The following procedure is valid in both scheme r6rs and Racket:
;; create a list of all the numbers from 1 to n
(define (make-nums n)
(do [(x n (- x 1)) (lst (list) (cons x lst))]
((= x 0)
lst)))
I've tested it for both r6rs and Racket and it does work properly, but I only know that for sure for DrRacket.
My question is if it is guaranteed that the step expressions ((- x 1) and (cons x lst) in this case) will be evaluated in order. If it's not guaranteed, then my procedure isn't very stable.
I didn't see anything specifying this in the standards for either language, but I'm asking here because when I tested it was evaulated in order.
They're generally not guaranteed to be evaluated in order, but the result will still be the same. This is because there are no side-effects here -- the loop doesn't change x or lst, it just rebinds them to new values, so the order in which the two step expressions are evaluated is irrelevant.
To see this, start with a cleaner-looking version of your code:
(define (make-nums n)
(do ([x n (- x 1)] [lst null (cons x lst)])
[(zero? x) lst]))
translate to a named-let:
(define (make-nums n)
(let loop ([x n] [lst null])
(if (zero? x)
lst
(loop (- x 1) (cons x lst)))))
and further translate that to a helper function (which is what a named-let really is):
(define (make-nums n)
(define (loop x lst)
(if (zero? x)
lst
(loop (- x 1) (cons x lst))))
(loop n null))
It should be clear now that the order of evaluating the two expressions in the recursive loop call doesn't make it do anything different.
Finally, note that in Racket evaluation is guaranteed to be left-to-right. This matters when there are side-effects -- Racket prefers a predictable behavior, whereas others object to it, claiming that this leads people to code that implicitly relies on this. A common small example that shows the difference is:
(list (read-line) (read-line))
which in Racket is guaranteed to return a list of the first line read, and then the second. Other implementations might return the two lines in a different order.
Related
I want to implement the sorting function in common-lisp with this INSERT function
k means cons cell with number & val, and li means list where I want insert k into.
with this function, I can make a list of cell
(defun INSERT (li k) (IF (eq li nil) (cons (cons(car k)(cdr k)) nil)
(IF (eq (cdr li) nil)
(IF (< (car k)(caar li)) (cons (cons(car k)(cdr k)) li)
(cons (car li) (cons (cons(car k)(cdr k)) (cdr li)) )
)
(cond
( (eq (< (caar li) (car k)) (< (car k) (caadr li)) )
(cons (car k) (cons (cons (car k) (cdr k)) (cdr li)) ) )
(t (cons (car li) (INSERT (cdr li) k)) )))))
and what I want is the code of this function below. it has only one parameter li(non sorted list)
(defun Sort_List (li)(...this part...))
without using assignment, and using the INSERT function
Your insert function is very strange. In fact I find it so hard to read that I cn't work out what it's doing except that there's no need to check for both the list being null and its cdr being null. It also conses a lot of things it doesn't need, unless you are required by some part of the specification of the problem to make copies of the conses you are inserting.
Here is a version of it which is much easier to read and which does not copy when it does not need to. Note that this takes its arguments in the other order to yours:
(defun insert (thing into)
(cond ((null into)
(list thing))
((< (car thing) (car (first into)))
(cons thing into))
(t (cons (first into)
(insert thing (rest into))))))
Now, what is the algorithm for insertion sort? Well, essentially it is:
loop over the list to be sorted:
for each element, insert it into the sorted list;
finally return the sorted list.
And we're not allowed to use assignment to do this.
Well, there is a standard trick to do this sort of thing, which is to use a tail-recursive function with an accumulator argument, which accumulates the results. We can either write this function as an explicit auxiliary function, or we can make it a local function. I'm going to do the latter both because there's no reason for a function which is only ever used locally to be globally visible, and because (as I'm assuming this is homework) it makes it harder to submit directly.
So here is this version of the function:
(defun insertion-sort (l)
(labels ((is-loop (tail sorted)
(if (null tail)
sorted
(is-loop (rest tail) (insert (first tail) sorted)))))
(is-loop l '())))
This approach is fairly natural in Scheme, but not very natural in CL. An alternative approach which does not use assignment, at least explicitly, is to use do. Here is a version which uses do:
(defun insertion-sort (l)
(do ((tail l (rest tail))
(sorted '() (insert (first tail) sorted)))
((null tail) sorted)))
There are two notes about this version.
