I want to write an equal function that tell that any two numbers are equal, and that a symbol is equal only to itself. Also two lists should be equal if all their elements are recursively equal. I wrote a small equal function below, but it doesn't exactly achieve what I want.
(defun equal (x y)
(if (atom x)
(eql x y)
(if (atom y)
'nil
(if (equal (car x) (car y))
(equal (cdr x) (cdr y))
'nil))))
Some of the tests cases that my function should satisfy are:
(my-equal 2.2 3) -> T
(my-equal 2.2 'a) -> NIL
(my-equal '(b 4 (c d (2.2)) e) '(b 2 (c d (0)) e)) -> T
(my-equal '(b 4 (c d (2.2)) e) '(b 2 (5 d (2.2)) e)) -> NIL
Please note that I want to only use the basic Lisp operations, the existing equal functions, reverse, append, null, nil, listp, atom, cdr, car, etc. No stuff like map, or other more sophisticated functions.
Here is a possible solution (assuming the name of the new function is my-equal):
(defun my-equal (x y)
(cond ((numberp x) (numberp y))
((symbolp x) (eq x y))
((and (consp x) (consp y))
(and (my-equal (car x) (car y)) (my-equal (cdr x) (cdr y))))
(t nil)))
(my-equal 2.2 3) ;; => T
(my-equal 2.2 'a) ;; => NIL
(my-equal '(b 4 (c d (2.2)) e) '(b 2 (c d (0)) e)) ;; => T
(my-equal '(b 4 (c d (2.2)) e) '(b 2 (5 d (2.2)) e)) ;; => NIL
The last case could be replaced by something like (t (primitive-equal-function x y)) (or other cases) if you care for some test of equality when x and y are strings, characters, etc.
Related
I'm trying to create a LISP function that creates from a list all possible pairs.
Example of what I'm trying to achieve: (a b c d) --> ((a b) (a c) (a d) (b c) (b d) (c d))
Any advice please? I'm not sure how to approach this problem
Here is a simple solution:
(defun make-couples (x l)
"makes a list of couples whose first element is x and the second is each element of l in turn"
(loop for y in l collect (list x y)))
(defun all-pairs (l)
"makes a list of all the possible pairs of elements of list l"
(loop for (x . y) on l nconc (make-couples x y)))
A recursive solution is:
(defun make-couples (x l)
"makes a list of couples whose first element is x and the second is each element of l in turn"
(if (null l)
nil
(cons (cons x (first l)) (make-couples x (rest l)))))
(defun all-pairs (l)
"makes a list of all the possible pairs of elements of list l"
(if (null l)
nil
(nconc (make-couples (first l) (rest l))
(all-pairs (rest l)))))
Here is a version (this is quite closely related to Gwang-Jin Kim's) which has two nice properties:
it is tail recursive;
it walks no list more than once;
it allocates no storage that it does not use (so there are no calls to append and so on);
it uses no destructive operations.
It does this by noticing that there's a stage in the process where you want to say 'prepend a list of pairs of this element with the elements of this list to this other list' and that this can be done without using append or anything like that.
It does return the results in 'reversed' order, which I believe is inevitable given the above constraints.
(defun all-pairs (l)
(all-pairs-loop l '()))
(defun all-pairs-loop (l results)
(if (null (rest l))
results
(all-pairs-loop (rest l)
(prepend-pairs-to (first l) (rest l) results))))
(defun prepend-pairs-to (e them results)
(if (null them)
results
(prepend-pairs-to e (rest them) (cons (list e (first them))
results))))
the simplest tail recursive variant without explicit loops / mapcar could also look like this:
(defun pairs (data)
(labels ((rec (ls a bs res)
(cond
((null ls) (nreverse res))
((null bs) (rec
(cdr ls)
(car ls)
(cdr ls)
res))
(t (rec
ls
a
(cdr bs)
(cons (cons a (car bs)) res))))))
(rec data nil nil nil)))
CL-USER> (pairs (list 1 2 3 4))
;; ((1 . 2) (1 . 3) (1 . 4) (2 . 3) (2 . 4) (3 . 4))
Tail call recursive solution:
(defun pairs (lst &key (acc '()))
(if (null (cdr lst))
(nreverse acc)
(pairs (cdr lst)
:acc (append (nreverse
(mapcar #'(lambda (el)
(list (car lst) el))
(cdr lst)))
acc))))
Both nreverses are there just for aesthetics (for a nicer looking output). They can be left out.
