cyclic structures

puzzle

• complete
  (define x …)
  
• so that
  (car x) ==> 1
  
  (car (cdr x)) ==> 1
  (car (cdr (cdr x))) ==> 1
  (car (cdr (cdr (cdr x)))) ==> 1
  … , forever
• and (eq? x (cdr x)) ==> #t

solution

• code
  (define x(let ((c (cons 1 nil))) (set-cdr! c c) c))
  
• property
  (eq? c (cdr c)) ==> #t
  

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