I'm stuck on the question for Exercise 17.1.2 from HTDP. This is what I've got so far:
(define (cross alist1 alist2)
(cond
[(empty? alist1) empty]
[else (append (list (first alist1)(first alist2))
(cons (first alist1)(list (first (rest alist2))))
(cross (rest alist1) alist2))]))
(cross '(a b c) '(1 2))
;correctly outputs (list 'a 1 'a 2 'b 1 'b 2 'c 1 'c 2)
This works for the test case, but when the second list has more than 2 elements, the function falls apart.
(cross '(a b c) '(1 2 3))
;outputs (list 'a 1 'a 2 'b 1 'b 2 'c 1 'c 2)
The problem seems to be the second line after the append, because it's only cons'ing up to two elements from the second list. How should I go about resolving this? Thanks for any insight. :)
It only works for two elements in list two because you only specified it to work for two elements in list two. We need to harness the power of abstraction.
If we were working in imperative languages, then we'd use nested for-loops on this problem. You start on the first element of alist1 and match with all the elements of alist2. Then you move on to the second element of alist1 and match with all the elements of alist2. Since we're working in a functional language (Scheme) we'll use nested functions instead of nested for-loops.
You want to write a function that takes 'a and '(1 2 3) and produces '(a 1 a 2 a 3) and then another function to call the first one with varying values of 'a. Relevant code that you should ignore if you don't want the solution spoiled for you below.
(define (cross alist1 alist2)
(cond
((null? alist1) '())
(else
(append (innercross (car alist1) alist2)
(cross (cdr alist1) alist2)))))
(define (innercross a1 alist2)
(cond
((null? alist2) '())
(else
(cons a1 (cons (car alist2) (innercross a1 (cdr alist2)))))))
Related
Pretty straightforward question. My initial approach was to define another procedure to find the last element of lst within first-last. After finding the last element I appended it with the first element of lst (car lst).
This is how append works.
(append list1 list2)
e.g., (append '(1 2 3) '(2 1 5)) -> (1 2 3 2 1 5)
I'm wondering if the problem is just with my syntax but I am not sure.
(define (first-last lst)
(define (last lst)
(cond ((null? (cdr lst))(car lst))
(else (last (cdr lst)))))
(append(car lst)(last lst)))
The error occurs in the
(append(car lst)(last lst)))
"mcar: contract violation
expected: mpair?
given: 1"
This is my first question on stack, so I'm sorry if the question is not presented in the correct way.
append is only for joining two or more lists. Here, though, you're not joining existing lists, but building a list from two elements. For that, use list:
(list (car lst) (last lst))
If you can use match, a neat solution is possible:
(define first-last
(lambda (x)
(match x
((first rest ... last)
(list first last))
((only) (list only only))
(_ #f))))
Of course, you could return something other than #f in the catch-all clause.
In ISL, how would you create a recursive append function that takes two lists and returns a list of all highest position elements of the first list with the highest position elements of the second list (without using lambda or append)?
Basically a function that would hold for these check expects:
(check-expect (append-test '(a b c) '(d e f g h)) (list 'a 'b 'c 'd 'e 'f 'g 'h))
(check-expect (append-test '() '(7 2 0 1 8 3 4)) (list 7 2 0 1 8 3 4))
I feel like it would definitely use map, since that's what we've been focusing on lately. Here's what I have, which does work, but I was wondering if there was a way to simplify this with map, foldr, foldl, filter, or something like that.
Here's what I have so far:
(define (append-test lst1 lst2)
(cond
[(and (empty? lst1)(empty? lst2)) '()]
[(empty? lst1) lst2]
[(empty? lst2) lst1]
[else (cons (first (first (list lst1 lst2)))
(append-test (rest lst1) lst2))]))
It is much simpler than that.
(define (append-test lhs rhs)
(if (empty? lhs)
rhs
(cons (first lhs) (append-test (rest lhs) rhs))))
This is the function that removes the last element of the list.
(define (remove-last ll)
(if (null? (cdr ll))
'()
(cons (car ll) (remove-last (cdr ll)))))
So from my understanding if we cons a list (eg. a b c with an empty list, i.e. '(), we should get
a b c. However, testing in interaction windows (DrScheme), the result was:
If (cons '() '(a b c))
(() a b c)
If (cons '(a b c) '())
((a b c))
I'm like what the heck :(!
Then I came back to my problem, remove all elements which have adjacent duplicate. For example,
(a b a a c c) would be (a b).
(define (remove-dup lst)
(cond ((null? lst) '())
((null? (cdr lst)) (car lst))
((equal? (car lst) (car (cdr lst))) (remove-dup (cdr (cdr lst))))
(else (cons (car lst) (car (cdr lst))))
)
)
It was not correct, however I realize the answer have a . between a b. How could this happen?
