[SciPy-Dev] License compatibility SciPy -> MIT -> GPL

[Robert Kern]

On Sun, Aug 25, 2013 at 9:00 AM, Saullo Castro <saullogiovani at gmail.com>
wrote:
>
> Well, we've been discussion the license combatility for the cubature
module.
>
> From the links @Warren gave me:
> http://wiki.scipy.org/License_Compatibility
> http://matplotlib.org/devel/license.html#license-discussion
>
> it states that the MIT license is compatible with SciPy, and in the MIT
description in Wikipedia (http://en.wikipedia.org/wiki/MIT_License) it
states that this license is compatible with the GPL. So... is there a crack
here where we could place a GPL software in SciPy?

No. License compatibility is not commutative. A license is GPL-compatible
if it adds no restrictions above those in the GPL. A license is
BSD-compatible (and thus scipy-compatible) if it adds no restrictions above
those in the BSD license. The GPL adds restrictions above those in the BSD
license.

We have decided as a matter of policy for the scipy project to only include
code that does not add more restrictions than roughly that of the BSD
license. This isn't really an issue of license compatibility. You can
combine BSD scipy code and GPL code just fine, but we have a policy against
including GPL code into scipy because we want to keep the licensing
situation simple and understandable and not copylefted. If some scipy
packages were BSD-licensed and some were GPL-licensed, everyone will have
to pay close attention to which ones they were using to be sure that they
are obeying the terms of the GPL license. That's really annoying. It is
much easier for everyone involved to keep the GPL code in an entirely
separate package.

I appreciate your effort in learning about these issues, but please
consider it an exercise in personal education rather than trying to find a
workaround to include GPL code in scipy. There isn't one.

--
Robert Kern
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