I am making a Common Lisp function to print the first N prime numbers. So far I've managed to write this code:
;globals
(setf isprime 1) ;if 1 then its a prime, 0 if not.
(setf from 1) ;start from 1
(setf count 0) ;should act as counter to check if we have already
; N primes printed
;function so far.
(defun prime-numbers (to)
(if (> count to) nil(progn
(is-prime from from)
(if (= isprime 1) (print from)(setf count (+ count 1)))
(setf isprime 1)
(setf from (+ from 1))
(prime-numbers to)))
(if (>= count to)(setf count 0) (setf from 1)))
;code to check if a number is prime
(defun is-prime(num val)
(if (< num 3) nil
(progn
(if (= (mod val (- num 1)) 0) (setf isprime 0))
(is-prime (- num 1) val))))
My problem is, it does not print N primes correctly.
If I call >(prime-numbers 10),
results are:
1
2
3
5
7
11
13
17
19
1,
i.e. it printed only 9 primes correctly.
but then if i call >(prime-numbers 2)
the results are: 1
2
3
5
7
1
what am I doing wrong here?? this is my first time to code in LISP.
UPDATE:
(defparameter from 1)
(defparameter count 0)
(defun prime-numbers (to)
(if (> count to)nil
(progn
(when (is-prime from)
(print from)
(setf count (+ count 1)))
(setf from (+ from 1))
(prime-numbers to)))
(when (>= count to)
(setf count 0)
(setf from 1)))
(defun is-prime (n)
(cond ((= 2 n) t)
((= 3 n) t)
((evenp n) nil)
(t
(loop for i from 3 to (isqrt n) by 2
never (zerop (mod n i))))))
works fine. but outputs a NIL at the end.
First, there's no need to use globals here, at all.
Use true/false return values. That would allow your is-prime function to be something like:
(defun is-prime (n)
(cond ((= 2 n) t) ;; Hard-code "2 is a prime"
((= 3 n) t) ;; Hard-code "3 is a prime"
((evenp n) nil) ;; If we're looking at an even now, it's not a prime
(t ;; If it is divisible by an odd number below its square root, it's not prime
(loop for i from 3 to (isqrt n) by 2
never (zerop (mod n i))))))
That way, the function is not relying on any external state and there's nothing that can confuse anything.
Second, the last 1 you see is (probably) the return value from the function.
To check that, try:
(progn (prime-numbers 10) nil)
Third, re-write your prime-numbers function to not use global variables.
Fourth, never create global variables with setf or setq, use either defvar or defparameter. It's also (mostly, but some disagree) good style to use *earmuffs* on your global (really, "special") variables.
To expand on Vatines answer:
A possible rewrite of the prime-numbers function, using the same algoritm but avoiding globals is
(defun prime-numbers (num &optional (from 2))
(cond ((<= num 0) nil)
((is-prime from) (cons from (prime-numbers (1- num) (1+ from))))
(t (prime-numbers num (1+ from)))))
This function also returns the primes instead of printing them.
The problem with this recursive solution is it consumes stack for each prime found/tested. Thus stack space may be exhausted for large values of num.
A non-recursive variant is
(defun prime-numbers (num &optional (start 2))
(loop for n upfrom start
when (is-prime n)
sum 1 into count
and collect n
until (>= count num)))
Related
Exercise 3.52,
(define sum 0)
(define (accum x)
(set! sum (+ x sum))
sum)
;1: (define seq (stream-map accum (stream-enumerate-interval 1 20)))
;2: (define y (stream-filter even? seq))
;3: (define z (stream-filter (lambda (x) (= (remainder x 5) 0))
; seq))
;4: (stream-ref y 7)
;5: (display-stream z)
Step 1:
;1: ==> (cons-stream 1 (stream-map proc (stream-cdr s)) (Assume stream-cdr is evaluated only when we force the cdr of this stream)
sum is now 1
Step 2:
1 is not even, hence (also memoized so not added again), it calls (stream-filter pred (stream-cdr stream)).
