I was assigned to write a merge sort in Scheme but I have some issues with it. I showed it my professor and he said there is one simple mistake. Can someone help me?
Plzz!
(define msort
(lamdba(1st)
(cond
((null?? 1st) 1st)
((null? (cdr 1st)) 1st)
(#t ((letrec ((half (quotient (lenght 1st) 2))
(merge (lamdba (a b result)
(cond ((null? a) (apped (reserve a) result))
((null? b) (append (reserve a) result))
((> (car a) (car b) (merge a (cdr b) (cons (car b) result))
(#t (merge (cdr a) b (cons (car a) result)))))))
(merge (msort (take 1st half)) (msort (drop 1st half)) '()))))))
One simple mistake? He probably referred to #1, but even after fixing that you have some identifiers and parenthesis to fix:
lambda, null?, length, append, and reverse is spelled incorrectly.
letrec result gets applied since you have excess parenthesis around it.
cond in merge where you compare elements are missing parenthesis two places.
It's obvious you need help with parenthesis matching so you should download a decent IDE to write code in. I use DrRacket for Scheme development (#!R5RS, #!R6RS and #!racket) and it idents (just press CTRL+i to get it reidented after pasting in code) and indicate where function names are written wrong when you hit RUN.
Making merge a global function in the beginning and perhaps move it to a letrec later (if you have to) might ease development. Eg. you could find errors by testing stuff like (merge '(3 2 1) '()).
This is no guarantee the program will work since I only address syntax here. You need to debug it! DrRacket has a debugger too!
I think it is useful to implement first a function that allow to merge two ordered lists:
(define (merge l1 l2)
(if (empty? l1)
l2
(if (empty? l2)
l1
(if (< (car l1) (car l2))
(cons (car l1) (merge (cdr l1) l2))
(cons (car l2) (merge l1 (cdr l2)))))))
Now assume we have a function (get ls pos) capable to return the element of ls in position pos:
(define (get ls pos)
(if (= pos 1)
(car ls)
(get (cdr ls) (- pos 1))))
Finally, we can implement mergesort function:
(define (mergesort l p r)
(if (= p r)
(cons (get l p) empty)
(merge (mergesort l p (floor (/ (+ p r) 2))) (mergesort l (+ (floor (/ (+ p r) 2)) 1) r))))
Related
I'm new to Scheme and Lisp in general, and upon learning I've stumbled upon a cryptic syntax used in local procedure binding:
(define mock
(lambda (s)
;; this is what I don't understand
(let splice ([l '()] [m (car s)] [r (cdr s)])
(append
(map (lambda (x) (cons m x)) r)
(if (null? r) '()
(splice (cons m l) (car r) (cdr r)))))))
It took me a while to grasp that splice is a scoped procedure with 3 arities. Rewriting this in an ML-esque style it seems to produce similar output:
(define mock2
(lambda (s)
;; define `splice` first
(define splice
(lambda (la lb lc)
(append
(map (lambda (x) (cons lb x)) lc)
(if (null? lc) '()
(splice (cons lb la) (car lc) (cdr lc))))))
;; bind `splice` and its arguments together and call it with them
(let ([sp splice] [l '()] [m (car s)] [r (cdr s)])
(splice l m r))))
The second version is a bit longer and somewhat look more imperative, but defining splice as a normal procedure inside the scope before binding it in parallel with the arguments (or just chuck them in as-is) and calling it looks saner.
Question is are these two versions replaceable? If yes, could you help explain the first version's syntax of binding local variables (l, m, and r) within the splice binding form?
Calling splice is like re-entering a loop, which is what it is there for. A tail call is a goto anyway. It is often named loop, instead of thinking up some special name for it.
"looks saner" is debatable, and actually with Schemers you'll lose this one, because this is a very popular Scheme construct, called "named let". It is usually re-written with letrec btw, if/when one wants to rewrite it, to understand it better. Internal define can be used as well, but then, why not use (define (mock s) ... in the first place.
So, the usual way to re-write this
(define mock ; or: (define (mock s) ...
