I am trying to display a Fibonacci sequence less than an upper bound, but I'm having trouble printing out the series that is constrained by that upper bound.
Without using imperative programming principles with declaring variables such as setf, setq, and set, etc. How can I solve this problem?
So far I have
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (< a n)
nil)
(cons a (fibonacci (1- n) b (+ a b)))))
Expected output of (fibonacci 100): (0 1 1 2 3 5 8 13 21 34 55 89). However, what I am getting is (0 1 1 2 3 5 8 13 21 34 55).
You are decreasing upper bound for no reason. When you remove 1- and move one ) to right place, it works as expected:
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (> a n))
nil
(cons a (fibonacci n b (+ a b)))))
Tests:
CL-USER 4 > (fibonacci 10)
(0 1 1 2 3 5 8)
CL-USER 5 > (fibonacci 100)
(0 1 1 2 3 5 8 13 21 34 55 89)
What you meant is
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (> a n))
nil
(cons a (fibonacci n b (+ a b)))))
You had a ) placement typo / error, and the flipped test. Also, n is the upper limit, not a count. So there's no reason to decrease it.
Related
I am trying to count how many times the first number of the list of list called one, two or three is repeated, the idea of the code that I made is for example to choose the list of list called "one" and then take the first letter that has "a" inside and then compare in this case the first value that would be "10" with the others in this list , in this case I should get it to say that the "10" was repeated once, then it would continue with the other two "b" and "c", but when doing the following code
(define a (list 10 2 3 54 6 9 7 10))
(define b (list 5 1 8 6 5 5 4 77 8 6))
(define c (list 80 80 80))
(define e (list 99 156 54 48 99))
(define d (list 16 94 75 30 56 16 8 16))
(define one (list a b c))
(define two (list c e b a))
(define three (list b c d e))
; receives 'one', 'two' or 'three' and finds how many times the first number of the list is repeated in the list of lists
(define (find-repeated-number list)
(define (find-repeated-number-helper list)
(if (null? list)
0
(if (equal? (car list) (car (cdr list)))
(+ 1 (find-repeated-number-helper (cdr list)))
; displays the number of times the first number of the list is repeated in the list of lists
(find-repeated-number-helper (cdr list)))))
(find-repeated-number-helper list))
(find-repeated-number one)
I get the following error, why is it? how can i get what i want to do? I would appreciate any help
*** ERROR: pair required, but got ()
While loading "./jdoodle.sc" at line 19
Stack Trace:
_______________________________________
0 (car (cdr list))
at "./jdoodle.sc":14
1 (equal? (car list) (car (cdr list)))
at "./jdoodle.sc":14
Command exited with non-zero status 70
I'm still not sure if the main point of this exercise is to write low-level recursive code or use some functions provided by the standard library (in this case, count or a combination of filter and length), but here's my wild guess:
(define a (list 10 2 3 54 6 9 7 10))
(define b (list 5 1 8 6 5 5 4 77 8 6))
(define c (list 80 80 80))
(define e (list 99 156 54 48 99))
(define d (list 16 94 75 30 56 16 8 16))
(define one (list a b c))
(define two (list c e b a))
(define three (list b c d e))
(define (find-count lst element)
(count (lambda (n) (= n element))
lst))
(define (print-count lst)
(let ((number-count (find-count (cdr lst) (car lst))))
(write (string-append
"number of times the first number is repeated: "
(number->string (car lst))
" is "
(number->string number-count)
(if (= number-count 1)
" time"
" times")))
(newline)))
(for-each print-count one)
Output:
"number of times the first number is repeated: 10 is 1 time"
"number of times the first number is repeated: 5 is 2 times"
"number of times the first number is repeated: 80 is 2 times"
The task I'm given is, in Racket, to "write a function, countIncreases, which takes a list of numbers and returns how many times the consecutive numbers increase in value. For example, countIncreases '(1 3 2 4 5 1) should return 3 because there are three increases: 1 3, 2 4, 4 5."
I've written a recursive function with the base case being if the list is empty, return 0. I believe my problem is properly comparing one value to the next. Should I be using a standard library list iteration function like foldr or map to accomplish this?
