I am practicing for my programming paradigms exam and working through problem sets I come to this problem. This is the first problem after reversing and joining lists recursively, so I suppose there is an elegant recursive solution.
I am given a list of lists and a permutation. I should permute every list including a list of lists with that specified permutation.
I am given an example:
->(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
->((1 3 2) (5 7 6) (a c b))
I have no idea even how to start. I need to formulate the problem in recursive interpretation to be able to solve it, but I can not figure out how.
Well, let's see how we can break this problem down. We are given a list of lists, and a list of numbers, and we want to order each list according to the order specified by the list of numbers:
=>(permute '((1 2 3) (4 5 6)) '(3 2 1))
'((3 2 1) (6 5 4))
We can see that each list in the list of lists can be handled separately, their solutions are unrelated to each other. So we can have a helper permute1 that handles the case of one list, then use map to apply this function to each of the lists (with the same ordering each time):
(define (permute lists ordering)
(map (lambda (xs) (permute1 xs ordering))
lists))
(define (permute1 items ordering)
...)
Now, to calculate (permute1 '(4 5 6) '(3 2 1)), what we mean is:
The first item of the new list will be the 3rd item of items, because the first number in ordering is 3.
The rest of the items of the new list will be determined by using the rest of the numbers in the ordering.
If the ordering is the empty list, return the empty list.
This forms the base case (3), the recursive case (1), and the steps to recur deeper (2). So a sketch of our solution would look like:
(define (permute1 items ordering)
(if (empty? ordering)
'()
(let ([next-item ???])
(??? next-item
(permute1 items (rest ordering))))))
Where the ???s represent getting the item based on the first number in ordering and combining this item with the remainder of the calculation, respectively.
Here's another option, using higher-order functions. This is the idiomatic way to think about a solution in a functional language - we split the problem in sub-problems, solve each one using existing procedures and finally we compose the answer:
(define (atom? x)
(and (not (null? x))
(not (pair? x))))
(define (perm lst order)
(foldr (lambda (idx acc)
(cons (list-ref lst (sub1 idx)) acc))
'()
order))
(define (permute lst order)
(if (atom? lst)
lst
(perm (map (lambda (x) (permute x order)) lst)
order)))
We start by defining atom?, a generic predicate and perm, a helper that will reorder any given list according to the ordering specified in one of its parameters. It uses foldr to build the output list and list-ref to access elements in a list, given its 0-based indexes (that's why we subtract one from each index).
The main permute function takes care of (recursively) mapping perm on each element of an arbitrarily nested input list, so we can obtain the desired result:
(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
=> '((1 3 2) (5 7 6) (a c b))
I am given an example:
(permute ('(1 2 3) '(a b c) '(5 6 7)) '(1 3 2))
((1 3 2) (5 7 6) (a c b))
The syntax you've given isn't correct, and will cause an error, but it's fairly clear what you mean. You want that
(permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
;=> ((1 3 2) (5 7 6) (a c b))
Now, it's not clear how you're indicating the permutation. Is '(1 3 2) a permutation because it has some (1-based) indices, and indicates the way to rearrange elements, or is it because it is actually a permutation of the elements of the first list of the first list? E.g., would
(permute '((x y z) (a b c) (5 6 7)) '(1 3 2))
;=> ((x z y) (5 7 6) (a c b))
work too? I'm going to assume that it would, because it will make the problem much easier.
I have no idea even how to start. I need to formulate the problem in
recursive interpretation to be able to solve it, but I can not figure
out how.
You need to write a function that can take a list of indices, and that returns a function that will perform the permutation. E.g,.
(define (make-permutation indices)
…)
such that
((make-permutation '(3 1 2)) '(a b c))
;=> (c a b)
One you have that, it sounds like your permute function is pretty simple:
(define (permute lists indices)
(let ((p (make-permutation indices)))
(p (map p lists))))
That would handle the case you've given in your example, since (map p lists) will return ((1 3 2) (a b c) (5 7 6)), and then calling p with that will return ((1 3 2) (5 7 6) (a c b)). If you need to be able to handle more deeply nested lists, you'll need to implement a recursive mapping function.
