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From Matthew Foley <ma...@yahoo-inc.com>
Subject Re: Defining Hadoop Compatibility -revisiting-
Date Mon, 16 May 2011 21:17:57 GMT
It's important to distinguish between the name "Hadoop", which is protected by trademark law,
and the Hadoop implementation, which is licensed as opensource under copyright law.

The term "derivative work" is, I believe, only relevant under copyright law, not trademark
(N.B., I'm not a lawyer -- and this email is my opinion, not my employer's.)  Since the Apache
explicitly allows derivative works, I don't think it's a useful term for this discussion.

However, the ASF, and by delegation the Hadoop PMC, has a lot of control over the name,
and how we allow it to be used, under trademark law.  But to keep our rights under that
law, we have to enforce the trademark consistently.  So it's good that we're having this discussion,
and it's important to reach a conclusion, document it, and enforce it consistently.

There are a lot of subtleties; for instance, if I recall correctly from my days with Adobe
PostScript(R), someone who has not licensed a trademark "X" can still claim "compatible with
as long as they ALSO make clear that the product is NOT, itself, an "X".  But you really need
a lawyer to get into that stuff.


On May 16, 2011, at 5:00 AM, Segel, Mike wrote:

But Cloudera's release is a bit murky.

The math example is a bit flawed...

X represents the set of stable releases.
Y represents the set of available patches.
C represents the set of Cloudera releases.

So if C contains a release X(n) plus a set of patches that is contained in Y,
Then does it not have the right to be considered Apache Hadoop?
It's my understanding is that any enhancement to Hadoop is made available to Apache and will
eventually make it into a later release...

So while it may not be 'official' release X(z), all of it's components are in Apache.
(note: I'm talking about the core components and not Cloudera's additional toolsets that encompass

Cloudera is clearly a derivative work.
And IMHO is the only one which can say ... 'Includes Apache Hadoop'.

That doesn't mean that others can't, depending on how they implemented their changes.
Based on EMC marketing material, they've done a rip and replace of HDFS.
So it wouldn't be a superset since it doesn't contain a complete subset, but contains code
that implements the API... So they can't say 'Includes Apache Hadoop',but they can say it's
a derivative work based on Apache Hadoop and then go on to show how and why, in their opinion
their product is better.(that's marketing for you...)

Clearly there are others out there...
Hadoop on Cassandra as an example...

Fragmentation of Hadoop will occur. It's inevitable. Too much money is on the table...

But because Apache's licensing is so open, Apache will have a hard time controlling derivative
I believe that Steve is incorrect in his assertion concerning potential loss of any patent
protection. Again Apache's licensing is very open and as long as they follow Apache's Ts and
Cs, they are covered.

Note: because I am sending this from my email address at my client, I am obliged to say that
this email is my opinion and does not reflect on the opinion of my client...
(you know the rest....)

Sent from a remote device. Please excuse any typos...

Mike Segel

On May 16, 2011, at 6:02 AM, "Steve Loughran" <stevel@apache.org<mailto:stevel@apache.org>>

On 13/05/11 23:57, Allen Wittenauer wrote:

On May 13, 2011, at 3:53 PM, Ted Dunning wrote:

But "distribution Z includes X" kind of implies the existence of some such
that X != Y, Y != empty-set and X+Y = Z, at least in common usage.

Isn't that the same as a non-trunk change?

So doesn't this mean that your question reduces to the question of what
happens when non-Apache changes are made to an Apache release?  And isn't
that the definition of a derived work?

  Yup. Which is why I doubt *any* commercial entity can claim "includes Apache Hadoop" (including

but they can claim it is a derivative work, which CDH clearly is,
(Though if we were to come up with a formal declaration of what a
derivative work is, we'd have to handle the fact that it is a superset.
Even worse, you may realise a release is the ordered application of a
sequence of patches, and if the patches are applied in a different order
you may end up with a different body of source code...)

Something that implements the APIs may not be a derivative work,
depending on how much of the original code is in there. You could look
at the base classes and interfaces and produce a clean room
implementation (relying on the notion that interfaces are a list of
facts and not copyrightable in the US), but whoever does that may
encounter the issue that Google's donation of the right to use their MR
patent may not apply to such implementations.

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