Detecting repeated subsequences of identical items

[Chris Angelico]

On Fri, Apr 22, 2016 at 12:30 AM, Oscar Benjamin
<oscar.j.benjamin at gmail.com> wrote:
> 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.

Right, good point. Forgot about that. So this situation can be
triggered by anywhere up to <cycle length> lines up. Not as helpful an
optimization now. Oh well.

ChrisA