-starting-from n)
(stream-cons n (integers-starting-from (+ n 1))))
(define (prng seed)
(define xn (remainder (+ c (* a seed)) m))
(define prn (/ (remainder (+ c (* a seed)) m) m))
(stream-cons prn
(prng xn)))
When I run this code my current output is
(stream->list (prng 3) 5)
> (0.015852607786655426 0.4954120593611151 0.998752823099494 0.7253396362066269 0.03071586787700653)
But Output has to be
(stream->list (prng 3) 5)
> (0.01585 0.4954 0.9988 0.7253 0.0307)
How do I make output to ten-thousandth place value?
Here's one way, if you're using Racket:
(define (prng seed)
(define xn (remainder (+ c (* a seed)) m))
(define prn (/ (remainder (+ c (* a seed)) m) m))
(stream-cons (truncate prn 4)
(prng xn)))
(define (truncate num precision)
(string->number (~r num #:precision precision)))
Now the output will be:
'(0.0159 0.4954 0.9988 0.7253 0.0307)

What's wrong with my iterative expmod implementation?

The following program is designed to calculate base^expo mod m.
(define (expmod base expo m)
(define (square n)
(* n n))
(define (even? n)
(= (remainder n 2) 0))
(define (expmod-iter base expo m result)
(cond ((= expo 0) result)
((even? expo)
(expmod-iter base
(/ expo 2)
m
(remainder (square result) m)))
(else
(expmod-iter base
(- expo 1)
m
(remainder (* base result) m)))))
(expmod-iter base expo m 1))
In fact, I'm trying to convert a tail-recursive program from SICP to its iterative equivalent. Here is the original program:
(define (expmod base exp m)
(cond ((= exp 0) 1)
((even? exp)
(remainder (square (expmod base (/ exp 2) m))
m))
(else
(remainder (* base (expmod base (- exp 1) m))
m))))
The result of (expmod 42 1000000007 1000000007) is 270001056, but according to Fermat's Little Theorem, since 1000000007 is prime, the result should be 42.
What am I doing wrong?
This is my implementation of an iterative expmod:
(define (expmod base exp mod)
(let loop ((base base)
(exp exp)
(result 1))
(cond ((zero? exp) result)
((odd? exp) (loop base (sub1 exp) (modulo (* result base) mod)))
(else (loop (modulo (sqr base) mod) (quotient exp 2) result)))))
Tested in Racket with your sample input. You'll need to replace sub1 and sqr with suitable implementations if you're not using Racket.
Note that, while you do have to square the base for an even exponent, you can actually mod the result of that, as you can see in my code. So it doesn't get too massive.

How to find amicable pairs in scheme?

I am new in scheme.
How to find "amicable pais"?
(define (SumCD n)
(define s 1 )
(set! m (quotient n 2))
(while (<= i m)
(if (=(modulo n i) 0)
(set! s (+ s i)))
(set! i (+ i 1))
)
)
And in main program I want to check (if (m=SumCD n) and (n=SumCD m)) then m and n is a amicable pair.
How can I do this?
Excessive use of set! indicates an imperative style of programming, which is usually discouraged in Scheme. Here's a Racket-specific implementation of sum-of-divisors that does not use set! at all.
(define (sum-of-divisors n)
(define-values (q r) (integer-sqrt/remainder n))
(for/fold ((sum (if (and (zero? r) (> q 1)) (add1 q) 1)))
((i (in-range 2 q))
#:when (zero? (modulo n i)))
(+ sum i (quotient n i))))
Equivalent version in standard R6RS/R7RS Scheme, if you're not using Racket:
(define (sum-of-divisors n)
(define-values (q r) (exact-integer-sqrt n))
(let loop ((sum (if (and (zero? r) (> q 1)) (+ q 1) 1))
(i 2))
(cond ((>= i q) sum)
((zero? (modulo n i))
(loop (+ sum i (quotient n i)) (+ i 1)))
(else (loop sum (+ i 1))))))
Note that this is not equivalent to the set!-based version you have. What this code actually does is create an inner function, loop, that gets tail-called with new arguments each time.
