This procedure is supposed to replace the values in vec1 according to the procedure given. So, if the procedure was +, then it would replace each value in vec1 with the sum of each element. For example:
~ (define v (vector 1 2 3 4 5 6))
~ (vector-join v + v)
~ v
#(2 4 6 8 10 12)
I know there's an issue with the recursion but I don't know how to fix it. I've only learned how to do recursion with cons which I don't think is the right thing to do in this kind of problem.
Here's my code so far:
(define v (vector 1 2 3 4 5 6))
(define (vector-join vec1 pre vec2)
(define (help v1 proc v2 i)
(if (null? v1) v1
(if (null? v2) v1
(if (>= i (vector-length v1)) v1
(cons (vector-set! v1 i (proc (vector-ref v1 i) (vector-ref v2 i)))
(help v1 proc v2 (add1 i)))))))
(help vec1 pre vec2 0))
When I input this:
(vector-join v + v)
It returns this:
(#<void> #<void> #<void> #<void> #<void> #<void> . #(2 4 6 8 10 12))
The last part is the right answer, but I don't know why the #voids are coming up. Any help?
Notice that cons, null? have nothing to do with this answer, here we're not dealing with lists. Also, you're modifying one of the vectors received as parameter, that's not the best idea but let's ignore that fact for the time being. As usual, I'll give you the general structure of the solution, so you can work out the details:
(define (vector-join vec1 pre vec2)
(define (help v1 proc v2 i)
(cond (<???> v1) ; what's the exit condition?
(else
<???> ; set the current value at position `i`
<???>))) ; advance the recursion, no consing here!
(help vec1 pre vec2 0))
The exit condition will be "when i is outside of the vector", and in the normal case we take care of updating the current position in the vector (denoted by i) before moving to the next position. The recursion will advance over the indexes; given that we're modifying v1, that's what we'll return at the end of the procedure. It works as expected:
(define v (vector 1 2 3 4 5 6))
(vector-join v + v)
=> '#(2 4 6 8 10 12)
This procedure is written in a way that reminds of a solution in an imperative language. The unusual part (for an Scheme program, that is) is the fact that in the second <???> you're mutating a data structure (a vector in this case) but you do so just for the effect, not for the value - the vector-set! operation doesn't return an useful value, that explains all the #<void> that were showing up in the code in the question.
Related
I am new to scheme and am trying to gain an understanding of how the following piece of code is able to recursively multiply two numbers provided (how is the function stack occurring here):
;Recursive Arithmetic
(define increment
(lambda (x)
(+ x 1)))
(define decrement
(lambda (x)
(- x 1)))
(define recursive-add
(lambda (x y)
(if (zero? y)
x
(recursive-add (increment x) (decrement y)))))
"Define Multiplication Recursively"
(define recursive-mult
(lambda (x y)
(if (zero? y)
0
(recursive-add x (recursive-mult x (decrement y)))))) ;else
(recursive-mult 9 5)
My first doubt is what part of the else section is executed first?
In the else section, if (decrement y) is executed first, then y is first decremented by 1, then the(recursive-mult x section runs next by which function calls itself, so the section (recursive-add x never gets called in this same line of code.
Now, if in the else section, the recursive-add x is executed first, x gets added first, then the recursive-mult x runs which runs the the whole function again and the (decrement y) is never reached.
Now I know either of my two assumptions above, or maybe both are wrong. Would someone be able to point me in the right direction please? I apologize if my question displays a high level of inadequacy in this subject but I really want to learn scheme properly. Thank you.
This is an answer in Common Lisp, in which I am more familiar, but the approach is the same. Here is a translation of your code:
(defpackage :so (:use :cl))
(in-package :so)
(defun add (x y)
(if (zerop y)
x
(add (1+ x) (1- y))))
(defun mul (x y)
(if (zerop y)
0
(add x (mul x (1- y)))))
The increment and decrement functions are not defined, I'm using 1+ and 1- which are equivalent.
In the else section, if (decrement y) is executed first, then y is first decremented by 1, then the(recursive-mult x section runs next by which function calls itself, so the section (recursive-add x never gets called in this same line of code.
You can use the substitution model to evaluate recursive functions.
(add (add 1 3) 1)
Let's evaluate first the first argument of the top-most and:
(add (add 2 2) 1)
(add (add 3 1) 1)
(add (add 4 0) 1)
(add 4 1)
Here the call eventually returned and the first argument is 4. Once the execution is resumed, the rest of the code for (add (add 1 3) 1) is executed, which is equivalent to:
(add 5 0)
And finally:
5
Each time you invoke the function a new frame is pushed on the call stack, which is what is shown in a trace:
(trace mul add)
The macro above in Common Lisp makes it so the mul and add functions are traced.
Now when I run a program, a trace is printed when they are entered and exited.