Here's my take, which seems to be shorter than the previous examples:
(define (permute lst ord)
(define ord-1 (map sub1 ord)) ; change from 1-based to 0-based indexes
(define (perm elts) ; recursive sub-procedure
(if (list? elts)
(map perm (map (curry list-ref elts) ord-1)) ; list -> recurse
elts)) ; else return unchanged
(perm lst)) ; initial call
testing
> (permute '((1 2 3) (a b c) (5 6 7)) '(1 3 2))
'((1 3 2) (5 7 6) (a c b))
> (permute '((1 (i permute did) 3) (a b (scheme cool is)) (5 6 7)) '(1 3 2))
'((1 3 (i did permute)) (5 7 6) (a (scheme is cool) b))
Related
I'm trying to get the lowest integer out of a vector only containing numbers. I know how to do it with lists. You compare the first two values of the list and depending on which is larger you either save your value to output it later or call the function again with the rest of the list (all elements except the first) using the cdr procedure.
But with vectors I'm completely lost. My guess would be that the way of thinking about the solution would be the same for lists and vectors. I've been reading on the racket-lang website but haven't been able to come up with a solution to the problem. The procedures I've been experimenting most with are vector-ref and vector-length as they seem to be the most useful in this problem (but this is my first time working with vectors so what do I know).
So my two questions are:
How can we get all values except the first from a vector? Is there a procedure like cdr but for vectors?
If you were working with lists you would use cons to save the values you would want to output. But is there a similar way of doing it when working with vectors?
Thanks!
The simplest solution is to use a variant of for called for/fold.
I thought there were an for/min but alas.
#lang racket
(define v (vector 11 12 13 4 15 16))
(for/fold ([m +inf.0]) ([x (in-vector v)])
(min m x))
If you like a more explicit approach:
(define (vector-min xs)
(define n (vector-length xs))
(let loop ([i 0] ; running index
[m +inf.0]) ; minimum value so far
(cond
[(= i n) ; if the end is reached
m] ; return the minimum
[else ; else
(define x (vector-ref v i)) ; get new element in vector
(loop (+ i 1) ; increment index
(min m x))]))) ; new minimum
UPDATE
(let loop ([x 1] [y 10])
(loop (+ x 1) (- y 1))
is the same as:
(let ()
(define (loop (x y)
(loop (+ x 1) (- y 1)))
(loop 1 10))
Vectors are O(1) access and indexed so it is a completely different data structure, however you have SEFI-43 which is like the SRFI-1 List library, but for vectors.
#lang racket
(require srfi/43)
(define (min-element lst)
(vector-fold min (vector-ref lst 0) lst))
(max-element #(7 8 1 2 3 4 5 12))
; ==> 1
The racket/vector module has vector-argmin for finding the minimum element of a vector (Well, the minimum after feeding the elements through a transformation function). Combine that with a function like identity from racket/function and it's trivial:
(vector-argmin identity '#(5 4 3 2 1 6))
I am learning to use scheme and practicing by creating functions from practice problems I saw in a book. This one is called zipper.
I have already made the zipper function using recursion with only cons, car, and cdr.
Now I am trying to make this same function again but also using map or fold.
The function takes in two lists and 'zips' them together:
(zip '(1 2 3) '(4 5 6))
==> '((1 4) (2 5) (3 6))
How can I do this using either map or fold?
It may become clearer if the function to be used by map is declared separately:
(define (zip1 x y)
(list x y))
(map zip1 '(1 2 3) '(4 5 6)) ; corresponding elements of 2 lists will be sent to zip1
Since zip1 is only 'list', following also works:
(map list '(1 2 3) '(4 5 6))
Output for both:
'((1 4) (2 5) (3 6))
Hence to do (zip L1 L2), one does not need to write special zip function, one can just do (map list L1 L2).
This can also be used for any number of lists, e.g. (map list L1 L2 L3 L4) which may be an advantage over custom-made zip function made for 2 lists only.
The Racket documentation for map provides the following example:
> (map (lambda (number1 number2)
(+ number1 number2))
'(1 2 3 4)
'(10 100 1000 10000))
'(11 102 1003 10004)
Notice that this is very similar to what you are trying to achieve, but instead of mapping the + function between corresponding elements in two lists, you can map a list function, as follows:
> (map (lambda (number1 number2)
(list number1 number2))
'(1 2 3 4)
'(10 100 1000 10000))
'((1 10) (2 100) (3 1000) (4 10000))
and hence, a zip function becomes:
(define (zip lst1 lst2)
(map
(lambda(elem1 elem2)
(list elem1 elem2))
lst1 lst2))
I have a small exercise in Lisp:
Write a function test-delta with parameters delta and lst, which will
check if the difference between successive elements in lst is smaller than
delta. Write the function in two ways:
recursively
using a mapping function
I have no problem writing that function recursively, but I don't know which mapping function I should use. All the standard mapping functions work with only one element of the list at a time. reduce cannot be used either, because I do not have some operation to use between successive elements. What function could I use here?
