I have written the following program to calculate the sum of all multiples of 3 & 5 below 1000 in scheme. However, it gives me an incorrect output.
Any help would be much appreciated.
(define (multiples)
(define (calc a sum ctr cir)
(cond (> a 1000) (sum)
(= ctr 7) (calc (+ a (list-ref cir 0)) (+ sum a) 0 (list 3 2 1 3 1 2 3))
(else (calc (+ a (list-ref cir ctr)) (+ sum a) (+ 1 ctr) (list 3 2 1 3 1 2 3)))))
(calc 0 0 0 (list 3 2 1 3 1 2 3)))
You can simply port imperative style solution to functional Scheme by using an accumulator(sum parameter) and a target parameter to test when to stop summing:
(define (multiples)
(define (multiples-iter num sum target)
(if (> num target)
sum
(multiples-iter (+ 1 num)
(if (or (zero? (mod num 3)) (zero? (mod num 5)))
(+ sum num)
sum)
target)))
(multiples-iter 0 0 1000))
Here's my (Racket-specific) solution, which doesn't involve lots of (or, for that matter, any) modulo calls, and is completely general (so that you don't need to construct the (3 2 1 3 1 2 3) list that the OP has):
(define (sum-of-multiples a b limit)
(define (sum-of-multiple x)
(for/fold ((sum 0))
((i (in-range 0 limit x)))
(+ sum i)))
(- (+ (sum-of-multiple a) (sum-of-multiple b))
(sum-of-multiple (lcm a b))))
Test run:
> (sum-of-multiples 3 5 1000)
233168
If you're using Racket, there's a very compact way to do what you ask, using looping constructs:
(for/fold ([sum 0])
([i (in-range 1 1000)]
#:when (or (zero? (modulo i 3)) (zero? (modulo i 5))))
(+ sum i))
=> 233168
One problem is that your code is missing a pair of parentheses around the cond clauses.
In the line (cond (> a 1000) (sum) the condition is just> while a and 1000 are interpreted as forms to be evaluated if > is true (which it is), and thus 1000 will be returned as the result.
Two other problem (masked by the first one) is that you are initializing ctr to 0 when it reaches 7, while it should be set to the next value, i.e. 1, and that you are including 1000 in the result.
The corrected version of your function is
(define (multiples)
(define (calc a sum ctr cir)
(cond ((>= a 1000) sum)
((= ctr 7) (calc (+ a (list-ref cir 0)) (+ sum a) 1 (list 3 2 1 3 1 2 3)))
(else (calc (+ a (list-ref cir ctr)) (+ sum a) (+ 1 ctr) (list 3 2 1 3 1 2 3)))))
(calc 0 0 0 (list 3 2 1 3 1 2 3)))
The same algorithm can also be defined as a non-recursive function like this:
(define (multiples)
(do ((cir (list 3 2 1 3 1 2 3))
(ctr 0 (+ ctr 1))
(a 0 (+ a (list-ref cir (modulo ctr 7))))
(sum 0 (+ sum a)))
((>= a 1000) sum)))
(require-extension (srfi 1))
(define (sum-mod-3-5 upto)
(define (%sum-mod-3-5 so-far generator-position steps)
(let ((next (car generator-position)))
(if (> (+ steps next) upto)
so-far
(%sum-mod-3-5 (+ so-far steps)
(cdr generator-position)
(+ steps next)))))
(%sum-mod-3-5 0 (circular-list 3 2 1 3 1 2 3) 0)) ; 233168
For this particular task, it will do on average half the operations then you would do if incrementing the counter by one, also, one less if condition to check.
Also, modulo (as being division in disguise, probably) is more expensive then summation.
EDIT: I'm not a pro on modular system in different dialects of Scheme. The SRFI-1 extension here is only required to make it easier to create a circular list. I couldn't find an analogue to Common Lisp (#0=(3 2 1 3 1 2 3) . #0#), but perhaps, someone more knowledgeable will correct this.
If you absolutely want to use the "repeating pattern" method, you could go about it something like this.
This uses recursion on the list of intervals rather than relying on list-ref and explicit indexing.