(lambda (s)
(let splice ([l '()] [m (car s)] [r (cdr s)])
(append
(map (lambda (x) (cons m x)) r)
(if (null? r) '()
(splice (cons m l) (car r) (cdr r)))))))
is this:
(define mock
(lambda (s)
(letrec ([splice (lambda (l m r) ; or: (define (splice l m r) ...
(append
(map (lambda (x) (cons m x)) r)
(if (null? r) '()
(splice (cons m l) (car r) (cdr r)))))])
(splice '() (car s) (cdr s)))))
and writing it in the named let way saves one from having it defined in one place and called in another, potentially far away. A call enters its body from the start anyway, and named let better reflects that.
This is pretty self-explanatory. The transformation from one form to the other is purely syntactical, and both can be used interchangeably.
I'm trying to create a specific response for a given list if it has shared elements with another list. As in if I have a list that is (My name is John) and I have another list of (John Adam Jacob) I would want the first list to be able to see that John is in the second list, and be able to print something along the lines of (this is a known name) or something similiar.
The code I have thought of uses map, and member.
(define (specific-reply user-list)
(cond (member (map (lambda (user-list)) '(John Adam Jacob)))
(write (this is a known name))
(else
(write (this is not a known name)))))
I'm extremely knew to both racket and scheme however and I haven't really gotten it to compile yet so I think I'm largely off.
Any help would be greatly appreciated.
You don't need to complicate the problem if your task is to just find if a is a member of (a b c),
Here's a piece of Scheme code that can tell if a is a member of lat.
It's just a simple recursive function that compares each element of lat with a for a match.
(define member?
(lambda (a lat)
(cond
((null? lat) #f)
((eq? a lat) #t)
(else
(member? a (cdr lat))))))
If you want to take this further and find the intersection of two lists, we can do something like this!
(define intersect
(lambda (set1 set2)
(letrec
((I (lambda (set)
(cond
((null? set) (quote ()))
((member? (car set) set2)
(cons (car set)
(I (cdr set))))
(else (I (cdr set)))))))
(I set1))))
You can use this code as such. Tested from guile compiler
(begin
(display (intersect `(1 2 3) `(1 3 4 5 2)))
(newline))
>> (1 2)
EDIT
I recommend you read The Little Schemer and the The Seasoned Schemer to get more familiar with these kind of concepts
Why not use set in racket:
(define (list-intersect-2 lst1 lst2)
(set->list
(set-intersect (list->set lst1)
(list->set lst2))))
For a solution that takes one or more lists:
(define (list-intersect lst1 . lstn)
(set->list
(foldl set-intersect
(list->set lst1)
(map list->set lstn))))
(list-intersect '(1 2 3) '(2 3 4) '(3 4 8))
; ==> (3)
One can also use built-in functions filter and member to find intersection of 2 lists:
(define (intersection l1 l2)
(remove-duplicates
(filter (λ (x) (member x l1))
l2)))
Above checks each item of l2 to keep it only if it is a member of l1 also.
One can also use for/list to check each element and return a list of common items:
(define (intersect l1 l2)
(remove-duplicates
(for/list ((i l1)
#:when (member i l2))
i)))
Both above function remove duplicates. Just avoiding use of remove-duplicates may result in different result if simply the order of l1 and l2 is interchaged. If one wants that the repeated elements to come repeatedly in outcome list, one can use following function in which common items are removed before proceeding:
(define (intersection2 l1 l2)
(let loop ((l1 l1)
(l2 l2)
(ol '()))
(cond
[(empty? l1) (reverse ol)]
[(member (first l1) l2) ; first item of l1 is common
(loop (rest l1) ; loop with rest of l1
(remove (first l1) l2) ; remove common item from l2
(cons (first l1) ol))] ; add common item to outlist
[else
(loop (rest l1)
l2
ol)])))
Testing:
(intersection2 '(2 4 2 7 2 10) '(10 2 9 2 0 11))
Output:
'(2 2 10)
myList is a list with elements both as symbols or lists of the same type of myList.
For example: myList = '(a b (a d c) d ()) , etc.