First of all, although it's not explicitly using assignment it pretty clearly implicitly is doing so. I think that's probably cheating.
Secondly it's a bit subtle why it works: what, exactly, is the value of tail in (insert (first tail) sorted), and why?
A version which is clearer, but uses loop which you are probably not meant to know about, is
(defun insertion-sort (l)
(loop for e in l
for sorted = (insert e '()) then (insert e sorted)
finally (return sorted)))
This, however, is also pretty explicitly using assignment.
As Kaz has pointed out below, there is an obvious way (which I should have seen!) of doing this using the CL reduce function. What reduce does, conceptually, is to successively collapse a sequence of elements by calling a function which takes two arguments. So, for instance
(reduce #'+ '(1 2 3 4))
is the same as
(+ (+ (+ 1 2) 3) 4)
This is easier to see if you use cons as the function:
> > (reduce #'cons '(1 2 3 4))
(((1 . 2) . 3) . 4)
> (cons (cons (cons 1 2) 3) 4)
(((1 . 2) . 3) . 4)
Well, of course, insert, as defined above, is really suitable for this: it takes an ordered list and inserts a new pair into it, returning a new ordered list. There are two problems:
my insert takes its arguments in the wrong order (this is possibly why the original one took the arguments in the other order!);
there needs to be a way of 'seeding' the initial sorted list, which will be ().
Well we can fix the wrong-argument-order either by rewriting insert, or just by wrapping it in a function which swaps the arguments: I'll do the latter because I don't want to revisit what I wrote above and I don't want two versions of the function.
You can 'seed' the initial null value by either just prepending it to the list of things to sort, or in fact reduce has a special option to provide the initial value, so we'll use that.
So using reduce we get this version of insertion-sort:
(defun insertion-sort (l)
(reduce (lambda (a e)
(insert e a))
l :initial-value '()))
And we can test this:
> (insertion-sort '((1 . a) (-100 . 2) (64.2 . "x") (-2 . y)))
((-100 . 2) (-2 . y) (1 . a) (64.2 . "x"))
and it works fine.
So the final question the is: are we yet again cheating by using some function whose definition obviously must involve assignment? Well, no, we're not, because you can quite easily write a simplified reduce and see that it does not need to use assignment. This version is much simpler than CL's reduce, and in particular it explicitly requires the initial-value argument:
(defun reduce/simple (f list accum)
(if (null list)
accum
(reduce/simple f (rest list) (funcall f accum (first list)))))
(Again, this is not very natural CL code since it relies on tail-call elimination to handle large lists, but it makes the point that you can do this without assignment.)
And so now we can write one final version of insertion-sort:
(defun insertion-sort (l)
(reduce/simple (lambda (a e)
(insert e a))
l '()))
And it's easy to check that this works as well.
Define the function iota1(n, m) that takes positive integers n, m with n < m as input, and outputs the list (n,n+1,n+2,...,m)
I've tried switching the code around multiple times but cannot seem to get it to function and display a list the right way
(define (iota1 n m)
(if (eq? n 0)
'()
(append (iota1 (< n m) (+ n 1)) (list n))))
There's a few oddities to the code you provided, which I've formatted for readability:
(define (iota1 n m)
(if (eq? n 0)
'()
(append (iota (< n m) (+ n 1))
(list n))))
The first is that the expression (< n m) evaluates to a boolean value, depending on whether n is less than m or not. When you apply iota to (< n m) in the expression (iota (< n m) (+ n 1)), you are giving iota a boolean value for its first argument instead of a positive integer.
Secondly, the use of append is strange here. When constructing a list in Racket, it's much more common to use the function cons, which takes as arguments a value, and a list, and returns a new list with the value added to the front. For example,
(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)
It's a good idea to opt for using cons instead of append because it's simpler, and because it is faster, since cons does not traverse the entire list like append does.
Since this sounds a bit like a homework problem, I'll leave you with a "code template" to help you find the answer:
; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
(if (> n m) ; base case; no need to do work when n is greater than m
... ; value that goes at the end of the list
(cons ... ; the value we want to add to the front of the list
(iota1 ... ...)))) ; the call to iota, generating the rest of the list
Welcome to the racket world, my version is here:
#lang racket
(define (iota1 n m)
(let loop ([loop_n n]
[result_list '()])
(if (<= loop_n m)
(loop
(add1 loop_n)
(cons loop_n result_list))
(reverse result_list))))
Google Common Lisp Style Guide say Avoid modifying local variables, try rebinding instead
What does it mean? What does rebinding mean in that sentence?
