[chaley]

Originally Posted by kiwidude

The authors point is an interesting one. One issue which I didn't see you mention of having multiple author cache entries is replication of groups.

You are right. Hmmm...

I guess the question is whether this is a problem or not. If the user resolves their duplicates in order, the second group if identical would disappear automatically. If they skip through them with highlighting it may jump around a bit but still be valid.

Thinking on paper and (I think) agreeing with you: what you are saying is that adding a book to multiple buckets can create a situation where one group is a (possibly improper) subset of another. It seems to me that there isn't much point in showing both groups, at least in author mode. For example, why show a group containing (1,2,3) and another containing (2,3)?

Subsets can be removed rather easily, with performance that should be acceptable if there aren't thousands of groups. Something like this:

Spoiler:

def clean_dup_groups(dups):
    res = [set(d) for d in dups]
    res.sort(cmp=lambda x, y: cmp(len(x), len(y)))
    ans = []
    for i,a in enumerate(res):
        for b in res[i+1:]:
            if a.issubset(b):
                break
        else:
            ans.append(a)
    return ans


dups = [(1,2,3),(4,5)]
print dups
print clean_dup_groups(dups)

print '========================'
dups = [(1,2,3,4,5), (1,6,7)]
print dups
print clean_dup_groups(dups)

print '========================'
dups = [(1,2,3,4,5), (1,6,7), (1,6,7)]
print dups
print clean_dup_groups(dups)

print '========================'
dups = [(1,2,3,4,5), (1,6,7), (3,4), (6,7)]
print dups
print clean_dup_groups(dups)


with output:
[(1, 2, 3), (4, 5)]
[set([4, 5]), set([1, 2, 3])]
========================
[(1, 2, 3, 4, 5), (1, 6, 7)]
[set([1, 6, 7]), set([1, 2, 3, 4, 5])]
========================
[(1, 2, 3, 4, 5), (1, 6, 7), (1, 6, 7)]
[set([1, 6, 7]), set([1, 2, 3, 4, 5])]
========================
[(1, 2, 3, 4, 5), (1, 6, 7), (3, 4), (6, 7)]
[set([1, 6, 7]), set([1, 2, 3, 4, 5])]

If they added exemptions using mark all groups it would create some duplication I think but not a major drama.

Again, I don't see a reason to keep exemption groups that are subsets of another group. The same set cleanup would fix this, eliminating the subsets.