[OSM-talk] Licence of Facebook's derived road datasets? ODbL?

[Phil Wyatt]

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From: Nuno Caldeira <nunocapelocaldeira at gmail.com> 
Sent: Tuesday, 31 December 2019 5:50 AM
To: Christoph Hormann <osm at imagico.de>
Cc: OpenStreetMap talk mailing list <talk at openstreetmap.org>
Subject: Re: [OSM-talk] Licence of Facebook's derived road datasets? ODbL?

 

here's a interested case https://www.gislounge.com/gis-data-high-resolution-global-hydrography-dataset/amp/

are they allowed to share this on CC4? Shouldn't it only be ODbL? are they allowed to share only after a registration? anyone wanna try getting a copy of the derivated work as they need to without registration?

 

On Fri, Nov 15, 2019 at 4:20 PM Christoph Hormann <osm at imagico.de <mailto:osm at imagico.de> > wrote:

On Friday 15 November 2019, Martin Koppenhoefer wrote:
> > > there isn't OSM data in their dataset.
> >
> > And neither is there is my ocean data set - the OSM data set used
> > only contains land masses, my resulting data set (D2 in Rory's
> > terms) only contains oceans.  So no OSM data in it.
>
> I doubt this cheap trick would pass when contested in a trial.

Well - it is not my cheap trick, it is facebook's cheap trick.  I am 
just following the lead here.  There is no principal difference between 
what facebook does and what my scenario describes.

> > If the question is not "addition or subtraction" consider the
> > following scenario.  You create a data set using some AI and big
> > data process of 'potential restaurants' world wide and create a set
> > intersection between those and the restuarants in OSM would the
> > results be a derivative of OSM data?
>
> yes, if you look at the intersection (data in both sets), it would
> be. If you took only what is not in OSM, I guess it wouldn't (no data
> from OSM contained).

So the set operation chosen (difference or intersection or any other) 
decides on the legal status of the resulting data set?

You are aware that a difference is the same as an intersection with the 
complement, i.e. A \setminus B = A \cap B^c - see:

https://en.wikipedia.org/wiki/Complement_(set_theory)

-- 
Christoph Hormann
http://www.imagico.de/

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