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I want to test if a list is even, in . Like (evenatom '((h i) (j k) l (m n o)) should reply #t because it has 4 elements.

From Google, I found how to check for odd:

(define (oddatom lst)
  (cond
    ((null? lst)       #f)
    ((not (pair? lst)) #t)
    (else (not (eq? (oddatom (car lst)) (oddatom (cdr lst)))))))

to make it even, would I just swap the car with a cdr and cdr with car?

I'm new to Scheme and just trying to get the basics.

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Harold B
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    You say that "`(evenatom '((h i) (j k) l (m n o))` should reply #t because it has 4 elements." However, the `oddatom` that you cite would return false because the list contains 8 atoms, not because it has four elements. Do you want `(evenatom '((h i) (j k) l (m n o))` to return true because it has four elements, or because it has 8 atoms? – Joshua Taylor Apr 04 '14 at 15:16

4 Answers4

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No, swapping the car and cdr won't work. But you can swap the #f and #t.

Also, while the list you gave has 4 elements, what the function does is actually traverse into sublists and count the atoms, so you're really looking at 8 atoms.

C. K. Young
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  • Well, you _would_ know your code better than I would. :-) (For my version, it really is as simple as swapping the `#f` with `#t`.) – C. K. Young Apr 04 '14 at 18:36
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You found odd atom using 'Google' and need even atom. How about:

(define (evenatom obj) (not (oddatom obj)))

or, adding some sophistication,

(define (complement pred)
  (lambda (obj) (not (pred obj))))

and then

(define evenatom (complement oddatom))
GoZoner
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you are mixing a procedure to check if a list has even numbers of elements (not restricted to atoms) and a procedure that checks if there are an even number of atomic elements in the list structure. Example: ((a b) (c d e) f) has an odd number of elements (3) but an even number (6) of atoms.

If you had some marbles, how would you determine if you had an odd or even number of marbles? you could just count them as normal and check the end sum for evenness or count 1,0,1,0 or odd,even,odd,even so that you really didn't know how many marbles I had in the end, only if it's odd or even. Lets do both:

(define (even-elements-by-count x)
  (even? (length x)))

(define (even-elements-by-boolean x)
  (let loop ((x x)(even #t))
     (if (null? x)
         even
         (loop (cdr x) (not even)))))

now imagine that you had some cups in addition and that they had marbles to and you wondered the same. You'd need to count the elements on the floor and the elements in cups and perhaps there was a cup in a cup with elements as well. For this you should look at How to count atoms in a list structure and use the first approach or modify one of them to update evenness instead of counting.

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Sylwester
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The equivalent of the procedure you link to, for an even number of atoms, is

(define (evenatom lst)
  (cond
    ((null? lst)       #t)
    ((not (pair? lst)) #f)
    (else (eq? (evenatom (car lst)) (evenatom (cdr lst))))))

You need to swap #t and #f, as well as leave out the not clause of the last line.

uselpa
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  • Thank you for the reply what does actually pair do? – Harold B Apr 05 '14 at 00:25
  • It checks for a cons cell. See http://groups.csail.mit.edu/mac/ftpdir/scheme-7.5/7.5.17/doc-html/scheme_8.html – uselpa Apr 05 '14 at 06:43
  • Hi, new dilema What if I did a swap with a bogus argument (an atom, not a list): (swap 'a) How would I tell it 'USAGE': followed by an indication of the correct invocation. For example :"|USAGE:(thirds[list])"? ((NOT(list? 1st)) (newline) "USAGE: (swap[list])") is not working out to great – Harold B Apr 09 '14 at 04:34