Comparing strings from the back?

[Dan Goodman]

On 10/09/2012 18:07, Dan Goodman wrote:
> On 04/09/2012 03:54, Roy Smith wrote:
>> Let's assume you're testing two strings for equality.  You've already
>> done the obvious quick tests (i.e they're the same length), and you're
>> down to the O(n) part of comparing every character.
>>
>> I'm wondering if it might be faster to start at the ends of the strings
>> instead of at the beginning?  If the strings are indeed equal, it's the
>> same amount of work starting from either end.  But, if it turns out that
>> for real-life situations, the ends of strings have more entropy than the
>> beginnings, the odds are you'll discover that they're unequal quicker by
>> starting at the end.
>
>  From the rest of the thread, it looks like in most situations it won't
> make much difference as typically very few characters need to be
> compared if they are unequal.
>
> However, if you were in a situation with many strings which were almost
> equal, the most general way to improve the situation might be store a
> hash of the string along with the string, i.e. store (hash(x), x) and
> then compare equality of this tuple. Almost all of the time, if the
> strings are unequal the hash will be unequal. Or, as someone else
> suggested, use interned versions of the strings. This is basically the
> same solution but even better. In this case, your startup costs will be
> higher (creating the strings) but your comparisons will always be instant.

Just had another thought about this. Although it's unlikely to be 
necessary in practice since (a) it's rarely necessary at all, and (b) 
when it is, hashing and optionally interning seems like the better 
approach, I had another idea that would be more general. Rather than 
starting from the beginning or the end, why not do something like: check 
the first and last character, then the len/2 character, then the len/4, 
then 3*len/4, then len/8, 3*len/8, etc. You'd need to be a bit clever 
about making sure you hit every character but I'm sure someone's already 
got an efficient algorithm for this. You could probably even make this 
cache efficient by working on cache line length blocks. Almost certainly 
entirely unnecessary, but I like the original question and it's a nice 
theoretical problem.

Dan