I'm a newbie and I didn't understand very well the language. Could anyone please explain to me what this functions do?
First function:
(define (x l)
(cond
((null? l) 0)
((list? (car l))
(+ (x (car l)) (x (cdr l))))
(else (+ 1 (x (cdr l))))
))
Second function:
(define (x l)
(cond
((null? l) 0)
((list? (car l))
(+ (x (car l)) (x (cdr l))))
(else (+ (car l) (x (cdr l)))
))
I do understand the begining but the conditions I didn't understand. Any help?
I will call your second function y.
Writing in pseudocode,
x [] -> 0
x [a . b] -> x a + x b , if list a
x [a . b] -> 1 + x b , else, i.e. if not (list a)
y [] -> 0
y [a . b] -> y a + y b , if list a
y [a . b] -> a + y b , else, i.e. if not (list a)
So for example,
x [2,3] = x [2 . [3]]
= 1 + x [3]
= 1 + x [3 . []]
= 1 + (1 + x [])
= 1 + (1 + 0 )
and
y [2,3] = y [2 . [3]]
= 2 + y [3]
= 2 + y [3 . []]
= 2 + ( 3 + y [])
= 2 + ( 3 + 0 )
See? The first counts something in the argument list, the second sums them up.
Of course both functions could be called with some non-list, but then both would just cause an error trying to get (car l) in the second clause, (list? (car l)).
You might have noticed that the two are almost identical. They both accumulates (fold) over a tree. Both of them will evaluate to 0 on the empty tree and both of them will sum the result of the same procedure on the car and cdr when the car is a list?. The two differ when the car is not a list and in the first it adds 1 for each element in the other it uses the element itself in the addition. It's possible to write the same a little more compact like this:
(define (sum l)
(cond
((null? l) 0) ; null-value
((not (pair? l)) l) ; term
(else (+ (sum (car l)) (sum (cdr l)))))) ; combine
Here is a generalisation:
(define (accumulate-tree tree term combiner null-value)
(let rec ((tree tree))
(cond ((null? tree) null-value)
((not (pair? tree)) (term tree))
(else (combiner (rec (car tree))
(rec (cdr tree)))))))
You can make both of your procedures in terms of accumulate-tree:
(define (count tree)
(accumulate-tree tree (lambda (x) 1) + 0))
(define (sum tree)
(accumulate-tree tree (lambda (x) x) + 0))
Of course you can make a lot more than this with accumulate-tree. It doesn't have to turn into an atomic value.
(define (double tree)
(accumulate-tree tree (lambda (x) (* 2 x)) cons '()))
(double '(1 2 ((3 4) 2 3) 4 5)) ; ==> (2 4 ((6 8) 4 6) 8 10)
Related
I want to create a function in LISP
to count the number of 0 in given arguments
Ex
(count_number_of_0 '(1 0 5 9 0 0 0 7 1 0) )
Output : 5
Here is an implementation in Racket, which is a lisp-family language. It would be quite easy to translate into Common Lisp (but a little more verbose in CL):
(define make-counter
(λ (v same?)
(λ (l)
((λ (c)
(c c 0 l))
(λ (c a t)
(if (null? t)
a
(c c (if (same? (first t) v) (+ a 1) a) (rest t))))))))
(define count-zeros
(make-counter 0 =))
And now
> (count-zeros '(1 2 0 3 4 0))
2
one way is:
(defun count-number-of-0 (lst &optional (cnt 0)) ;counter starts at zero
(if lst
(if (and (numberp (car lst)) ;better verify that element is a number
(= 0 (car lst)))
(progn
(setq cnt (+ cnt 1))
(count-number-of-0 (cdr lst) cnt))
(count-number-of-0 (cdr lst) cnt))
cnt)) ;return counter
This should work in all implementations of common-lisp.
I am trying to implement a function called funPower, which takes a function f, an integer n and returns the function f^n. For example ((funPower sqrt 2) 16) should return 2, which is (sqrt (sqrt 16)).
This is what I have so far but it is not giving me correct output
(define (funPower f n)
(lambda(x) (if (<= n 1)
(f x)
(f (funPower f (- n 1)) x))))
First, you're missing one more pair of parens.
(define (funPower1 f n)
(lambda (x) (if (<= n 1)
(f x)
;; (f ( funPower1 f (- n 1)) x))))
(f ( ( funPower1 f (- n 1)) x)))) )
;; ^^^ ^^^
because (funPower1 f (- n 1)) returns a function to be called on x, the future argument value, as you show with the example, ((funPower sqrt 2) 16).
Second, it's <= 0, not <= 1, and the function f shouldn't be called at all in such a case:
(define (funPower2 f n)
(lambda (x) (if (<= n 0)
;; (f x) ^^^
x
(f ( ( funPower2 f (- n 1)) x)))) )
Now that it's working, we see that it defers the decisions to the final call time, of ((funPower f n) x). But it really could do all the decisions upfront -- the n is already known.