Let's try with small values:
(mul 2 3)
The trace is printed as follows (don't mind the SO:: prefix, this is part of the fully-qualified name of symbols):
0: (SO::MUL 2 3)
1: (SO::MUL 2 2)
2: (SO::MUL 2 1)
3: (SO::MUL 2 0)
3: MUL returned 0
3: (SO::ADD 2 0)
3: ADD returned 2
2: MUL returned 2
2: (SO::ADD 2 2)
3: (SO::ADD 3 1)
4: (SO::ADD 4 0)
4: ADD returned 4
3: ADD returned 4
2: ADD returned 4
1: MUL returned 4
1: (SO::ADD 2 4)
2: (SO::ADD 3 3)
3: (SO::ADD 4 2)
4: (SO::ADD 5 1)
5: (SO::ADD 6 0)
5: ADD returned 6
4: ADD returned 6
3: ADD returned 6
2: ADD returned 6
1: ADD returned 6
0: MUL returned 6
The difference with Scheme is that Scheme does not define the order by which function arguments are evaluated (it is left unspecified), so maybe your code would not exactly behave as above, but it should still compute the same answer (because there are no side effects).
Scheme has eager evaluation so if you have
(op1 expr1 (op2 1 (op3 expr2 expr3)))
Then op1 form will depend on expr1 and the form with op2 to be complete before op1 can be applied. If we have an implementation that
does left to right evaluation it will do it like this:
(let ((vexpr1 expr1)
(vexpr2 expr2)
(vexpr3 expr3))
(let ((vop3 (op3 vexpr2 vexpr3)))
(let ((vop2 (op2 1 vop3)))
(op1 vexpr1 vop2))))
So to answer your question the order in the same left to right Scheme will be:
(let ((vdy (decrement y)))
(let ((vrm (recursive-mult x vdy)))
(recursive-add x vrm)))
Same in CPS:
(decrement& y
(lambda (vdy)
(recursive-mult& x
vy
(lambda (vrm)
(recursive-add& x vrm continuation)))))
So in practice the application for the first round doesn't happen before the whole recursive-mult for the smaller expression happens first. Thus (recursive-mult 3 3) turns into
(recursive-add 3 (recursive-add 3 (recursive-add 3 0)))
And as you can see the last one is being done first, then the second one and the last addition to be performed is the one of the first round before recursion.
According to the book, this is what the function definition is,
The function scramble takes a non-empty tuple in which no argument is greater than its own index and returns a tuple of same length. Each number in the argument is treated as a backward index from its own position to a point earlier in tuple. The result at each position is obtained by counting backward from the current position according to this index.
And these are some examples,
; Examples of scramble
(scramble '(1 1 1 3 4 2 1 1 9 2)) ; '(1 1 1 1 1 4 1 1 1 9)
(scramble '(1 2 3 4 5 6 7 8 9)) ; '(1 1 1 1 1 1 1 1 1)
(scramble '(1 2 3 1 2 3 4 1 8 2 10)) ; '(1 1 1 1 1 1 1 1 2 8 2)
Here is the implementation,
(define pick
(λ (i lat)
(cond
((eq? i 1) (car lat))
(else (pick (sub1 i)
(cdr lat))))))
(define scramble-b
(lambda (tup rev-pre)
(cond
((null? tup) '())
(else
(cons (pick (car tup) (cons (car tup) rev-pre))
(scramble-b (cdr tup)
(cons (car tup) rev-pre)))))))
(define scramble
(lambda (tup)
(scramble-b tup '())))
This is a case where using a very minimal version of the language means that the code is verbose enough that understanding the algorithm is not perhaps easy.
One way of dealing with this problem is to write the program in a much richer language, and then work out how the algorithm, which is now obvious, is implemented in the minimal version. Let's pick Racket as the rich language.
Racket has a function (as does Scheme) called list-ref: (list-ref l i) returns the ith element of l, zero-based.
It also has a nice notion of 'sequences' which are pretty much 'things you can iterate over' and a bunch of constructs whose names begin with for for iterating over sequences. There are two functions which make sequences we care about:
in-naturals makes an infinite sequence of the natural numbers, which by default starts from 0, but (in-naturals n) starts from n.
in-list makes a sequence from a list (a list is already a sequence in fact, but in-list makes things clearer and there are rumours also faster).
And the iteration construct we care about is for/list which iterates over some sequences and collects the result from its body into a list.
Given these, then the algorithm is almost trivial: we want to iterate along the list, keeping track of the current index and then do the appropriate subtraction to pick a value further back along the list. The only non-trivial bit is dealing with zero- vs one-based indexing.
