#16866: Radical difference families
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Reporter: vdelecroix | Owner:
Type: enhancement | Status: needs_info
Priority: major | Milestone: sage-6.4
Component: combinatorial | Resolution:
designs | Merged in:
Keywords: | Reviewers:
Authors: Vincent Delecroix | Work issues:
Report Upstream: N/A | Commit:
Branch: | 721af75ec2b2c6ca904f2feaf86e75e054cb089d
u/vdelecroix/16866 | Stopgaps:
Dependencies: #16863 |
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Comment (by ncohen):
Yo!
> Too bad. If you tell me where, I can be clearer in the comments.
Well, for a start I had no idea at first what was this "A" that you
defined, least of all why you have this `-1`. Then you seemed to say that
all differences belonged to different cosets of `H^{mt}` (I don't
understand what this set is), but it seems to mean that you can only
consider the differences obtained with 1.
I don't get why you have this `A.append(K.one())`, I would have expected
`-K.one()` (i.e. the difference 0-1). Then I do not understand the cut
with the `len(set(logA))`.
> What do you mean by "the same time". There is the same number of tiles
in the packing at the end but if you ignore the symmetries there are much
more possible translations.
I do not get exactly what your reduction of the ground set means with
respect to the partition instance. What I mean is that an instance with
the A0 U A0', A1 U A1', ... above is totally equivalent (in terms of
resolution, and in terms of time) to an instance with A0, A1, ...
I was wondering if that was the difference that there is, in this branch,
between considering the cosets that you mention and not considering them.
> That my code works and that it is good to have smaller instances of the
problem to solve.
It is not necessarily an improvement to reduce the number of vertices,
e.g. the example above. And so far I do not understand how your code
works.
Nathann
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Ticket URL: <http://trac.sagemath.org/ticket/16866#comment:22>
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