All standard functions are working only with one element at time.
Reduce function cannot be use either
because i do not have some operation to use between to elements.
There's already an answer by uselpa showing that you can do this with reduce, but it feels a bit awkward to me to bend reduce to this case.
It's much more natural, in my opinion, to recognize that the standard mapping functions actually let you work with multiple lists. I'll show mapcar and loop first, and then every, which I think is the real winner here. Finally, just for completeness, I've also included maplist.
mapcar
The standard mapcar can take more than one list, which means that you can take elements from two different lists at once. Of particular note, it could take a list and (rest list). E.g.,
(let ((list '(1 2 3 4 5 6)))
(mapcar 'cons
list
(rest list)))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
loop
You can use loop to do the same sort of thing:
(loop
with l = '(1 2 3 4 5 6)
for a in l
for b in (rest l)
collect (cons a b))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
There are some other variations on loop that you can use, but some of them have less conventient results. E.g., you could loop for (a b) on list, but then you get a (perhaps) unexpected final binding of your variables:
(loop for (a b) on '(1 2 3 4 5 6)
collect (list a b))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
This is similar to what maplist will give you.
every
I think the real winners here, though, are going to the be every, some, notevery, and notany functions. These, like mapcar can take more than one list as an argument. This means that your problem can simply be:
(let ((delta 4)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> T
(let ((delta 2)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> NIL
maplist
You could also do this with maplist, which works on successive tails of the list, which means you'd have access to each element and the one following. This has the same 6 NIL at the end that the second loop solution did, though. E.g.:
(maplist (lambda (tail)
(list (first tail)
(second tail)))
'(1 2 3 4 5 6))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
reduce can be used:
(defun testdelta (delta lst)
(reduce
(lambda (r e)
(if (< (abs (- r e)) delta)
e
(return-from testdelta nil)))
lst)
t)
or, without return-from (but possibly slower):
(defun testdelta (delta lst)
(and
(reduce
(lambda (r e)
(and r (if (< (abs (- r e)) delta) e nil)))
lst)
t))
I am having a problem using list-ref.
Is there a way to use list-ref to obtain size of an unknown list?
(... (if (number? (deref(+ array-ref index)))
(array-len array-ref (+ index 1))
0) )
#|(define (deref ref)
(list-ref the-store ref))
|#
If you want the size of a list use length, not list-ref. For example, in a list such as this:
(define lst '(1 2 3 4 5))
… The valid indexes will be between 0 and list's length minus one:
(list-ref lst 0)
=> 1
(list-ref lst (- (length lst) 1))
=> 5
However: in Scheme it's unusual to write code that depends on the index of an element in a list, that's how you'd think about a solution in a C-like language using an array, but Scheme lists are different, and normally you traverse a list using recursion - forget about indexes!
If I understand correctly, the list is a list of lists such as:
(define lst '((a b c)
(1 "hi")
((lambda(x)(+ 5 x)) 42 'a (14 7 12))
("hello" " world")))
A quick and dirty way to get the size of an element would be
(define third-of-a-size (third (map length lst)))
I want to write a function in lisp that reverses all elements from the list using map functions but I don't have any idea how to start this.. I think I have to use the built in reverse function somehow..
For example if I have the list (1 2 3 (4 5 6 (7 8 9))) I would get (((9 8 7) 6 5 4) 3 2 1)
or if I had the list(1 2 3 (4 5) (6 7)) I would get ((7 6) (5 4) 3 2 1) ..
Any help is appreciated!
Just a quick answer, not sure about efficiency/elegancy:
(defun reverse-deeply (list)
(mapcar #'(lambda (li)
(cond
((consp li) (reverse-deeply li))
(t li)))
(reverse list)))
(defun reverse-list (list)
(let ((result nil))
(dolist (e list result)
(push e result))))
Here is a version that works for me in Common-Lisp.
(defun reverse-list (list)
(if (atom list)
list ;; Not actually a list, return the atom
(reverse (mapcar #'reverse-list list)))
;; Testing it out
(reverse-list '((1 2 3) (4 5 (3 6))))
Output:
(((6 3) 5 4) (3 2 1))
Mapcar is a function that takes another function as its first parameter and a list as its second parameter. It then calls that function on each element of the list. It returns a list of all of the answers. So after I use 'mapcar' to reverse all of the sublists, I call 'reverse' again to reverse the bigger list.
The function that it calls on each sublist is 'reverse-list'. This checks if the list is an atom. If it is, then it returns itself. If it's a list, then it calls mapcar again on each element in the list, then reverses the result.