I want to write a function in Scheme which would just traverse it (eventually I will replace the symbols with other values).
I wrote this function:
(define traversal (lambda (myList)
(if (null? myList) '()
(if (and (list? (car myList)) (not (null? (car myList))))
(list (traversal (car myList)) (traversal (cdr myList)))
; else if car is an empty list
(if (null? (car myList))
(list (traversal (cdr myList)))
; else car is a symbol
(append (list (car myList)) (traversal (cdr myList))))))))
It gives correct results for some configuration of myList, but definitely it is not the one.
For example,
(display (traversal '((f) h (r t b) (x m b m y) b (c (d)))))
adds additional paranthesis which I don't need.
What would be a correct way to display such a list?
You're testing null? in so many places, where one test is generally enough.
You rarely use list in these traversals, but simply cons.
Also, append is best avoided, and not needed here.
Repetitive use of (car ...) is optimised with a let form.
The simplified form of your code would be:
(define traversal
(lambda (myList)
(if (null? myList)
'()
(let ((c (car myList)))
(cons (if (list? c) (traversal c) c)
(traversal (cdr myList)))))))
EDIT
While this procedure works well for proper lists, it doesn't correctly work for improper lists (although it appears to). The following is a more general approach that works for every kind of S-expression, including proper lists, and I recommend this over the previous code:
(define traversal
(lambda (sexp)
(cond
((null? sexp) '())
((pair? sexp) (cons (traversal (car sexp))
(traversal (cdr sexp))))
(else sexp))))
You are close to the solution. Here are a few hints:
Instead of nested ifs try using the cond form, it is more readable.
The expression (and (list? (car myList)) (not (null? (car myList)))) is correct, but you may use (pair? (car myList)) which is shorter and does almost the same thing.
traversal should return a list but using list with list arguments here
(list (traversal (car myList)) (traversal (cdr myList)))
will return a list of lists. E.g. (list '(a) '(b)) will return ((a) (b)) instead of (a b). In these cases you should use append (append '(a) '(b)) -> (a b).
If a value is not a list but you want to add it to an existing list, use the cons procedure.
(cons 'a '(b c)) -> (a b c).
I have a deep reverse for a basic tree data structure in Scheme
(define (deep-reverse t)
(cond ((null? t) '())
((not (pair? t)) t)
(else (cons (deep-reverse (cdr t)) (deep-reverse (car t))))))
(define stree (cons (list 1 2) (list 3 4)))
1 ]=> (deep-reverse stree)
;Value: (((() . 4) . 3) (() . 2) . 1)
I feel like a cleaner, better result would be:
(4 3 (2 1))
Can anyone provide some guidance as to where I'm going wrong in my deep-reverse function? Thank you.
It's better to split the task into simple operations instead of trying to do all at once. What you want to achieve can be described like this: Reverse the current list itself, then deep-reverse all sublists in it (or the other way round, the order of the two steps doesn't really matter. I choose this order because it results in nicer formatting of the source code).
Now, there already is a function in the standard library for simply reversing a list, reverse. So all you need to do is to combine that with the recursion on those elements that are sublists:
(define (deep-reverse t)
(map (lambda (x)
(if (list? x)
(deep-reverse x)
x))
(reverse t)))
Try this:
(define (deep-reverse t)
(let loop ((t t)
(acc '()))
(cond ((null? t) acc)
((not (pair? t)) t)
(else (loop (cdr t)
(cons (loop (car t) '()) acc))))))
Call it like this:
(define stree (cons (list 1 2) (list 3 4)))
(deep-reverse stree)
> (4 3 (2 1))
For creating a reversed list, one technique is to accumulate the answer in a parameter (I usually call it acc). Since we're operating on a list of lists, the recursion has to be called on both the car and the cdr part of the list. Lastly, I'm using a named let as a shorthand for avoiding the creation of an extra function, but the same result could be obtained by defining a helper function with two parameters, the tree and the accumulator:
(define (deep-reverse t)
(aux t '()))
(define (aux t acc)
(cond ((null? t) acc)
((not (pair? t)) t)
(else (aux (cdr t)
(cons (aux (car t) '()) acc)))))
I think it better to reverse a list based on its element count:
an empty list is reverse, a single element list is also reverted, more than 1 element is concatenation of the reverse of tail and head.