Now, we can define amicable? and perfect? accordingly:
(define (amicable? n)
(define sum (sum-of-divisors n))
(and (not (= n sum))
(= n (sum-of-divisors sum))))
(define (perfect? n)
(= n (sum-of-divisors n)))
If you really want to test two numbers to see if they are an amicable pair, you can do this:
(define (amicable-pair? a b)
(and (not (= a b))
(= a (sum-of-divisors b))
(= b (sum-of-divisors a))))
Update for OP's new question about how to use this to find amicable pairs between m and n. First, let's define a variant of amicable? that returns a number's amicable "peer":
(define (amicable-peer n)
(define sum (sum-of-divisors n))
(and (not (= n sum))
(= n (sum-of-divisors sum))
sum))
If you're using Racket, use this:
(define (amicable-pairs-between m n)
(for*/list ((i (in-range m (add1 n)))
(peer (in-value (amicable-peer i)))
#:when (and peer (<= m peer n) (< i peer)))
(cons i peer)))
If you're not using Racket, use this:
(define (amicable-pairs-between m n)
(let loop ((result '())
(i n))
(if (< i m)
result
(let ((peer (amicable-peer i)))
(if (and peer (<= m peer n) (< i peer))
(loop (cons (cons i peer) result) (- i 1))
(loop result (- i 1)))))))
The way this works, is that because lists are built from right-to-left, I've decided to count downward from n through to m, keeping only numbers that have an amicable peer, and where the peer is within range. The (< i peer) check is to ensure that the amicable pair only appears once in the results.
Example:
> (amicable-pairs-between 0 10000)
((220 . 284) (1184 . 1210) (2620 . 2924) (5020 . 5564) (6232 . 6368))
More OP updates (wherein he asked what the difference between a recursive version and an accumulative version is). The version of amicable-pairs-between I wrote above is accumulative. A recursive version would look like this:
(define (amicable-pairs-between m n)
(let recur ((i m))
(if (> i n)
'()
(let ((peer (amicable-peer i)))
(if (and peer (<= m peer n) (< i peer))
(cons (cons i peer) (recur (+ i 1)))
(recur (+ i 1)))))))
Note that there is no result accumulator this time. However, it's not tail-recursive any more.
Your program doesn't work: i is never initialized. And it's very poor style; proper Scheme programs seldom use while or set!. Let's go back to the beginning.
A perfect number is equal to the sum of its proper divisors; for instance, the divisors of 28 are 1, 2, 4, 7, and 14, and 1 + 2 + 4 + 7 + 14 = 28, so 28 is a perfect number. Two numbers m and n form an amicable pair if the sum of the divisors of m equals n and the sum of the divisors of n equals m; for instance, 220 has divisors 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 which sum to 284, and 284 has divisors 1, 2, 4, 71, 142 which sum to 220, so 220 and 284 form an amicable pair.
A simple way to compute the divisors of a number n is try each integer from 1 to ⌊n/2⌋ and see if it divides n:
(define (divisors n)
(let loop ((i 1) (ds (list)))
(cond ((< n (+ i i)) (reverse ds))
((zero? (modulo n i))
(loop (+ i 1) (cons i ds)))
(else (loop (+ i 1) ds)))))
> (divisors 220)
(1 2 4 5 10 11 20 22 44 55 110)
> (divisors 284)
(1 2 4 71 142)
> (divisors 36)
(1 2 3 4 6 9 12 18)
Note that we are excluding n from the list of divisors of n; that's what we want when computing amicable pairs, but in some cases you might want to add n to the list of divisors of n. Instead of making a list of divisors, we can compute their sum:
(define (sum-div n)
(let loop ((i 1) (s 0))
(cond ((< n (+ i i)) s)
((zero? (modulo n i))
(loop (+ i 1) (+ s i)))
(else (loop (+ i 1) s)))))
> (sum-div 220)
284
> (sum-div 284)
220
> (sum-div 36)
55
Instead of counting up to ⌊n/2⌋, it is faster to note that divisors appear in pairs, so it is only necessary to count up to the square root of n; be careful when n is a perfect square to include exactly one instance of the square root in the sum:
(define (divisors n)
(let loop ((i 2) (ds (list 1)))
(cond ((<= n (* i i))
(sort < (if (= n (* i i)) (cons i ds) ds)))
((zero? (modulo n i))
(loop (+ i 1) (cons i (cons (/ n i) ds))))
(else (loop (+ i 1) ds)))))
(define (sum-div n)
(let loop ((i 2) (s 1))
(cond ((<= n (* i i))
(if (= n (* i i)) (+ i s) s))
((zero? (modulo n i))
(loop (+ i 1) (+ s i (/ n i))))
(else (loop (+ i 1) s)))))
> (divisors 220)
(1 2 4 5 10 11 20 22 44 55 110)
> (divisors 284)
(1 2 4 71 142)
> (divisors 36)
(1 2 3 4 6 9 12 18)
> (sum-div 220)
284
> (sum-div 284)
220
> (sum-div 36)
55
If you know the prime factorization of n, it is easy to find the divisors of n: simply take the products of the members of the powerset of the factor of n, eliminating duplicates.
