But MD5 is order-dependent, isn't it? Otherwise, MD5 would say that
documents containing "hello, world" and "world, hello" are the same.
- Gordon
From: flexcoders@yahoogroups.com [mailto:[EMAIL PROTECTED] On
Behalf Of andrii_olefirenko
Sent: Sunday, February 24,
I agree that it is O(1) in the sense that you describe, because the O(n)
work has been amortized to be imperceptible.
Do you actually know of a good hash function for this case?
- Gordon
From: flexcoders@yahoogroups.com [mailto:[EMAIL PROTECTED] On
Behalf Of a
True. However, andrii's solution does amortize the cost of the
comparison across all array modifications, so you don't incur it all at
once when doing the comparison. Depending on the actual requirements of
the situation (and assuming you control array modifications, and are not
just trying to comp
> Of course, modifying an array would become more expensive operation,
but it's still O(1) operation
The overhead when modifying each element is O(1), so the total overhead
you've incurred is O(n).
Gordon Smith
Adobe Flex SDK Team
From: flexcoders@yahoogroups.
A more specific use case may result in a more elegant solution
On Sun, Feb 24, 2008 at 10:18 AM, andrii_olefirenko <[EMAIL PROTECTED]>
wrote:
> The trick is to keep hash value of all elements of the array.
> Prerequisites are following:
> 1. Hash function H(index, object) should return unique v
Can you give more detail? I don't believe there is any O(1) algorithm
for this. O(1) means that comparing two 100,000-element arrays would
take the same time as comparing two 100-element arrays.
Gordon Smith
Adobe Flex SDK Team
From: flexcoders@yahoogroups.com [
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