# palindrome iteration

Matteo Landi landimatte at gmail.com
Sun Aug 29 12:45:54 CEST 2010

```I thought they reached you. Here they are again:

def palindrome(str, i=0, j=-1):
try:
if str[i] == str[j]:
return palindrome(str, i + 1, j - 1)
return False
except IndexError:
return True

def palindrome(str, i=0, j=-1):
try:
while True:
if str[i] != str[j]:
return False
i, j = i + 1, j - 1
return True
except IndexError:
return True

On Sun, Aug 29, 2010 at 12:36 PM, Arnaud Delobelle
> Matteo Landi <landimatte at gmail.com> writes:
>
>> Well, I tried the also the solution posted above (recursive w/o
>> slicing and iterative), and I discovered they were the slowest..
>>
>> is_palindrome_recursive 2.68151649808
>> is_palindrome_slice 0.44510699381
>> is_palindrome_list 1.93861944217
>> is_palindrome_reversed 3.28969831976
>> is_palindrome_recursive_no_slicing 6.78929775328
>> is_palindrome_iterative 4.88826141315
>
> What are the last two functions?
>
> I suggest another:
>
> def is_palindrome(s):
>    return all(map(str.__eq__, s, reversed(s)))
>
> :)
>
>> Nothing to say about the iterative function, but the benchmark of the
>> recursive was unexpected, at least for my point of view: do you think
>> it is due to the try/except overhead?
>>
>> On Sun, Aug 29, 2010 at 8:53 AM, Josh English
>> <joshua.r.english at gmail.com> wrote:
>>> This whole conversation got interesting, so I thought I'd run some
>>> speed tests:
>>>
>>> The code:
>>> from timeit import Timer
>>>
>>> def is_palindrome_recursive(s):
>>>    if len(s) <= 1:
>>>        return True
>>>    if s[0] != s[-1]:
>>>        return False
>>>    else:
>>>        return is_palindrome(s[1:-1])
>
> This should be return is_palindrome_recursive(s[1:-1]).  If this is
> copy-pasted, then you may call a different is_palindrome function and
> invalidate the timings!
>
> [...]
>
> --
> Arnaud
> --
> http://mail.python.org/mailman/listinfo/python-list
>

--
Matteo Landi
http://www.matteolandi.net/

```