(define (but-last xs)
(if (null? xs) (error 'but-last "empty list")
(reverse (cdr (reverse xs)))))
(define (unique eql? xs)
(cond ((null? xs) '())
((null? (cdr xs)) xs)
((eql? (car xs) (cadr xs)) (unique eql? (cdr xs)))
(else (cons (car xs) (unique eql? (cdr xs))))))
(define (power-set xs)
(if (null? xs) (list (list))
(let ((rest (power-set (cdr xs))))
(append (map (lambda (x) (cons (car xs) x)) rest) rest))))
(define (divisors n)
(but-last (unique = (sort <
(map (lambda (xs) (apply * xs))
(power-set (factors n)))))))
> (divisors 220)
(1 2 4 5 10 11 20 22 44 55 110)
> (divisors 284)
(1 2 4 71 142)
> (divisors 36)
(1 2 3 4 6 9 12 18)
It is even easier to find the sum of the divisors of n if you know the prime factorization of n by examining the multiplicities of the factors of n:
(define (sum-div n)
(define (div f x) (/ (- (expt f (+ x 1)) 1) (- f 1)))
(let ((fs (factors n)))
(let loop ((f (car fs)) (fs (cdr fs)) (x 1) (s 1))
(cond ((null? fs) (- (* s (div f x)) n))
((= (car fs) f) (loop f (cdr fs) (+ x 1) s))
(else (loop (car fs) (cdr fs) 1 (* s (div f x))))))))
> (sum-div 220)
284
> (sum-div 284)
220
> (sum-div 36)
55
A simple method to find the factors of a number n uses a prime wheel; this is slow if n is a large prime or semi-prime but reasonable otherwise:
(define (factors n)
(define (last-pair xs) (if (null? (cdr xs)) xs (last-pair (cdr xs))))
(define (cycle . xs) (set-cdr! (last-pair xs) xs) xs)
(let ((wheel (cons 1 (cons 2 (cons 2 (cycle 4 2 4 2 4 6 2 6))))))
(let loop ((n (abs n)) (f 2) (wheel wheel) (fs (list)))
(cond ((< n (* f f)) (if (= n 1) fs (reverse (cons n fs))))
((zero? (modulo n f)) (loop (/ n f) f wheel (cons f fs)))
(else (loop n (+ f (car wheel)) (cdr wheel) fs))))))
Given all this, it is easy to determine if a number n is perfect, or if it is part of an amicable pair:
(define (perfect? n)
(= n (sum-div n)))
(define (amicable? n)
(let ((s (sum-div n)))
(and (< 1 s) (= (sum-div s) n))))
> (perfect? 6)
#t
> (perfect? 28)
#t
> (amicable? 220)
#t
> (amicable? 284)
#t
It is also easy to find the perfect numbers and amicable pairs less than some limit:
(define (perfect limit)
(let loop ((n 2) (ps (list)))
(cond ((< limit n) (reverse ps))
((= n (sum-div n))
(loop (+ n 1) (cons n ps)))
(else (loop (+ n 1) ps)))))
(define (amicable limit)
(let loop ((n 2) (as (list)))
(if (< limit n) (reverse as)
(let ((s (sum-div n)))
(if (and (< n s) (= n (sum-div s)))
(loop (+ n 1) (cons (list n s) as))
(loop (+ n 1) as))))))
> (perfect 10000)
(6 28 496 8128)
> (amicable 10000)
((220 284) (1184 1210) (2620 2924) (5020 5564) (6232 6368))
Instead of factoring each number up to a limit, it is much faster to find the sums of the divisors of all numbers up to a limit by sieving: Make a vector from 1 to the limit, each item initialized to 1. Then, for each i from 2 to the limit, add i to each multiple of i:
(define (make-sum-divs n)
(let ((s (make-vector (+ n 1) 0)))
(do ((i 1 (+ i 1))) ((< n i) s)
(do ((j (+ i i) (+ j i))) ((< n j))
(vector-set! s j (+ i (vector-ref s j)))))))
(define max-sum-div 1000)
(define sum-divs (make-sum-divs max-sum-div))
Given the sieve, it is easy to find perfect numbers and amicable pairs:
(define (perfect limit)
(when (< max-sum-div limit)
(set! max-sum-div limit)
(set! sum-divs (make-sum-divs max-sum-div)))
(let loop ((n 2) (ps (list)))
(cond ((< limit n) (reverse ps))
((= n (vector-ref sum-divs n))
(loop (+ n 1) (cons n ps)))
(else (loop (+ n 1) ps)))))
(define (pairs limit)
(when (< max-sum-div limit)