Try it with:
(pairs '(a b c d))
;; => ((A B) (A C) (A D) (B C) (B D) (C D))
General Combinations
(defun pair (el lst)
"Pair el with each element of lst."
(mapcar (lambda (x) (cons el x)) lst))
(defun dedup (lst &key (test #'eql))
"Deduplicate a list of lists by ignoring order
and comparing the elements by test function."
(remove-duplicates lst :test (lambda (x y) (null (set-difference x y :test test)))))
(defun comb (lst &key (k 3) (acc '()) (test #'eql))
"Return all unique k-mer combinations of the elements in lst."
(labels ((%comb (lst &key (k k) (acc '()) (test #'eql) (total lst))
(let ((total (if total total lst)))
(cond ((or (null (cdr lst)) (zerop k)) (nreverse acc))
((= k 1) (mapcar #'list lst))
(t (let* ((el (car lst))
(rst (remove-if (lambda (x) (funcall test x el)) total)))
(dedup (%comb (cdr lst)
:k k
:total total
:test test
:acc (append (pair el (comb rst :k (1- k) :test test))
acc)))))))))
(%comb lst :k k :acc acc :test test :total lst)))
The number of combinations are calculatable with the combinations formula:
(defun fac (n &key (acc 1) (stop 1))
"n!/stop!"
(if (or (= n stop) (zerop n))
acc
(fac (1- n) :acc (* acc n) :stop stop)))
(defun cnr (n r)
"Number of all r-mer combinations given n elements.
nCr with n and r given"
(/ (fac n :stop r) (fac (- n r))))
We can test and count:
(comb '(a b c d) :k 2)
;; => ((A D) (B D) (B A) (C D) (C B) (C A))
(comb '(a b c d e f) :k 3)
;; => ((B A F) (C B A) (C B F) (C A F) (D C A) (D C B)
;; => (D C F) (D B A) (D B F) (D A F) (E D A) (E D B)
;; => (E D C) (E D F) (E C A) (E C B) (E C F) (E B A)
;; => (E B F) (E A F))
(= (length (comb '(a b c d e f) :k 3)) (cnr 6 3)) ;; => T
(= (length (comb '(a b c d e f g h i) :k 6)) (cnr 9 6)) ;; => T
I want a predicate as a parameter of a function.
(DEFUN per (F L)
(cond ((F L) 'working)
(T 'anything)))
(per 'numberp 3)
as result it raises an error:
Undefined operator F in form (F L).
As explained in Technical Issues of Separation in Function Cells and Value Cells,
Common Lisp is a Lisp-2, i.e., you
need funcall:
(defun per (F L)
(if (funcall F L)
'working
'other))
(per #'numberp 3)
==> WORKING
(per #'numberp "3")
==> OTHER
See also apply.
Late to the party, but here's another example:
(defun strip-predicate (p list)
(cond ((endp list) nil)
((funcall p (first list)) (strip-predicate (rest list)))
( T (cons (first list) (strip-Predicate p (rest list))))))
This could be used on predicates such as atom or numberp:
(strip-predicate 'numberp '(a 1 b 2 c 3 d))
(a b c d)
or:
(strip-predicate 'atom '(a (a b) b c d))
((a b))
I am confused about the difference between '(()) and (cons null null) in scheme.
The code below show that b and c are completely the same thing.
(define (dup2 x)
(let ((d '(())))
(set-car! d (car x))
(set-cdr! d (cdr x))
d))
(define a '(1 2))
(define b (dup2 a))
(define c (dup2 a))
(set-car! b 2)
> c ;; --> (2 2)
However, when I used dup instead of dup2:
(define (dup x)
(let ((d (cons null null)))
(set-car! d (car x))
(set-cdr! d (cdr x))
d))
(define a '(1 2))
(define b (dup a))
(define c (dup a))
(set-car! b 2)
> c ;; --> (1 2)
Variable b and c are different. I have done some experiments, but I haven't understand yet.