`(a . b)`
There was only one call to cons in my code above, I couldn't see which part could generate this .. Any idea?
Thanks,
cons build pairs, not lists. Lisp interpreters uses a 'dot' to visually separate the elements in the pair. So (cons 1 2) will print (1 . 2). car and cdr respectively return the first and second elements of a pair. Lists are built on top of pairs. If the cdr of a pair points to another pair, that sequence is treated as a list. The cdr of the last pair will point to a special object called null (represented by '()) and this tells the interpreter that it has reached the end of the list. For example, the list '(a b c) is constructed by evaluating the following expression:
> (cons 'a (cons 'b (cons 'c '())))
(a b c)
The list procedure provides a shortcut for creating lists:
> (list 'a 'b 'c)
(a b c)
The expression (cons '(a b c) '()) creates a pair whose first element is a list.
Your remove-dup procedure is creating a pair at the else clause. Instead, it should create a list by recursively calling remove-dup and putting the result as the second element of the pair. I have cleaned up the procedure a bit:
(define (remove-dup lst)
(if (>= (length lst) 2)
(if (eq? (car lst) (cadr lst))
(cons (car lst) (remove-dup (cddr lst)))
(cons (car lst) (remove-dup (cdr lst))))
lst))
Tests:
> (remove-dup '(a b c))
(a b c)
> (remove-dup '(a a b c))
(a b c)
> (remove-dup '(a a b b c c))
(a b c)
Also see section 2.2 (Hierarchical Data and the Closure Property) in SICP.
For completeness, here is a version of remove-dup that removes all identical adjacent elements:
(define (remove-dup lst)
(if (>= (length lst) 2)
(let loop ((f (car lst)) (r (cdr lst)))
(cond ((and (not (null? r))(eq? f (car r)))
(loop f (cdr r)))
(else
(cons (car lst) (remove-dup r)))))
lst))
Here in pseudocode:
class Pair {
Object left,
Object right}.
function cons(Object left, Object right) {return new Pair(left, right)};
So,
1. cons('A,'B) => Pair('A,'B)
2. cons('A,NIL) => Pair('A,NIL)
3. cons(NIL,'A) => Pair(NIL,'A)
4. cons('A,cons('B,NIL)) => Pair('A, Pair('B,NIL))
5. cons(cons('A 'B),NIL)) => Pair(Pair('A,'B),NIL)
Let's see lefts and rights in all cases:
1. 'A and 'B are atoms, and whole Pair is not a list, so (const 'a 'b) gives (a . b) in scheme
2. NIL is an empty list and 'A is an atom, (cons 'a '()) gives list (a)
3. NIL and 'A as above, but as left is list(!), (cons '() 'a) gives pair (() . a)
4. Easy case, we have proper list here (a b).
5. Proper list, head is pair (a . b), tail is empty.
Hope, you got the idea.
Regarding your function. You working on LIST but construct PAIRS.
Lists are pairs (of pairs), but not all pairs are lists! To be list pair have to have NIL as tail.
(a b) pair & list
(a . b) pair not list
Despite cons, your function has errors, it just don't work on '(a b a a c c d). As this is not related to your question, I will not post fix for this here.
So I have to remove the last element of a list in scheme.
For example, let's say I have a list (1 2 3 4). I need to return:
(1 2 3)
My idea:
reverse(list)
car(list)
reverse(list)
Is there a reverse function in scheme(racket)?
You wrote: "reverse, car, reverse". I believe you meant to write "reverse, cdr, reverse". There's nothing wrong with this solution; it's linear in the size of the list, just like any solution to this that uses the standard lists.
As code:
;; all-but-last: return the list, not including the last element
;; list? -> list?
(define (all-but-last l) (reverse (cdr (reverse l))))
If the multiple traversal of the list or the needless construction of another list copy bothers you, you can certainly avoid it, by writing the thing directly.
Given your almost-solution, I'm going to assume that this isn't homework.
Here's what it would look like, in racket:
#lang racket
(require rackunit)
;; all-but-last : return the list, except for the last element
;; non-empty-list? -> list?
(define (all-but-last l)
(cond [(empty? l) (error 'all-but-last "empty list")]
[(empty? (rest l)) empty]
[else (cons (first l) (all-but-last (rest l)))]))
(check-equal? (all-but-last '(3 4 5))
'(3 4))
There is a reverse, but using it would not be very efficient. I suggest the following recursive function.
(define (remove-last lst)
(if (null? (cdr lst))
'()
(cons (car lst) (remove-last (cdr lst)))))
(remove-last '(1 2 3 4)) ; returns '(1 2 3)
The if checks whether it is at the last element of the list.
