Detecting repeated subsequences of identical items

[Oscar Benjamin]

On 21 April 2016 at 15:12, Chris Angelico <rosuav at gmail.com> wrote:
> On Fri, Apr 22, 2016 at 12:01 AM, Oscar Benjamin
> <oscar.j.benjamin at gmail.com> wrote:
>> In the recursive stack overflow case what you'll usually have is
>>
>> 1) A few frames leading up to the start of recursion
>> 2) A long repetitive sequence of frames
>> 3) A few frames at the end showing how the exception was ultimately triggered.
>>
>> You just need to find the cycle that makes that big long sequence.
>
> If the stack got overflowed, there won't usually be a part 3, as part
> 2 is the bit that hits sys.recursionlimit (unless increasing the
> recursion limit by a finite number would solve the problem). For other
> exceptions, yes, this is what you'd see.

If you have:

def f(x):
    return g(x+1)

def g(x):
    x = h(x)  # <-- stack can overflow inside here
    return f(x+1)

# etc.

So you have a long sequence that goes f, g, f, g but at the end the
stack can overflow while (not recursively) calling h leaving a small
non-cyclic part at the end.

--
Oscar