The value of d in the first implementation is literal data, and is modified with undefined consequences. To highlight what's happening, consider the following code:
(define (incorrect-list-null-and-x x)
(let ((l '(()))) ; a list of the form (() . ())
(set-cdr! l (cons x (cdr l))) ; (cdr l) is (), so (cons x (cdr l)) should be (x . ()) == (x), right?
; and now l should be (() . (x . ())) == (() x), right?
l))
The expected result is that (incorrect-list-null-and-x n) should return a list of the form (() n), and it does the first time, but successive calls are still accessing the same data:
(incorrect-list-null-and-x 1) ;=> (() 1)
(incorrect-list-null-and-x 2) ;=> (() 2 1)
(incorrect-list-null-and-x 3) ;=> (() 3 2 1)
(incorrect-list-null-and-x 4) ;=> (() 4 3 2 1)
The same problem manifests itself a bit differently in your dup2. Every value returned from dup2 is actually the same pair:
(let* ((x (dup2 (cons 1 2)))
(y (dup2 (cons 3 4))))
(display x)
(display y))
outputs:
(3 . 4)(3 . 4)
because the call (dup2 (cons 3 4)) modifies the same structure that was previously returned by (dup2 (cons 1 2)).
Data literals, like '(()), are meant to be read-only, and modifying it using set-car! or set-cdr! has undefined behaviour. For predictable behaviour, use the (cons '() '()) version if you want to use set-car! or set-cdr! on it.
In particular, cons creates a new cons cell, whereas a data literal usually won't.
Still, for the purposes of implementing dup, why are you using set-car! and set-cdr! anyway? Just use cons directly:
(define (dup x)
(cons (car x) (cdr x)))
In your first code snippet you use (d '(())) which ends up binding a literal to d. You then modify the literal which is generally undefined. In your second code snippet you use (d (cons null null)) which binds d to a newly created 'cons cell' which you then modify. There is no problem modifying that.
Note: you've not defined null. Perhaps you meant '()?
Please, help me with one simple exercise on the Scheme.
Write function, that return count of atoms on the each level in the
list. For example:
(a (b (c (d e (f) k 1 5) e))) –> ((1 1) (2 1) (3 2) (4 5) (5 1))
My Solution:
(define (atom? x)
(and (not (pair? x)) (not (null? x))))
(define (count L)
(cond ((null? L) 0)
((pair? (car L))
(count (cdr L)))
(else
(+ 1 (count (cdr L))))))
(define (fun L level)
(cons
(list level (count L))
(ololo L level)))
(define (ololo L level)
(if (null? L)
'()
(if (atom? (car L))
(ololo (cdr L) level)
(fun (car L) (+ level 1)))))
(fun '(a (b (c (d e (f) k 1 5) e))) 1)
It's work fine, but give not correctly answer for this list:
(a (b (c (d e (f) (k) 1 5) e)))
is:
((1 1) (2 1) (3 2) (4 4) (5 1))
But we assume that 'f' and 'k' on the one level, and answer must be:
((1 1) (2 1) (3 2) (4 4) (5 2))
How should I edit the code to make it work right?
UPD (29.10.12):
My final solution:
(define A '(a (b (c (d e (f) k 1 5) e))))
(define (atom? x)
(and (not (pair? x)) (not (null? x))))
(define (unite L res)
(if (null? L) (reverse res)
(unite (cdr L) (cons (car L) res))))
(define (count-atoms L answ)
(cond ((null? L) answ)
((pair? (car L))
(count-atoms (cdr L) answ))
(else
(count-atoms (cdr L) (+ answ 1)))))
(define (del-atoms L answ)
(cond ((null? L) answ)
((list? (car L))
(begin
(del-atoms (cdr L) (unite (car L) answ))))
(else
(del-atoms (cdr L) answ))))
(define (count L)
(define (countme L level answ)
(if (null? L) (reverse answ)
(countme (del-atoms L '()) (+ level 1) (cons (cons level (cons (count-atoms L 0) '())) answ))))
(countme L 1 '()))
(count A)
What can you say about this?
Do you know what you get if you run this?
(fun '(a (b (c (d e (f) k 1 5) e)) (a (b (c)))) 1)
You get this:
((1 1) (2 1) (3 2) (4 5) (5 1))
The whole extra nested structure that I added on the right has been ignored. Here is why...