This leads to
evaluation of cdr hence materializing 2 which is even, hence it should call: (cons-stream 2 (stream-cdr stream)).
According to this answer should be 1+2 = 3 , but it is 6
Can someone help with why the cdr's car is materialized before the current cdr is called?
Using Daniel P. Friedman's memoizing tail
#lang r5rs
(define-syntax cons-stream
(syntax-rules ()
((_ h t) (cons h (lambda () t)))))
(define (stream-cdr s)
(if (and (not (pair? (cdr s)))
(not (null? (cdr s))))
(set-cdr! s ((cdr s))))
(cdr s))
we observe:
> sum
0
> (define seq (stream-map accum (stream-enumerate-interval 1 20)))
> sum
1
> seq
(mcons 1 #<procedure:friedmans-tail.rkt:21:26>)
> (define y (stream-filter even? seq))
> sum
6
> seq
(mcons
1
(mcons
3
(mcons 6 #<procedure:friedmans-tail.rkt:21:26>)))
> y
(mcons 6 #<procedure:friedmans-tail.rkt:21:26>)
>
stream-filter? needs to get to the first element of the stream it is constructing in order to construct it. A stream has its head element already forced, calculated, so it must be already present.
In the list of accumulated sums of the enumerated interval from 1 to 20, the first even number is 6:
1 = 1
1+2 = 3
1+2+3 = 6
...
I wrote a procedure that gets a valid prefix list for subtraction (e.g, "(- 6 5)" for what we know as "6-5"). Here is my code:
(define parse-diff-list
(lambda (datum)
(cond
((number? datum) (const-exp datum)) ;; if datum is a number, return const-exp
((pair? datum) ;; if datum is a pair:
(let ((sym (car datum))) ;; let sym be the first of the pair
(cond
((eqv? sym '-) ;; if sym is minus:
(let ((lst1 (parse-diff-list (cdr datum)))) ;; parse second element of subtraction
(let ((lst2 (parse-diff-list (cdr lst1)))) ;; parse first element of subtraction
(cons (diff-exp (car lst1) (car lst2)) (cdr lst2))))) ;; "perform" the subtraction
((number? sym) ;; if sym is number:
(cons (const-exp sym) (cdr datum))) ;; return const-exp with the remainder of the list, yet to be processed
(else (eopl:error 'parse-diff-list "bad prefix-expression, expected - ~s" sym)))))
(eopl:error 'parse-diff-list "bad prefix-expression ~s" datum))))
(define parse-prefix
(lambda (lst)
(car (parse-diff-list lst))))
It works fine logically, but I don't understand the logic of the indentation in printing. For the input:
(parse-prefix '(- - 1 2 - 3 - 4 5))
It prints:
#(struct:diff-exp
#(struct:diff-exp #(struct:const-exp 1) #(struct:const-exp 2))
#(struct:diff-exp #(struct:const-exp 3) #(struct:diff-exp #(struct:const-exp 4) #(struct:const-exp 5)))
While I would want the following print style:
#(struct:diff-exp
#(struct:diff-exp
#(struct:const-exp 1)
#(struct:const-exp 2))
#(struct:diff-exp
#(struct:const-exp 3)
#(struct:diff-exp
#(struct:const-exp 4)
#(struct:const-exp 5)))
It's more than a petty question for me, as it does create indentations but I don't know how it does it.
Thanks a lot!
Take a look at racket/pretty the pretty printing library.
In particular note the parameter (pretty-print-columns) which
you can set like this:
`(pretty-print-columns 40)`
in order to avoid long lines.
http://docs.racket-lang.org/reference/pretty-print.html
(I am guessing you are using DrRacket based on the way the structures are printing)
I have a SCHEME function is-sexy? which takes one parameter, n, and returns true if n is part of a pair of sexy primes and false otherwise, and a SCHEME function, sexy-primes, which takes an integer, n, as a parameter and returns a list of pairs of prime numbers whose difference is 6 and whose smaller number is less than or equal to n.