(define (mults limit)
(define steps '(3 2 1 3 1 2 3))
(define (mults-help a sum ls)
(cond ((>= a limit) sum)
((null? ls) (mults-help a sum steps))
(else (mults-help (+ a (car ls))
(+ a sum)
(cdr ls)))))
(mults-help 0 0 steps))
Related
Is there a way I can generate multiple values in for/list during each iteration and have the results "flattened"?
For instance:
(for/list ([i (range n)]) (values i (+ i 1)))
I'd like the result to be (list 0 1 1 2 2 3 3 4 ...).
This question is very much relevant to https://github.com/racket/racket/pull/2483. In the link, you will find:
An unmerged PR that lets you write (for/append-list ([i (range n)]) (list i (+ i 1))).
My proposal to let you write (for/list* ([i (range n)]) (values i (+ i 1))) (with caveat, see the link for more details).
But since these don't exist in Racket yet, the easiest way to get what you want is:
(append* (for/list ([i (range n)]) (list i (+ i 1))))
I'm trying to write a program that returns the Pell numbers sequence based on a given number.
For example (pellNumb 6) should return a list (0 1 2 5 12 29 70)
This is my code so far.
I am able of calculating the numbers, but I am not able of skipping the double recursion.
(defun base (n)
(if (= n 0)
0
(if (= n 1)
1)))
(defun pellNumb (n)
(if (or (= n 0) (= n 1))
(base n)
(let ((x (pellNumb (- n 2))))
(setq y (+ (* 2 (pellNumb (- n 1))) x))
(print y))))
The output for (pellNumb 4) is 2 2 5 12, and this is because i'm recursing to (pellNumb 2) twice.
Is there a way to skip that, and store these values in a list ?
Thanks!
Get the nth number
Yes, there is a way - use multiple values:
(defun pell-numbers (n)
"Return the n-th Pell number, n-1 number is returned as the 2nd value.
See https://oeis.org/A000129, https://en.wikipedia.org/wiki/Pell_number"
(check-type n (integer 0))
(cond ((= n 0) (values 0 0))
((= n 1) (values 1 0))
(t (multiple-value-bind (prev prev-1) (pell-numbers (1- n))
(values (+ (* 2 prev) prev-1)
prev)))))
(pell-numbers 10)
==> 2378 ; 985
This is a standard trick for recursive sequences which depend on several previous values, such as the Fibonacci.
Performance
Note that your double recursion means that (pell-numbers n) has exponential(!) performance (computation requires O(2^n) time), while my single recursion is linear (i.e., O(n)).
Moreover, Fibonacci numbers have a convenient property which allows a logarithmic recursive implementation, i.e., taking O(log(n)) time.
Get all the numbers up to n in a list
If you need all numbers up to the nth, you need a simple loop:
(defun pell-numbers-loop (n)
(loop repeat n
for cur = 1 then (+ (* 2 cur) prev)
and prev = 0 then cur
collect cur))
(pell-numbers-loop 10)
==> (1 2 5 12 29 70 169 408 985 2378)
If you insist on recursion:
(defun pell-numbers-recursive (n)
(labels ((pnr (n)
(cond ((= n 0) (list 0))
((= n 1) (list 1 0))
(t (let ((prev (pnr (1- n))))
(cons (+ (* 2 (first prev)) (second prev))
prev))))))
(nreverse (pnr n))))
(pell-numbers-recursive 10)
==> (0 1 2 5 12 29 70 169 408 985 2378)
Note that the recursion is non-tail, so the loop version is probably more efficient.
One can, of course, produce a tail recursive version:
(defun pell-numbers-tail (n)
(labels ((pnt (i prev)
(if (= i 0)
prev ; done
(pnt (1- i)
(cond ((null prev) (list 0)) ; n=0
((null (cdr prev)) (cons 1 prev)) ; n=1
(t
(cons (+ (* 2 (or (first prev) 1))
(or (second prev) 0))
prev)))))))
(nreverse (pnt (1+ n) ()))))
(pell-numbers-tail 10)
==> (0 1 2 5 12 29 70 169 408 985 2378)
I would like to code a program that given a list and a percentage, splits the list in two different size lists. It should have random pick of the elements, that way the created lists are always different.