Each recursion of your function does two things:
Count all the atoms at the current "level"
Move down the level till you find an s-expression that is a pair (well, not an atom)
Once it finds a nested pair, it calls itself on that. And so on
What happens in oLoLo when fun returns from the first nested pair? Why, it returns! It does not keep going down the list to find another.
Your function will never find more than the first list at any level. And if it did, what would you to do add the count from the first list at that level to the second? You need to think carefully about how you recur completely through a list containing multiple nested lists and about how you could preserve information at each level. There's more than one way to do it, but you haven't hit on any of them yet.
Note that depending on your implementation, the library used here may need to be imported in some other way. It might be painstakingly difficult to find the way it has to be imported and what are the exact names of the functions you want to use. Some would have it as filter and reduce-left instead. require-extension may or may not be Guile-specific, I don't really know.
(require-extension (srfi 1))
(define (count-atoms source-list)
(define (%atom? x) (not (or (pair? x) (null? x))))
(define (%count-atoms source-list level)
(if (not (null? source-list))
(cons (list level (count %atom? source-list))
(%count-atoms (reduce append '()
(filter-map
(lambda (x) (if (%atom? x) '() x))
source-list)) (1+ level))) '()))
(%count-atoms source-list 1))
And, of course, as I mentioned before, it would be best to do this with hash-tables. Doing it with lists may have some didactic effect. But I have a very strong opposition to didactic effects that make you write essentially bad code.
(define (sum-two-sqrt a b c)
(cond ((and (<= c a) (<= c b)) sqrt-sum(a b))
((and (<= a b) (<= a c)) sqrt-sum(b c))
((and (<= b a) (<= b c)) sqrt-sum(a c))
)
)
(define (sqrt-sum x y)
(+ (* x x) (*y y))
)
(define (<= x y)
(not (> x y))
(sum-two-sqrt 3 4 5)
This is my code
Please help me to fix the problem. :)
I just start studing Lisp today.
learned some C before but the two language is QUITE DIFFERENT!
This is the question
Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.
If you have better algorithm
POST IT!
Thank you :)
There's no need to define <=, it's a primitive operation. After fixing a couple of typos:
sqrt-sum: you were incorrectly invoking the procedure; the opening parenthesis must be written before the procedure name, not after.
sqrt-sum: (*y y) is incorrect, you surely meant (* y y); the space(s) after an operator matter.
This should work:
(define (sqrt-sum x y)
(+ (* x x) (* y y)))
(define (sum-two-sqrt a b c)
(cond ((and (<= c a) (<= c b)) (sqrt-sum a b))
((and (<= a b) (<= a c)) (sqrt-sum b c))
((and (<= b a) (<= b c)) (sqrt-sum a c))))
Or another alternative:
(define (sum-two-sqrt a b c)
(let ((m (min a b c)))
(cond ((= a m) (sqrt-sum b c))
((= b m) (sqrt-sum a c))
(else (sqrt-sum a b)))))
Following up on a suggestion by #J.Spiral and seconded by #River, the following Racket code reads nicely to me:
#lang racket
(define (squares-of-larger l)
(define two-larger (remove (apply min l) l))
(for/sum ([i two-larger]) (* i i)))
(squares-of-larger '(3 1 4)) ;; should be 25
Please note that this solution is entirely functional, since "remove" just returns a new list.
Also note that this isn't even in the same neighborhood with HtDP; I just wanted to express this concisely, and show off for/sum.
I didn't have Scheme interpreter here, but below seems to be shorter then other suggestions :) So it's in CL, but should look very similar in Scheme.
(defun sum-two-sqrt (a b c)
(let ((a (max a b))
(b (max (min a b) c)))
(+ (* a a) (* b b))))
In Scheme this would translate to:
(define (sum-two-sqrt a b c)
(let ((a (max a b))
(b (max (min a b) c)))
(+ (* a a) (* b b))))
the algorithm seems to work, just turn
*y
to
* y
whitespace is important here, else you're telling the interpreter you want to usethe function *y
add a close paren after
(define (<= x y) (not (> x y))
sqrt-sum(a b)
turns to
(sqrt-sum a b)
and ditto for the other sqrt-sum calls
edit: also a possibility:
(define (square a) (* a a))
(define (square-sum a b c)
(- (+ (square a)
(square b)
(square c))
(square (min a b c))))