To achieve that, we need to swap the (lambda and the (funPower, to push the lambda "in". When we do, it'll become an additional argument to such augmented funPower:
(define (funPower3 f n)
(if (<= n 0) (lambda (x)
x )
(funPower3 f (- n 1) (lambda (x) (f x)))) )
Now this is completely out of sync. Where's that third argument?
(define (funPower4 f n fun)
(if (<= n 0) fun
(funPower4 f (- n 1) (lambda (x) (fun (f x)))) ))
That's a little bit better, but what's the fun, originally? Where does it come from? It must always be (lambda (x) x) at first or else it won't be right. The solution is to make it an internal definition and use that, supplying it the correct argument the first time we call it:
(define (funPower5 f n)
(define (loop n fun)
(if (<= n 0) fun
(loop (- n 1)
(lambda (x) (fun (f x))))))
(loop n (lambda (x) x)))
This kind of thing would normally be coded as a named let,
(define (funPower5 f n)
(let loop ((n n)
(fun (lambda (x) x)))
(if (<= n 0) fun
(loop (- n 1)
(lambda (x) (fun (f x)))))))
We could also try creating simpler functions in the simpler cases. For instance, we could return f itself if n is 1:
(define (funPower6 f n)
(cond
((zero? n) .....)
((= n 1) .....)
((< n 0) .....)
(else
(let loop ((n n)
(fun .....))
(if (= n .....) fun
(loop (- n 1)
(lambda (x) (fun (f x)))))))))
Complete it by filling in the blanks.
More substantive further improvement is to use exponentiation by repeated squaring -- both in constructing the resulting function, and to have it used by the function we construct!
try this:
(define funpow
(lambda (f n)
((lambda (s) (s s n (lambda (x) x)))
(lambda (s n o)
(if (zero? n)
o
(s s (- n 1)
(lambda (x)
(o (f x)))))))))
(define sqrt_2 (funpow sqrt 2))
(define pow2_2 (funpow (lambda (x) (* x x)) 2))
(sqrt_2 16)
(pow2_2 2)
Is there way to change - (minus) function to + (plus) function?
My homework is to implement sin calculation on Macluaurin series
sin(x) = x-(x^3/3!)+(x^5/5!) -(x^7/7!)+(x^9/9!)-...
Each article has different sign. This is my Lisp code
(defun sinMac (x series n plusminus)
(cond ((= series 0) 0)
(t (funcall plusminus
(/ (power x n) (factorial n))
(sinMac x (- series 1) (+ n 2) plusminus)))))
Is it possible to change plusminus to exchange sign? if I get '+ function send '- to next recursive call. From that call (got '-) I call '+ and so on...
You could do it with a circular list. Like so:
(defun sin-mac (x series n plus-minus)
(cond ((zerop series) 0)
(t (funcall (car plus-minus)
(/ (power x n) (factorial n))
(sin-mac x (1- series) (+ n 2) (cdr plus-minus))))))
(sin-mac x series 1 '#0=(+ - . #0#))
Or even better, wrap up the initial arguments using labels:
(defun sin-mac (x series)
(labels ((recur (series n plus-minus)
(cond ((zerop series) 0)
(t (funcall (car plus-minus)
(/ (power x n) (factorial n))
(recur (1- series) (+ n 2) (cdr plus-minus)))))))
(recur series 1 '#0=(+ - . #0#))))
If the function is a symbol, this is easy:
(defun next-function (function)
(ecase function
(+ '-)
(- '+)))
(defun sinMac (x series n plusminus)
(cond ((= series 0) 0)
(t (funcall plusminus
(/ (power x n) (factorial n))
(sinMac x
(- series 1)
(+ n 2)
(next-function plusminus))))))
I would not swap the function but just the sign. Using a loop for this also seems clearer to me (and is most likely more efficient, although there is still plenty of opportunity for optimization):
(defun maclaurin-sinus (x n)
"Calculates the sinus of x by the Maclaurin series of n elements."
(loop :for i :below n
:for sign := 1 :then (- sign)
:sum (let ((f (1+ (* 2 i))))
(* sign
(/ (expt x f)
(factorial f))))))
A few optimizations make this about 10 times faster (tested with n = 5):
(defun maclaurin-sinus-optimized (x n)
"Calculates the sinus of x by the Maclaurin series of n elements."
(declare (integer n))
(loop :repeat n
:for j :from 0 :by 2
:for k :from 1 :by 2
:for sign := 1 :then (- sign)
:for e := x :then (* e x x)
:for f := 1 :then (* f j k)
:sum (/ e f sign)))
I must create a function - (search-last x list), i made this:
(define (search-last o lst)
(let loop ((lst lst))
(if (eqv? (caar lst) o)
(cdar lst)
(if (pair? (cdr lst))
(loop (cdr lst))
o))))
but for ex.
(define l '((1 . 2) (2 . 5) (3 . 5) (2 . 1)))
should be 1 but
my output is 5 i know where i made mistake but i dont know how can i improve it.
I cant use expression with "!" and vector, for, while, set, sort, reverse, list-ref, list-tail, append, length.