(defun deep-reverse (tree)
(cond ((zerop (length tree)) nil)
((and (= 1 (length tree)) (atom (car tree))) tree)
((consp (car tree)) (append (deep-reverse (cdr tree))
(list (deep-reverse (car tree)))))
(t (append (deep-reverse (cdr tree)) (list (car tree))))))
The following worked for me:
(define (deep-reverse tree)
(define (deep-reverse-iter items acc)
(cond
((null? items) acc)
((not (pair? items)) items)
(else (deep-reverse-iter
(cdr items)
(cons (deep-reverse (car items)) acc)))))
(deep-reverse-iter tree ()))
(define x (list (list 1 2) (list 3 4 (list 5 6))))
(newline)
(display (deep-reverse x))
It prints (((6 5) 4 3) (2 1)) as expected and uses the minimum of standard library functions: pair? to check if the tree is a cons and null? to check for an empty tree/list.
This solution for trees is a generalization of the reverse function for lists:
(define (reverse items)
(define (reverse-iter items acc)
(cond
((null? items) acc)
((not (pair? items)) items)
(else (reverse-iter (cdr items) (cons (car items) acc)))))
(reverse-iter items ()))
the difference being that deep-reverse is also applied to car items
how can I write a function to take the last element of the list?
find the last of a list:
(define (last l)
(cond ((null? (cdr l)) (car l))
(else (last (cdr l)))))
use map to map last to a list:
(map last '((a b) (c d) (e f)))
==> (b d f)
so a new function:
(define (last-list l)
(map last l)
)
(last-list '((a b) (c d) (e f)))
==> (b d f)
May not be the most efficient, but certainly one of the simplest:
(define (last lst)
(car (reverse lst)))
Examples:
(last '(1 2 3 4)) => 4
(last '((a b) (b c) (d e))) => (d e)
The code you've written - to take the last element of a list - is correctly returning the last element of the list. You have a list of lists. There is an outer list
(x y z)
where
x = (a b)
y = (c d)
z = (e f)
So you're getting the last element of the list, z, which is (e f)
Did you want your last function to do something different? If you want it to return the last element of the last nested list, you need to change your base case. Right now you return the car. Instead, you want to check if the car is a list and then call your nested-last function on that.
Make sense?
Your last function is good, but you have to think about what you want to do with it now.
You have a list of lists, and you want to take the last of all those.
So recurse down your list applying it each time:
(define (answer lst)
(cond ((null? (cdr l)) null)
(else (cons (last (car l)) (answer (cdr lst))))
Yet another possibility:
(define (last thelist)
(if
(null? (cdr thelist)) (car thelist)
(last (cdr thelist))))
(define (all-last lists) (map last lists))
Edit: just saw that you don't know map, and want a solution without it:
(define (all-last lists)
(if
(null? lists) `()
(cons (last (car lists)) (all-last (cdr lists)))))
As far as getting an empty list goes, I'd guess you're trying to use this map-like front-end with your original definition of last, whereas it's intended to work with the definition of last I gave above. Try the following definitions:
(define (last thelist) (if
(null? (cdr thelist)) (car thelist)
(last (cdr thelist))))
(define (all-last lists) (if
(null? lists) `()
(cons (last (car lists)) (all-last (cdr lists)))))
and running a quick test:
(all-last `((a b) (c d) (e f)))
The result should be:
(b d f)
(define last
(lambda (ls)
(list-ref ls (- (length ls) 1))))
I like short, sweet, fast, tail-recursive procedures.
Named let is my friend.
This solves the original problem and returns #f if the list has no last element.
(define (last L) (let f ((last #f) (L L)) (if (empty? L) last (f (car L) (cdr L)))))
The best way to get what you want:
(define (last lst)
(cond [(empty? lst) empty]
[(empty? (rest lst)) (first lst)]
[else (last (rest lst))]))