How do I define a stream of sexy prime pairs?
(define (is-sexy? n)
(define (is-prime? x)
(define (is-prime?-aux x k)
(cond ((< x 1) #f)
((= x k) #t)
(else
(if (= (remainder x k) 0) #f
(is-prime?-aux x (+ k 1))))))
(cond ((= x 1) #t)
((= x 2) #t)
(else (is-prime?-aux x 2))))
(if (and (is-prime? n)
(or (is-prime? (- n 6)) (is-prime? (+ n 6)))) #t
#f))
(define (sexy-primes n)
(if (= n 0) '()
(if (is-sexy? n) (cons n (sexy-primes (- n 1)))
(sexy-primes (- n 1)))))
This works:
(define (sexyprimes-from k)
(if (is-sexy? k) (cons (cons k (+ k 6)) (delay (sexyprimes-from (+ k 1))))
(sexyprimes-from (+ k 1))))
(define sexy-primes (sexyprimes-from 5))
I can print n-numbers as list with this code below:
(define (print-first-n stream1 n)
(cond((= n 0) '())
(else(cons(stream-car stream1) (print-first-n (stream-cdr stream1) (- n 1))))))
But I have no idea about how to add commas.
You can't print a comma in a normal list, but we can build a string with the contents of the stream, separated by commas. This will work, assuming that the string contains numbers:
(define (print-first-n stream1 n)
(cond ((= n 1)
(number->string (stream-car stream1)))
(else
(string-append
(number->string (stream-car stream1)) ", "
(print-first-n (stream-cdr stream1) (- n 1))))))
The above solution is fine for a small value of n, but terribly inefficient for large values (lots of temporary strings will be created, with O(n^2) complexity for the append operation). For a more efficient implementation, consider using SRFI-13's concatenation procedures, like this:
(require srfi/13)
(define (print-first-n stream1 n)
(let loop ((strm stream1) (n n) (acc '()))
(if (= n 1)
(string-concatenate-reverse
(cons (number->string (stream-car strm)) acc))
(loop (stream-cdr strm)
(sub1 n)
(list* ", " (number->string (stream-car strm)) acc)))))
Either way: let's say that integers is an infinite stream of integers starting at 1, this is how it would look:
(print-first-n integers 5)
=> "1, 2, 3, 4, 5"
If the stream contains some other data type, use the appropriate procedure to convert each element to a string.
If your function just prints the stream contents, and doesn't need to build a string (like Óscar's answer), here's my take on it (uses SRFI 41 streams):
(define (print-first-n stream n)
(stream-for-each (lambda (delim item)
(display delim)
(display item))
(stream-cons "" (stream-constant ", "))
(stream-take n stream)))
Example:
> (define natural (stream-cons 1 (stream-map (lambda (x) (+ x 1)) natural)))
> (print-first-n natural 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
To output to a string (like Óscar's answer), just wrap the whole thing in a string port:
(define (print-first-n stream n)
(call-with-output-string
(lambda (out)
(stream-for-each (lambda (delim item)
(display delim out)
(display item out))
(stream-cons "" (stream-constant ", "))
(stream-take n stream)))))
I have a list of ((a)(b)(f(x))). What i would like to get is a linked list structure of((a)(b)(f(x1))(a)(b)(f(x2))(a)(b)(f(x3)))). That is, repeatively appending the list the on the basis of requirement of the user and the value of the variable is chaning so that its value will be unique from each other. How can i implement it in LISP?
? (let ((list '((a) (b) (f (x))))
(n 3))
(flet ((copier (l n)
(setf l (copy-tree l))
(let ((sym (first (second (third l)))))
(setf (first (second (third l)))
(intern (format nil "~a~a"
(symbol-name sym)
n))))
l))
(loop for i from 1 upto n
nconc (copier list i))))
((A) (B) (F (X1)) (A) (B) (F (X2)) (A) (B) (F (X3)))