My code and tests are below, and a screenshot of error messages is attached here
(define (countIncreases aList)
(if (empty? aList)
0
(if (< (first aList) (rest aList))
(+ 1 (countIncreases (rest aList)))
(countIncreases (rest aList)))))
(check-expect (countIncreases '(1 3 2 4 5 1)) 3)
(check-expect (countIncreases '()) 0)
(check-expect (countIncreases '(1 2 3 4 5)) 4)
(check-expect (countIncreases '(5 4 3 2 1 2)) 1)
I try to implement a function that computes the Lagrange's four-square theorem with a natural number n in scheme. However I haven't any idea to do that... Can someone give me a example/code please?
For example, a^2 + b^2 + c^2 + d^2 = n, where n is the input of the function.
You can do this with two functions (main & auxiliary):
If n is natural, (lagrange n) gives the list of all the quadruplets (a, b, c, d) such as a^2 + b^2 + c^2 + d^2 = n.
(define lagrange
(lambda (n)
(lagrange-aux 4 0 n)))
(define lagrange-aux
(lambda (size m sum)
(cond ((and (zero? size) (zero? sum)) '(()))
((or (zero? size) (> (* m m) sum)) '())
(else (append (map (lambda (x) (cons m x))
(lagrange-aux (- size 1) 0 (- sum (* m m))))
(lagrange-aux size (+ m 1) sum))))))
Example:
(lagrange 13) ==>
((0 0 2 3) (0 0 3 2) (0 2 0 3) (0 2 3 0) (0 3 0 2) (0 3 2 0)
(1 2 2 2) (2 0 0 3) (2 0 3 0) (2 1 2 2) (2 2 1 2) (2 2 2 1)
(2 3 0 0) (3 0 0 2) (3 0 2 0) (3 2 0 0))
I'm stuck 5 days with this task in Racket, does anybody know how can I approach it?
Given a function of arity 2 and a list of n elements, return the evaluation of the string function of all the elements, for example:
>(reduce + '(1 2 3 4 5 6 7 8 9 10))
55
> (reduce zip '((1 2 3) (4 5 6) (7 8 9)))
'((1 (4 7)) (2 (5 8)) (3 (6 9)))
Here you go.
(define (reduce func list)
(assert (not (null? list)))
(if (null? (cdr list))
(car list)
(func (car list) (reduce func (cdr list)))))
Tests:
> (reduce + '(1 2 3 4 5 6 7 8 9 10))
55
> (reduce zip '((1 2 3) (4 5 6) (7 8 9)))
((1 (4 7)) (2 (5 8)) (3 (6 9)))
For completeness, an implementation for zip (one that assumes two lists and that your lists are all the same length) is:
(define (zip l1 l2) (map list l1 l2))
You can express it in terms of foldl:
(define (reduce f xs)
(and (not (empty? xs)) (foldl f (first xs) (rest xs))))
(define reduce
(λ (f init ls)
(if (empty? ls)
init
(reduce f (f init (first ls)) (rest ls)))))
I'd like to understand the idiomatic way with which to operate over collections of different sizes in clojure. Is there a way I can tell the function 'map' to pad the rest of a collection with some default value?
As an example, suppose I have 3 vectors:
(def x [1 2 3 4])
(def y [1 2 3 4 5])
(def z [1 2 3 4 5 6 7])
(map + x y z) ; yields (3 6 9 12)
In this case, how can I pad x and y with zeroes and have this yield:
(3 6 9 12 10 6 7)
map doesn't do it itself, but you can use a combination of concat and repeat to obtain the desired result:
(def x [1 2 3 4])
(def y [1 2 3 4 5])
(def z [1 2 3 4 5 6 7])
(map +
(concat x (repeat 0))
(concat y (repeat 0))
z) ; => (3 6 9 12 10 6 7)
Here's the API documentation for concat, and for repeat.
And here's a sketch of how you could abstract this away a bit, so you don't need to know which of the collections is longest. (In the snippet above, if you concat all the collections to (repeat 0) you'll have an infinite sequence).
(defn map-longest
[f default & colls]
(lazy-seq
(when (some seq colls)
(cons
(apply f (map #(if (seq %) (first %) default) colls))
(apply map-longest f default (map rest colls))))))
(map-longest +
0
[1 2 3 4]
[1 2 3 4 5]
[1 2 3 4 5 6 7]) ; => (3 6 9 12 10 6 7)
You can see a couple other approaches as answers to this previous question on Stack Overflow.
You can merge vector of maps with variable lengths by the following function:
(defn merge-maps
[& args]
(let [max-count (apply max (map #(count %1) args))
items (map #(take max-count (concat %1 (repeat nil))) args)]
(apply map merge items)))
This function makes the solution more generalized and can take any length of vector of maps.