Re: Comparing two arrays

Here’s another way to do it: scan the arrays to convert them to a vector using some form of “run-length encoding.” For instance, a list containing the position of the next '1' and the number of '1's that occur. Now for the magic: use a Digest algorithm of some sort ... could be CRC, SHA1, could be anything ... to convert that vector into a single value that can be used as a hash.

Actually, if you use a truly-decent message digest algorithm, like SHA1, you probably won’t have to do anything else, because one and only one vector will map to the same hash value. You have to preprocess each array one time to compute its message-digest hash, but from there on out you might not have to examine the array contents at all. A comparison of the digests ought to be sufficient. Aside from the overhead of passing through each array one time to calculate its digest, the overhead of laborious comparison ... disappears.

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Re^2: Comparing two arrays by roboticus (Chancellor) on Dec 16, 2013 at 13:43 UTC
sundialsvc4: If you're looking for duplicate vectors, digesting can greatly reduce the number of comparisons, but it won't let you escape them altogether: http://en.wikipedia.org/wiki/Pigeonhole_principle. Update: s/While/If you're looking for duplicate vectors/ and changed conjunction so the sentence still reads well. ...roboticus When your only tool is a hammer, all problems look like your thumb.	[reply]
Re^3: Comparing two arrays by BrowserUk (Patriarch) on Dec 16, 2013 at 13:53 UTC
While digesting can greatly reduce the number of comparisons That would only be true if the OP was looking for exact matches. He isn't. He's looking for the best matches, where 'best' is defined in terms of the number of set bits in matching positions. No hashing, digesting nor sorting approach to this problem is possible. Every X must be fully compared against every Y. With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday' Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply]
Re^4: Comparing two arrays by roboticus (Chancellor) on Dec 16, 2013 at 16:19 UTC
BrowserUk: That's certainly true, for the original problem. I was intending to refute the statement "because one and only one vector will map to the same hash value", but I didn't take the rest of the thread into context. (I corrected the node accordingly.) However, you needn't compare each X fully against each Y either, either. Just like your Bloom filter project a while ago, there may be ways to transform the problem so we don't have to explicitly compare vectors against each other. I've been working a bit on one, but I haven't posted it because the performance is currently worse than direct comparisons, and most of the changes I make to it slow it down even further. ...roboticus When your only tool is a hammer, all problems look like your thumb.	[reply]
Re^5: Comparing two arrays by BrowserUk (Patriarch) on Dec 16, 2013 at 17:21 UTC
Re^6: Comparing two arrays by roboticus (Chancellor) on Dec 16, 2013 at 21:19 UTC
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Re^5: Comparing two arrays by BrowserUk (Patriarch) on Dec 16, 2013 at 16:30 UTC
Re^2: Comparing two arrays by choroba (Cardinal) on Dec 16, 2013 at 13:21 UTC
one and only one vector will map to the same hash value Do you mean there are no hash collisions for SHA1? لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply]
Re^2: Comparing two arrays by BrowserUk (Patriarch) on Dec 16, 2013 at 15:23 UTC
the overhead of laborious comparison ... disappears. Total garbage. With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday' Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply]
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Re^2: Comparing two arrays by GotToBTru (Prior) on Dec 31, 2013 at 21:38 UTC
This method was considered and discarded in the OP: I was thinking about computing a hash key (a checksum of come sort) but then i would be able to identify only identical (x,y) pairs.	[reply]


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