It means that you should create new variables instead of changing the value of old ones. For example, let's take the following code:
(defun foo (x)
(when (minusp x)
(setq x (- x)))
do something with x)
Instead, one should create a new binding and use that one instead:
(defun foo (x)
(let ((xabs (if (minusp x)
(- x)
x)))
do something with xabs)
The reason for this is that you will always know what a variable contains, since it will never change. If you want the new value, simply use the variable that holds that new value.
Now you might ask why this is so important? Well, some people have a stronger preference for this than others. Especially people who prefer to emphasise the functional aspect of Lisp will advocate this style. However, regardless of preference, it can be very useful to always be able to rely on the fact that variables doesn't change. Here's an example where this can be important:
(defun foo (x)
(let ((function #'(lambda () (format t "the value of x is ~a~%" x))))
(when (minusp x)
(setq x (- x)))
(other-function x)
function))
Then, the return value of FOO is a function that when called with print the value of x. But, the value will be that of x later in the function, the absolute value. This can be very surprising if the function is large and complicated.
I don't know Common Lisp well enough to answer with how to do this in Common Lisp, so I'm using Scheme for my example below. Suppose you're writing a function to return the factorial of a number. Here's a "modify local variables" approach to that function (you'll have to define your own while macro, but it's not hard):
(define (factorial n)
(define result 1)
(while (> n 0)
(set! result (* result n))
(set! n (- n 1)))
result)
Here's a "rebind local variables" approach to that function:
(define (factorial n)
(let loop ((n n)
(result 1))
(if (zero? n)
result
(loop (- n 1) (* result n)))))
In this case, loop is called with new values to rebind with each time. The same function can be written using the do macro (which, by the way, also uses rebinding, not modifying, at least in Scheme):
(define (factorial n)
(do ((n n (- n 1))
(result 1 (* result n)))
((zero? n) result)))
I apologize for the bad English..
I have a task to write a function called "make-bag" that counts occurences of every value in a list
and returns a list of dotted pairs like this: '((value1 . num-occurences1) (value2 . num-occurences2) ...)
For example:
(make-bag '(d c a b b c a))
((d . 1) (c . 2) (a . 2) (b . 2))
(the list doesn't have to be sorted)
Our lecturer allows us to us functions MAPCAR and also FILTER (suppose it is implemented),
but we are not allowed to use REMOVE-DUPLICATES and COUNT-IF.
He also demands that we will use recursion.
Is there a way to count every value only once without removing duplicates?
And if there is a way, can it be done by recursion?
First of, I agree with Mr. Joswig - Stackoverflow isn't a place to ask for answers to homework. But, I will answer your question in a way that you may not be able to use it directly without some extra digging and being able to understand how hash-tables and lexical closures work. Which in it's turn will be a good exercise for your advancement.
Is there a way to count every value only once without removing duplicates? And if there is a way, can it be done by recursion?
Yes, it's straight forward with hash-tables, here are two examples:
;; no state stored
(defun make-bag (lst)
(let ((hs (make-hash-table)))
(labels ((%make-bag (lst)
(if lst
(multiple-value-bind (val exists)
(gethash (car lst) hs)
(if exists
(setf (gethash (car lst) hs) (1+ val))
(setf (gethash (car lst) hs) 1))
(%make-bag (cdr lst)))
hs)))
(%make-bag lst))))
Now, if you try evaluate this form twice, you will get the same answer each time:
(gethash 'a (make-bag '(a a a a b b b c c b a 1 2 2 1 3 3 4 5 55)))
> 5
> T
(gethash 'a (make-bag '(a a a a b b b c c b a 1 2 2 1 3 3 4 5 55)))
> 5
> T
And this is a second example:
;; state is stored....