SRFI 1 (activate in Racket using (require srfi/1)) has a drop-right function:
(drop-right '(1 2 3 4) 1) ; => (1 2 3)
I would do a recursive function that goes down the list and attaches the element (using cons) if the element after it is not the last, and appends nothing if it isn't.
I haven't done scheme for years though so that's as far as I can go.
Someone can run with how to implement it (unless it's homework then they probably shouldn't!)
I've done something simpler than: reverse(list), car(list), reverse(list) to get the last element, check out:
(define (last-one liste)
(if(null? (cdr liste))
null
(cons (car liste) (last-one (cdr liste)))
)
)
Those who are looking for another way can check this out:
(define (removing-last xx)
(remove (list-ref xx (- (length xx) 1)) xx))
I would write a simple recursion, altering the typical "empty? mylist" base case to "empty? (rest mylist)," so that I can return empty when the input list is only 1 element.
(define (removelast mylist)
(cond
[(empty? (rest mylist)) empty]
[(cons? mylist) (cons (first mylist) (removelast (rest mylist)))]))
(removelast (list 1 2 3 4 5))
By the way, this code is in Racket/PLT Scheme, a subset of Scheme.
I have the following items
(define itemslist
(list 'a1 'b2 'c3 (list 'z1 'z2) 'd5 'e6))
My method to find items is below
(define find-item
(lambda (item itemslist)
(cond ((null? itemslist) #f)
((list? (car itemslist))
(cond ((null? itemslist) #f)
(else (find-item item (car itemslist)))))
((equal? stn (car itemslist)) (display "found"))
(else (find-stn stn (cdr itemslist)))
)
)
)
With my method above I can find a1, b2, c3, z1, z2. But when I want to find d5 onwards, it returns nothing. It seems to have skip the stack. Btw, I am just starting to learn Scheme so simple explanation would be better.
One more qns, how about if I have this
(list 'a1 'b2 'c3 (list 'z1 (list 'y1 'y2) 'z2) 'd5 'e6)
does this works as well? Thanks!
Yes, you skip lists.
Example:
'(1 '(2 3) 4)
If (list? '(2 3)) is true, the else part (outer cond) wont be evaluated so 4 is skipped.
So, either you place the else part inside list? block or you redesign your code.
If the itemlist is a list/pair you can pass it right away in a recursive call so
you can avoid writing code like (car itemslist) inside predicates thus make the code simple.
Here is a redesigned (simplified) version:
(define (find-item item l)
(cond ((equal? item l) #t)
((pair? l) (or (find-item item (car l)) (find-item item (cdr l))))
(else #f)))
Tip: you can also can write lists with '(...) notation instead of (list ...), ie
(find-item 'z2 '('a1 'b2 'c3 '('z1 '('y1 'y2) 'z2) 'd5 'e6))
#t
(find-item 'e6 '('a1 'b2 'c3 '('z1 '('y1 'y2) 'z2) 'd5 'e6))
#t
To write more for the sake of writing more... This is usually called a "tree-walk" because a nested list like that is really a binary tree. Lists are really made up of binary pairs, so when you have a pair (l . r), l is the left branch of the tree and r is the right branch of the tree.
For example, '(1 (2 3) 4) is short for '(1 . ((2 . (3 . ())) . (4 . ()))) which can be drawn as
.
/ \
1 .
/ \
/ .
/ / \
. 4 ()
/ \
2 .
/ \
3 ()
and at any pair (l . r), car gets you the left tree and cdr gets you the right. So, when you write (pair? ls), you're really asking if you're at a branch in the tree, at which point you should recur on both the left branch (car) and the right branch (cdr). Hope that helps you understand lists.
Even though you got your [specific] answer, here's something that might help you with similar questions.
(define describe
(lambda (e)
(cond #;((list? e)
`(list ,#(map describe e)))
((pair? e)
`(cons ,(describe (car e))
,(describe (cdr e))))
((vector? e)
`(vector ,#(map describe (vector->list e))))
((or (null? e) (symbol? e)) `',e)
(else e))))
This procedure prints the code that generates a given sexpr. For example:
> (describe '(a 2 b 3))
(cons 'a (cons 2 (cons 'b (cons 3 '()))))
It will also place the quotes where needed, so it places them before symbols or () but not before numbers. When you are comfortable with nested pairs, you may want to remove the #; on the third line, to generate more compact output:
> (describe '(a 2 b 3))
(list 'a 2 'b 3)
This code can teach you many things about quote. For example:
> (describe ''''a)
(list 'quote (list 'quote (list 'quote 'a)))