(set! max-sum-div limit)
(set! sum-divs (make-sum-divs max-sum-div)))
(let loop ((n 2) (as (list)))
(if (< limit n) (reverse as)
(let ((s (vector-ref sum-divs n)))
(if (and (< s max-sum-div) (< n s)
(= n (vector-ref sum-divs s)))
(loop (+ n 1) (cons (list n s) as))
(loop (+ n 1) as))))))
> (perfect 1000000)
(6 28 496 8128)
> (pairs 1000000)
((220 284) (1184 1210) (2620 2924) (5020 5564) (6232 6368)
(10744 10856) (12285 14595) (17296 18416) (63020 76084)
(66928 66992) (67095 71145) (69615 87633) (79750 88730)
(100485 124155) (122265 139815) (122368 123152)
(141664 153176) (142310 168730) (171856 176336)
(176272 180848) (185368 203432) (196724 202444)
(280540 365084) (308620 389924) (319550 430402)
(356408 399592) (437456 455344) (469028 486178)
(503056 514736) (522405 525915) (600392 669688)
(609928 686072) (624184 691256) (635624 712216)
(643336 652664) (667964 783556) (726104 796696)
(802725 863835) (879712 901424) (898216 980984))
The sieving method is much faster than either of the other two methods. On my computer, it takes twelve seconds to compute the amicable pairs less than a million using trial division to find the divisors, and about the same amount of time for the factoring method, but only about a second-and-a-half to sieve the divisor sums to a million and another half-a-second to find the amicable pairs, a total of two seconds.
In addition to amicable pairs, there exist amicable chains that cycle back to the start after more than two items. For instance, the numbers 12496, 14288, 15472, 14536, and 14264 form an amicable chain of length 5, since sum-div(12496) = 14288, sum-div(14288) = 15472, sum-div(15472) = 14536, sum-div(14536) = 14264, and sum-div(14264) = 12496. The program to find amicable chains is a variant of the program to find amicable pairs:
(define (chain n limit)
(when (< max-sum-div limit)
(set! max-sum-div limit)
(set! sum-divs (make-sum-divs max-sum-div)))
(let loop ((s (vector-ref sum-divs n)) (cs (list n)))
(cond ((= s n) (reverse cs))
((not (< n s limit)) (list))
((member s cs) (list))
(else (loop (vector-ref sum-divs s) (cons s cs))))))
(define (chains limit)
(when (< max-sum-div limit)
(set! max-sum-div limit)
(set! sum-divs (make-sum-divs max-sum-div)))
(let loop ((n 2) (cs (list)))
(if (< limit n) (reverse cs)
(let ((c (chain n limit)))
(if (null? c) (loop (+ n 1) cs)
(loop (+ n 1) (cons c cs)))))))
> (sort (lambda (a b) (< (length a) (length b))) (chains 1000000))
((6) (28) (496) (8128) (220 284) (1184 1210) (2620 2924)
(5020 5564) (6232 6368) (10744 10856) (12285 14595)
(17296 18416) (63020 76084) (66928 66992) (67095 71145)
(69615 87633) (79750 88730) (100485 124155) (122265 139815)
(122368 123152) (141664 153176) (142310 168730)
(171856 176336) (176272 180848) (185368 203432)
(196724 202444) (280540 365084) (308620 389924)
(319550 430402) (356408 399592) (437456 455344)
(469028 486178) (503056 514736) (522405 525915)
(600392 669688) (609928 686072) (624184 691256)
(635624 712216) (643336 652664) (667964 783556)
(726104 796696) (802725 863835) (879712 901424)
(898216 980984) (12496 14288 15472 14536 14264)
(14316 19116 31704 47616 83328 177792 295488 629072 589786
294896 358336 418904 366556 274924 275444 243760 376736
381028 285778 152990 122410 97946 48976 45946 22976 22744
19916 17716))
The four perfect numbers form amicable chains of length 1, there are 40 amicable pairs, there is an amicable chain of length 5 mentioned above, and notice the spectacular amicable chain of length 28 that starts at 14316.
