This is a well studied problem. Robert Floyd came up with an algorithm to solve this in the 1960s. (Yes, the same Floyd of the from the more famous Floyd-Warshall algorithm.) More infomation at: http://en.wikipedia.org/wiki/Cycle_detection

People in the software industry seem to like to use this problem as an interview question. I personally have encountered this problem in at least two separate interviews for software jobs.

The following is an implementation of Floyd's idea:

(define (contains-cycle? lst) (define (safe-cdr l) (if (pair? l) (cdr l) '())) (define (iter a b) (cond ((not (pair? a)) #f) ((not (pair? b)) #f) ((eq? a b) #t) ((eq? a (safe-cdr b)) #t) (else (iter (safe-cdr a) (safe-cdr (safe-cdr b)))))) (iter (safe-cdr lst) (safe-cdr (safe-cdr lst)))) ; Tested with mzscheme implementation of R5RS: (define x '(1 2 3 4 5 6 7 8)) (define y '(1 2 3 4 5 6 7 8)) (set-cdr! (cdddr (cddddr y)) (cdddr y)) (define z '(1)) (set-cdr! z z) x ; (1 2 3 4 5 6 7 8) y ; (1 2 3 . #0=(4 5 6 7 8 . #0#)) z ; #0=(1 . #0#) (contains-cycle? x) ; #f (contains-cycle? y) ; #t (contains-cycle? z) ; #t

This runs in linear time and constant space. Note that the space is constant because the process is iterative and the parameters and a and b are pointers -- (cdr lst) is a pointer to, not a separate copy of the rest of the lst.

The "Turtle and Hare" algorithm won't work for this exercise. If we take the set of nodes (pairs) to be our set S then f: S -> S has to be a function. However, note that both the car and the cdr of a pair can be pointing to to another pair. This means that the "points to" relation cannot be a function from S to S, because a function can only produce one element, not two. This effectively breaks the preconditions of the algorithm and we can no longer assume that it works. In fact, the Turtle and Hare algorithm does *not* work as I am going to demonstrate.

Let's assume the following structure:

+---------------+ | | | _ _ V _ | |0|_|->|1|_|-----+ | | | _ _ | | | +->|2|X| | | | | | | +--------+ | | | | _ V _ V _ _ _ |6|_|<-- |3|_|->|4|_|->|5|_|-+ ^ | | | +------------------+

The nodes are numbered for reference, the numbers themselves are irrelevant. Node elements that point to a non-node don't have an arrow and the null element is an X. There is a cycle 3 -> 4 -> 5 -> 3.

The problem starts at the first node 0 already. We can send the turtle and the hare in two directions and obviously it only makes sense to send both in the same direction. Without loss of generality we will perform a car-first search. This means we send both towards 3, at which point we move car-first towards 6, then towards 1 and from there towards 2. At this point the hare would hit a wall and conclude that that path was a dead-end.

However, in order to find the cycle the two animals would have to backtrack to the last node with a fork in the path and resume the race again from there. In other words, the algorithm needs some sort of stack, a non-constant amount of memory. I know that one could have take another order for the search and have found the cycle, but for every order I can find a new graph where the hare will eventually run into a wall and miss the cycle.

The exercise says "Write a procedure that examines a list", so one could argue about what the term "list" in this context exactly means, but considering that the chapter has been referring to forked structures as lists as well I believe we must take forks into account as well. I don't know an about an algorithm that could handle my structure or if such an algorithm even exists. Maybe this was just an oversight by the authors and I'm overthinking the issue.

Brig Says: How about just change the values in the list to 'X'. Then all you have to do is check if the current value is x or not.

Joe Replies: Your suggestion is incorrect for (list 'a 'b 'c 'X) My solution did leverage a similar idea though, and is correct given the problem description (although it does trash the original list!):

```
(define (cycle-p! list)
(let ((flag (cons 'doesnt 'matter)))
(define (f pair)
(cond ((null? pair) #f)
((eq? flag (car pair)) #t)
(else (set-car! pair flag)
(f (cdr pair)))))
(f list)))
```

Solution that doesn't trash the original list:

```
(define (cycle? list)
(define first list)
(define (rec x)
(cond ((eq? first x) #t)
((not (pair? x)) #f)
(else (rec (cdr x)))))
(rec (cdr list)))
```

David says: This solution will get stuck in an infinite loop in the first element is not in the cycle. For example

(define x (list 'a 'b)) (set-cdr! (cdr x) (cdr x)) (cycle? x) ; Does not return

Solution that will not trash the original list(modified and recovered actually):

```
(define (cycle? x)
(let ((ret false) (header '(())))
(define (rec lst)
(cond ((null? lst) (set! ret false))
((eq? (cdr lst) header) (set! ret true))
(else
(let ((rest (cdr lst)))
(set-cdr! lst header)
(rec rest)
(set-cdr! lst rest)))))
(rec x)
ret))
```

gws says: I didnt know about Floyd solution and came up with a different algorithm that is correct and constant space but it is less efficient (time-wise) and less elegant:

```
(define (cycle-const-space? x)
(define (iter x cont elem num)
(cond ((null? (cdr x)) false)
((eq? x elem) true)
(else (if (= cont num)
(iter (cdr x) 0 x (+ 1 num))
(iter (cdr x) (+ cont 1) elem num)))))
(iter x 0 nil 0))
```

atrika says:

I didn't know about tortoise and hare too and came up with an algorithm similar to the one of gws. It selects a node, than advance n steps. If it didn't find the selected node, it selects the current node and advance n+1 steps. It keeps increasing the number of step until it hits the empty list (no cycle) or a selected element (cycle composed of n+1 elements)

```
(define (is-cycle? mlist)
(define (iter current tested-for remaining-steps max-steps)
(cond ((null? current) false)
((eq? current tested-for) true)
((= remaining-steps 0) (iter (mcdr current) current (+ max-steps 1) (+ max-steps 1)))
(else (iter (mcdr current) tested-for (- remaining-steps 1) max-steps))))
(iter (mcdr mlist) mlist 1 1))
```

sam says: Another way is to first run (mystery x) on the list. This reversal happens in place per the definition on page256 in the book and hence constant space (that is, no extra space). If there is a cycle, 'x' (after reversal) points to the same pair as the original argument list, else no. If needed,one can recover original list by running (mystery x) again.

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Takes a range of size m and compares each of the nth-cdr 's with the head. If no conclusions are reached, jump to the end of the range, double m, and start over.

I came up with this algorithm myself, but it appears to be similar, if not the exact same, as Brent's algorithm, which one can read about in the Cycle detection wikipedia article.

Certainly O(1) memory, and also linear time since sum 2^(lg n) = O(n)