(define (scramble l)
(for/list ([index (in-naturals)]
[element (in-list l)])
(list-ref l (+ (- index element) 1))))
And in fact if we cause in-naturals to count from 1 we can avoid the awkward adding-1:
(define (scramble l)
(for/list ([index (in-naturals 1)]
(element (in-list l)))
(list-ref l (- index element))))
Now looking at this code, even if you don't know Racket, the algorithm is very clear, and you can check it gives the answers in the book:
> (scramble '(1 1 1 3 4 2 1 1 9 2))
'(1 1 1 1 1 4 1 1 1 9)
Now it remains to work out how the code in the book implements the same algorithm. That's fiddly, but once you know what the algorithm is it should be straightforward.
If the verbal description looks vague and hard to follow, we can try following the code itself, turning it into a more visual pseudocode as we go:
pick i [x, ...ys] =
case i {
1 --> x ;
pick (i-1) ys }
==>
pick i xs = nth1 i xs
(* 1 <= i <= |xs| *)
scramble xs =
scramble2 xs []
scramble2 xs revPre =
case xs {
[] --> [] ;
[x, ...ys] -->
[ pick x [x, ...revPre],
...scramble2 ys
[x, ...revPre]] }
Thus,
scramble [x,y,z,w, ...]
=
[ nth1 x [x] (*x=1..1*)
, nth1 y [y,x] (*y=1..2*)
, nth1 z [z,y,x] (*z=1..3*)
, nth1 w [w,z,y,x] (*w=1..4*)
, ... ]
Thus each element in the input list is used as an index into the reversed prefix of that list, up to and including that element. In other words, an index into the prefix while counting backwards, i.e. from the element to the left, i.e. towards the list's start.
So we have now visualized what the code is doing, and have also discovered requirements for its input list's elements.
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))
A classic enumeration using unfold:
(unfold-left (lambda (x)
(if (> x 10)
(#;no values)
(+ x 1)))
#;from 0)
===> (0 1 2 3 4 5 6 7 8 9 10))
if limiting the scope is not needed is there any way to just write x without the lambda?
unfold is implemented like this:
(define (unfold p f g seed (tail-gen (λ (_) '())))
(let recur ((seed seed))
(if (p seed)
(tail-gen seed)
(cons (f seed)
(recur (g seed))))))
As you can see p, f, g, and tail-gen are all procedures since they get surrounded by parentheses in the implementation. If they are not procedures you will get an application: not a procedure error.
You are using unfold wrong. you need a procedure that takes the current value and return wether or not you are finished. Second is a procedure that takes the seed and return what value to collect and the third is a procedure to create the next seed. The optional tail-gen takes the seed and creates the tail where the empty list will be used if not provided. Here is how you make a list from 0 to 10:
#lang racket
(require srfi/1)
(require srfi/26)
(unfold (cut > <> 10) identity add1 0)
; ==> (0 1 2 3 4 5 6 7 8 9 10)
And of course, (range 11) gives the same answer.
I want to take an interval of a vector in Scheme. I know there is a procedure named vector->values, but seems like it returns each element separately, while I want to get the result as a vector. How can I achieve this?
> (vector->values (vector 1 2 3 4 5) 0 3)
1
2
3
while I need:
#(1 2 3)
If you're using PLT, you have a few easy ways to get this:
(define (subvector v start end)
(list->vector (for/list ([i (in-vector v start end)]) i)))
(define (subvector v start end)
(build-vector (- end start) (lambda (i) (vector-ref v (+ i start)))))
(define (subvector v start end)
(define new (make-vector (- end start)))
(vector-copy! new 0 v start end)
new)
The last one is probably going to be the fastest. The reason that there is no such operation that is built-in is that people usually don't do that. When you're dealing with vectors in Scheme, you're usually doing so because you want to optimize something so returning a vector and a range instead of allocating a new one is more common.
(And if you think that this is useful, please suggest it on the PLT mailing list.)
The Scheme R6RS standard has make-vector, vector-ref, vector-set! and vector-length. With that you can write your own function subvector, which does not seem to be part of R6RS (!). Some Scheme implementation have something like subvector already.
You can also switch to Common Lisp, which provides the function SUBSEQ in the standard.
Here is a portable R6RS version using SRFI 43:
#!r6rs
(import (rnrs base)
(prefix (srfi :43) srfi/43:))
(srfi/43:vector-copy (vector 1 2 3 4 5) 0 3)
#lang scheme
(define (my-vector-value v l h c)
(if (and (>= c l) (< c h))
(cons (first v) (my-vector-value (rest v) l h (add1 c)))
empty))
(list->vector (my-vector-value (vector->list (vector 1 2 3 4 5)) 0 3 0))
Ghetto? Yes, very. But it only took two minutes to write and gets the job done.
(I find it's generally easier to play with lists in Scheme)
you want subvector:
(subvector (vector 1 2 3 4 5) 0 3)