A tail recursive solution using named let
(define (search-last o lst)
(let loop ((lst lst) (match #f)) ; when not found match is #f
(cond ((null? lst) match) ; return last match at the end
((eqv? (caar lst) o) ; when found
(loop (cdr lst) (cdar lst))) ; we recurse with new match
(else (loop (cdr lst) match))))); or we keep the old match
You could have created your own reverse function and reverse the list in your named let loop. But my understanding is that this is not the point of the exercise.
This will work:
(define (search-last o lst)
(or
(let loop ((lst lst))
(cond
((null? lst) #f)
((eqv? (caar lst) o) (or (loop (cdr lst)) (cdar lst)))
(else (loop (cdr lst)))))
o))
The trick is that if you find a pair with the required first element, the or clause will search the rest of the list. If it returns #f, i.e. if there is no other such pair in the remaining list, then the clause will return (cdar lst). Otherwise it will return the result of the recursive search-last call. The outer or does the same with the final result, i.e. replacing #f with the argument o.
Testing:
> (search-last 1 '((1 . 2) (2 . 5) (3 . 5) (2 . 1)))
2
> (search-last 2 '((1 . 2) (2 . 5) (3 . 5) (2 . 1)))
1
> (search-last 3 '((1 . 2) (2 . 5) (3 . 5) (2 . 1)))
5
> (search-last 4 '((1 . 2) (2 . 5) (3 . 5) (2 . 1)))
4
FWIW, here's a tail-recursive version:
(define (search-last o lst)
(let loop ((lst lst) (res o))
(cond
((null? lst) res)
((eqv? (caar lst) o) (loop (cdr lst) (cdar lst)))
(else (loop (cdr lst) res)))))
I'm trying to implement an algorithm to multiply two bit-lists of 1s and 0s as a simulation to binary multiplication. It should return a like list, but I am having a hard time building on what I already have. Some help would be appreciated...
;;Function designed to accept two bit-list binary numbers (reverse order) and produce their product, a bitlist in reverse order.
;;Example: (multiply '(0 1 1 0 1) '(1 0 1)) should produce '(0 1 1 1 0 1 1)
(define (multiply x y)
(cond
;[(= null? y) 0]
[(zero? y) 0]
(#t (let ((z (multiply x (rest y )))) (cond
[(num_even? y) (cons 0 z)]
(#t (addWithCarry x (cons 0 z) 1)))))))
;This is to check if the current value of parameter x is the number 0
(define (zero? x)
(cond
((null? x) #t)
((=(first x) 1) #f)
(#t (zero? (rest x)))))
;This is to check if the current parameter x is 0 (even) or not.
(define (num_even? x)
(cond
[(null? x) #t]
[(=(first x) 0)#t]
[#t (num_even? (rest x))]))
;To add two binary numbers
(define(addWithCarry x y carry)
(cond
((and (null? x) (null? y)) (if (= carry 0) '( ) '(1)))
((null? x) (addWithCarry '(0) y carry))
((null? y) (addWithCarry x '(0) carry))
(#t (let ((bit1 (first x))
(bit2 (first y)))
(cond
((=(+ bit1 bit2 carry) 0) (cons 0 (addWithCarry (rest x)(rest y) 0)))
((=(+ bit1 bit2 carry) 1) (cons 1 (addWithCarry (rest x)(rest y) 0)))
((=(+ bit1 bit2 carry) 2) (cons 0 (addWithCarry (rest x)(rest y) 1)))
(#t (cons 1 (addWithCarry (rest x) (rest y) 1))))))))
Based on my previous answer for a base-10 multiplication, here's a solution that works for binary numbers (in the correct order):
(define base 2)
(define (car0 lst)
(if (empty? lst)
0
(car lst)))
(define (cdr0 lst)
(if (empty? lst)
empty
(cdr lst)))
(define (apa-add l1 l2) ; apa-add (see https://stackoverflow.com/a/19597007/1193075)
(let loop ((l1 (reverse l1))
(l2 (reverse l2))
(carry 0)
(res '()))
(if (and (null? l1) (null? l2) (= 0 carry))
res
(let* ((d1 (car0 l1))
(d2 (car0 l2))
(ad (+ d1 d2 carry))
(dn (modulo ad base)))
(loop (cdr0 l1)
(cdr0 l2)
(quotient (- ad dn) base)
(cons dn res))))))
(define (mult1 n lst) ; multiply a list by one digit
(let loop ((lst (reverse lst))
(carry 0)
(res '()))
(if (and (null? lst) (= 0 carry))
res
(let* ((c (car0 lst))
(m (+ (* n c) carry))
(m0 (modulo m base)))
(loop (cdr0 lst)
(quotient (- m m0) base)
(cons m0 res))))))
(define (apa-multi l1 l2) ; full multiplication
(let loop ((l2 (reverse l2))
(app '())
(res '()))
(if (null? l2)
res
(let* ((d2 (car l2))
(m (mult1 d2 l1))
(r (append m app)))
(loop (cdr l2)
(cons '0 app)
(apa-add r res))))))
so that
(apa-multi '(1 0 1 1 0) '(1 0 1))
=> '(1 1 0 1 1 1 0)