(let ((hs (make-hash-table)))
(defun make-bag (lst)
(if lst
(multiple-value-bind (val exists)
(gethash (car lst) hs)
(if exists
(setf (gethash (car lst) hs) (1+ val))
(setf (gethash (car lst) hs) 1))
(make-bag (cdr lst)))
hs)))
Now, if you try to evaluate this form twice, you will get answer doubled the second time:
(gethash 'x (make-bag '(x x x y y x z z z z x)))
> 5
> T
(gethash 'x (make-bag '(x x x y y x z z z z x)))
> 10
> T
Why did the answer doubled?
How to convert contents of a hash table to an assoc list?
Also note that recursive functions usually "eat" lists, and sometimes have an accumulator that accumulates the results of each step, which is returned at the end. Without hash-tables and ability of using remove-duplicates/count-if, logic gets a bit convoluted since you are forced to use basic functions.
Well, here's the answer, but to make it a little bit more useful as a learning exercise, I'm going to leave some blanks, you'll have to fill.
Also note that using a hash table for this task would be more advantageous because the access time to an element stored in a hash table is fixed (and usually very small), while the access time to an element stored in a list has linear complexity, so would grow with longer lists.
(defun make-bag (list)
(let (result)
(labels ((%make-bag (list)
(when list
(let ((key (assoc (car <??>) <??>)))
(if key (incf (cdr key))
(setq <??>
(cons (cons (car <??>) 1) <??>)))
(%make-bag (cdr <??>))))))
(%make-bag list))
result))
There may be variations of this function, but they would be roughly based on the same principle.
What is the exclusive or functions in scheme? I've tried xor and ^, but both give me an unbound local variable error.
Googling found nothing.
I suggest you use (not (equal? foo bar)) if not equals works. Please note that there may be faster comparators for your situiation such as eq?
As far as I can tell from the R6RS (the latest definition of scheme), there is no pre-defined exclusive-or operation. However, xor is equivalent to not equals for boolean values so it's really quite easy to define on your own if there isn't a builtin function for it.
Assuming the arguments are restricted to the scheme booleans values #f and #t,
(define (xor a b)
(not (boolean=? a b)))
will do the job.
If you mean bitwise xor of two integers, then each Scheme has it's own name (if any) since it's not in any standard. For example, PLT has these bitwise functions, including bitwise-xor.
(Uh, if you talk about booleans, then yes, not & or are it...)
Kind of a different style of answer:
(define xor
(lambda (a b)
(cond
(a (not b))
(else b))))
Reading SRFI-1 shed a new light upon my answer. Forget efficiency and simplicity concerns or even testing! This beauty does it all:
(define (xor . args)
(odd? (count (lambda (x) (eqv? x #t)) args)))
Or if you prefer:
(define (true? x) (eqv? x #t))
(define (xor . args) (odd? (count true? args)))
(define (xor a b)
(and
(not (and a b))
(or a b)))
Since xor could be used with any number of arguments, the only requirement is that the number of true occurences be odd. It could be defined roughly this way:
(define (true? x) (eqv? x #t))
(define (xor . args)
(odd? (length (filter true? args))))
No argument checking needs to be done since any number of arguments (including none) will return the right answer.
However, this simple implementation has efficiency problems: both length and filter traverse the list twice; so I thought I could remove both and also the other useless predicate procedure "true?".
The value odd? receives is the value of the accumulator (aka acc) when args has no remaining true-evaluating members. If true-evaluating members exist, repeat with acc+1 and the rest of the args starting at the next true value or evaluate to false, which will cause acc to be returned with the last count.
(define (xor . args)
(odd? (let count ([args (memv #t args)]
[acc 0])
(if args
(count (memv #t (cdr args))
(+ acc 1))
acc))))
> (define (xor a b)(not (equal? (and a #t)(and b #t))))
> (xor 'hello 'world)
$9 = #f
> (xor #f #f)
$10 = #f
> (xor (> 1 100)(< 1 100))
$11 = #t
I revised my code recently because I needed 'xor in scheme and found out it wasn't good enough...
First, my earlier definition of 'true? made the assumption that arguments had been tested under a boolean operation. So I change:
(define (true? x) (eqv? #t))
... for:
(define (true? x) (not (eqv? x #f)))
... which is more like the "true" definition of 'true? However, since 'xor returns #t if its arguments have an 'odd? number of "true" arguments, testing for an even number of false cases is equivalent. So here's my revised 'xor:
(define (xor . args)
(even? (count (lambda (x) (eqv? x #f)) args)))