These code is able to do that:
(define (clamp x a b)
(max (min x b) a))
(define (split pct xs)
(define pos (exact-round (* (clamp pct 0.0 1.0) (length xs))))
(split-at (shuffle xs) pos))
Here is an example:
(split 0.25 '(1 2 3 4 5 6 7 8 9))
'(6 2)
'(3 7 1 4 5 8 9)
But, instead of "shuffle" I would like to use this function to achieve the same:
(define (get-randomly-no-pair list)
(list-ref list (random (length list))))
so, get-randomly-no-pair takes one element randomly from the initial list. And all the elements are used to create both lists.
(define (shuffle-list lst)
(define indexes (shuffle (range (length lst))))
(lambda ()
(begin0
(list-ref lst (car indexes))
(set! indexes (cdr indexes)))))
(define gen (shuffle-list (list 10 12 14 16 18 20))
(gen) ; ==> 14 (e.g.)
Now I see you assume you need to pass the list then I would rather make a mapper:
(define (shuffle-accessor len)
(define indexes (list->vector (shuffle (range len))))
(lambda (lst index)
(list-ref lst (vector-ref indexes index))))
(define lst3-ref (shuffle-accessor 3))
(lst3-ref '(1 2 3) 0) ; ==> 3 (e.g.)
(lst3-ref '(6 7 8) 0) ; ==> 8
I'm attempting to create a function that creates copies of whatever list the user puts in by the desired number of copies.
User: (copy '(A) '(7))
Output: (A A A A A A A)
(defun copy (x y)
(cond ((-1 counter)
nil)
(T
(list (cons (car x) (cdr x)))
copy
(cdr x)))
I'm attempting to set up a counter and just create a new list into the current list by decrementing the counter. So far the counter is pseudo-code.
This is the counter I'm trying to figure out.
(defun count (y)
(let ((a y))
(- a 1)))
The error I get is that whatever I put into y isn't a number.
While I can understand why the first parameter is a list, the second must be a number. A very simple implementation might look like:
(defun copy (lst count)
(when (> count 0)
(append (copy-list lst) (copy lst (1- count)))))
Testing:
CL-USER> (copy '(A) 7)
(A A A A A A A)
CL-USER> (copy '(A B C) 7)
(A B C A B C A B C A B C A B C A B C A B C)
The usual caveats concerning the use of append and object copying apply.
I'd suggest passing your list as a &rest argument instead and use loop:
(defun repeat (n &rest items)
(loop repeat n append items))
Test
CL-USER> (repeat 10 0 1 2)
(0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2)
I am trying to write a function that takes one or more integers and returns a list of all the arguments that have the same even-odd parity as the first argument, for example
(same-parity 1 2 3 4 5 6 7)->(1 3 5 7)
(same-parity 2 3 4 5 6)->(2 4 6).
my code is
(define (same-parity g . w)
(define (iter-parity items)
(if (= (length items) 1)
(if (= (remainder items 2) (remainder g 2))
item
'())
(if (= (remainder g 2) (remainder (car items) 2))
(cons (car items) (iter-parity (cdr items)))
(iter-parity (cdr items)))))
(cons g (iter-parity w)))
when try this (same-parity (list 1 2 3 4)), I got an error message:
the object (), passed as the first argument to car, is not the correct type.
Can I somebody tell me what is going on?
Your code
Here's a refactoring proposal, keeping with your basic structure:
(define (same-parity g . w)
(define filter-predicate? (if (odd? g) odd? even?))
(define (iter-parity items)
(if (null? items)
'()
(if (filter-predicate? (car items))
(cons (car items) (iter-parity (cdr items)))
(iter-parity (cdr items)))))
(cons g (iter-parity w)))
Note that it is more idiomatic
to use the procedures odd? and even? rather than remainder
to have as a base case when the list is empty, not when it has only one item (in your code this clearly avoids repetition as a positive effect).
Also, since there is a built-in filter procedure in Scheme, you could express it as follows:
(define (same-parity g . w)
(cons g (filter (if (odd? g) odd? even?) w)))
Your question
As for your question regarding (same-parity (list 1 2 3 4)): you need either (as described in your specification) use your procedure like so
(same-parity 1 2 3 4)
or to use apply here:
> (apply same-parity (list 1 2 3 4))
'(1 3)
because apply will transform (same-parity (list 1 2 3 4)) (1 parameter, a list) into (same-parity 1 2 3 4) (4 parameters).