I just try to find amicable pairs between M and N
(define (find-amicable-pairs M N)
(< M N)
(define i M)
(define a 0)
(do ()
[(= i N)]
(set! a (sum-of-divisors i))
(if (and(= i (sum-of-divisors a)) (< i a))
(and (display i)
(display " and ")
(display a)
(newline))
#f)
(set! i (+ i 1))))
Thanks for your thoughts on this!

Translation of Scheme code for Sierpinski carpet

I found code for generating Sierpinski carpet at http://rosettacode.org/wiki/Sierpinski_carpet#Scheme - but it won't run in the DrRacket environment or WeScheme. Could someone provide solutions for either environments?
It looks like this code runs fine in DrRacket after prepending a
#lang racket
line indicating that the code is written in Racket. I can provide more detail if this is not sufficient.
I've translated the program to run under WeScheme. I've made a few changes: rather than use (display) and (newline), I use the image primitives that WeScheme provides to make a slightly nicer picture. You can view the running program and its source code. For convenience, I also include the source here:
;; Sierpenski carpet.
;; http://rosettacode.org/wiki/Sierpinski_carpet#Scheme
(define SQUARE (square 10 "solid" "red"))
(define SPACE (square 10 "solid" "white"))
(define (carpet n)
(local [(define (in-carpet? x y)
(cond ((or (zero? x) (zero? y))
#t)
((and (= 1 (remainder x 3)) (= 1 (remainder y 3)))
#f)
(else
(in-carpet? (quotient x 3) (quotient y 3)))))]
(letrec ([outer (lambda (i)
(cond
[(< i (expt 3 n))
(local ([define a-row
(letrec ([inner
(lambda (j)
(cond [(< j (expt 3 n))
(cons (if (in-carpet? i j)
SQUARE
SPACE)
(inner (add1 j)))]
[else
empty]))])
(inner 0))])
(cons (apply beside a-row)
(outer (add1 i))))]
[else
empty]))])
(apply above (outer 0)))))
(carpet 3)
Here is the modified code for WeScheme. WeScheme don't support do-loop syntax, so I use unfold from srfi-1 instead
(define (unfold p f g seed)
(if (p seed) '()
(cons (f seed)
(unfold p f g (g seed)))))
(define (1- n) (- n 1))
(define (carpet n)
(letrec ((in-carpet?
(lambda (x y)
(cond ((or (zero? x) (zero? y))
#t)
((and (= 1 (remainder x 3)) (= 1 (remainder y 3)))
#f)
(else
(in-carpet? (quotient x 3) (quotient y 3)))))))
(let ((result
(unfold negative?
(lambda (i)
(unfold negative?
(lambda (j) (in-carpet? i j))
1-
(1- (expt 3 n))))
1-
(1- (expt 3 n)))))
(for-each (lambda (line)
(begin
(for-each (lambda (char) (display (if char #\# #\space))) line)
(newline)))
result))))

Resources