set partition question

[pball.benetech]

dear pythonistas,

So imagine that we have a set of sets S. If we pick arbitrarily one
set, s0, what are all the combinations of other sets in S which when
combined by set operations result in s0?

s0 = set([1])
s1 = set([1,2])
s2 = set([2])
S = set([s0,s1,s2])
one answer we're searching for is s0 = s1 - s2

There may be arbitrarily many set elements (denoted by integers
1,2,3,...) and arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any operation or function which
takes and returns sets.

I think this problem is related to integer partitioning, but it's not
quite the same. The range of answers has a little combinatorial
explosion problem as S gains new members. In my problem, len(S) is
usually on the order of 1M, and in the worst case, 10M, and there are
on the order of 10k elements.

My attempts to come up with an algorithm have not succeeded. Any ideas
occur to you folks? -- PB.

--
Patrick Ball, Ph.D.
Chief Scientist
& Director, Human Rights Program
http://www.benetech.org
http://www.hrdag.org
http://www.martus.org

 

[Martin v. Löwis]

s0 = set([1])

s1 = set([1,2])
s2 = set([2])
S = set([s0,s1,s2])
one answer we're searching for is s0 = s1 - s2

There may be arbitrarily many set elements (denoted by integers
1,2,3,...) and arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any operation or function which
takes and returns sets.

In that case, there is always a trivial answer. As I can use any
function which takes and returns sets, and as I shall come up with
a function that returns s0, I just use the following function

def f(s):
return s0

To compute s0, just invoke f with any other of the sets, and you
will - get s0.

I think this problem is related to integer partitioning, but it's not
quite the same.

I think the problem is significantly underspecified. It would be a more
interesting problem if there was a restriction to a few selected set
operations, e.g. union, intersection, difference, and combinations
thereof.

Regards,
Martin

 

[pball.benetech]

I think the problem is significantly underspecified. It would be a more
interesting problem if there was a restriction to a few selected set
operations, e.g. union, intersection, difference, and combinations
thereof.

Ok, that's quite right -- when I just tried to define "any function,"
I found that all the solutions I came up with were combinations of the
set operations defined for immutable sets. Let me improve the spec as
the following:

There may be arbitrarily many set elements (denoted by integers
1,2,3,...), arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any of python's set operations
or combination of those operations.

Thanks for the clarification -- PB.

 

[Arnaud Delobelle]

dear pythonistas,

So imagine that we have a set of sets S. If we pick arbitrarily one
set, s0, what are all the combinations of other sets in S which when
combined by set operations result in s0?

s0 = set([1])
s1 = set([1,2])
s2 = set([2])
S = set([s0,s1,s2])
one answer we're searching for is s0 = s1 - s2

There may be arbitrarily many set elements (denoted by integers
1,2,3,...) and arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any operation or function which
takes and returns sets.

I think this problem is related to integer partitioning, but it's not
quite the same. The range of answers has a little combinatorial
explosion problem as S gains new members. In my problem, len(S) is
usually on the order of 1M, and in the worst case, 10M, and there are
on the order of 10k elements.

My attempts to come up with an algorithm have not succeeded. Any ideas
occur to you folks? -- PB.

Unless your sets have some sort of pattern, it sounds to me like this
problem is at least exponential... Good luck!

 

[miller.paul.w]

dear pythonistas,

I think this problem is related to integer partitioning, but it's not
quite the same. The range of answers has a little combinatorial
explosion problem as S gains new members. In my problem, len(S) is
usually on the order of 1M, and in the worst case, 10M, and there are
on the order of 10k elements.

If, by "integer partitioning," you mean the "subset sum
problem" (given a finite set S of integers, does S contain a subset
which sums up to some given integer k?), then you are right. I'm
reasonably sure there's a polynomial reduction from your problem to
the subset sum problem. That would make your problem NP-complete.

As for algorithms, I don't think there's an exact algorithm any better
than trying all the possibilities and stopping when one comes close.
Say, for example, we specify the problem to be: given sets s_0, s_1,
s_2, ..., s_n, with S defined to be the union of all the s_i's, return
all functions f which, using the sets s_1, s_2, ..., s_n and the
elementary set operations (union, intersection, difference), returns
the set s_0.

Now, you need to be a little more careful, because, given a function f
which satisfies the problem, I can always define f' = f(S) + s_1 - s_1
and get another function which satisfies the definition.

Like I said, because this looks related to the subset sum problem
(though harder, because you're asking for "all" such functions, for
some rigorous definition of "all," as per the previous sentence), I
suspect it's NP-complete. However, there are probably some good
heuristics, such as if s1 has a large intersection with s0, it's
probably a good idea to use s1 in whatever formula you come up with.

Other than that, I don't really have an idea. Can you say what the
application of this algorithm would be? Knowing the use case might
suggest a better approach.

 

[Martin v. Löwis]

There may be arbitrarily many set elements (denoted by integers

1,2,3,...), arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any of python's set operations
or combination of those operations.

That still allows for trivial solutions:

Given s0, and s1..sn, the following Python code outputs a Python
fragment that, when run, returns s0:

print "set()",
for e in s0:
print ".union(set([%s]))" % repr(e),

For s0=set([1,2,3]), I get

set() .union(set([1])) .union(set([2])) .union(set([3]))

Or, more trivially,

print repr(s0)

which gives me

set([1,2,3])

In either case, s0 is generated through "any of python's set
operations".

Regards,
Martin

 

[pball.benetech]

If, by "integer partitioning," you mean the "subset sum
problem" (given a finite set S of integers, does S contain a subset
which sums up to some given integer k?), then you are right.  I'm
reasonably sure there's a polynomial reduction from your problem to
the subset sum problem.  That would make your problem NP-complete.

Wow, that's helpful. http://en.wikipedia.org/wiki/Subset_sum clarifies
what I'm trying to do. Yes, it probably is NP-complete. Fortunately my
use case doesn't require that the process finish, just that it produce
a lot. I can exclude trivial cases (e.g., f' = f(S) + s_1 - s_1) with
some sort of simplifier.

Like I said, because this looks related to the subset sum problem
(though harder, because you're asking for "all" such functions, for
some rigorous definition of "all," as per the previous sentence), I
suspect it's NP-complete.  However, there are probably some good
heuristics, such as if s1 has a large intersection with s0, it's
probably a good idea to use s1 in whatever formula you come up with.

There are a few real-world constraints which make the problem easier,
but only by linear factors, I suspect. More on this below.

Other than that, I don't really have an idea.  Can you say what the
application of this algorithm would be?  Knowing the use case might
suggest a better approach.

This is a problem in statistical estimation. Given n records made up
of k variables, define a cell as the point in the cartesian product of
v_1 * v_2 * ... * v_k. I want to apply an estimator on the data in
each cell.

However, many cells are empty, and others are too sparse to support
estimation. One solution is to build lots of valid aggregations,
called "strata," for which estimates can be made for chunks of cells.
A "valid aggregation" is a sequence of *adjacent* cells (see below)
whose data are pooled (by whatever mechanism is relevant to the
estimator).

The cells are "elements" in my initial problem statement, and the
strata are the sets.

The constraints I mentioned are on the composition of strata: strata
can only contain sequences of adjacent cells, where adjacency is
defined for each variable. Two cells are adjacent if they are equal on
every variable except one, and on that one and by its definition, they
are adjacent.

Does this make it easier to understand, or more difficult? thanks --
PB.

 

[miller.paul.w]

This is a problem in statistical estimation. Given n records made up
of k variables, define a cell as the point in the cartesian product of
v_1 * v_2 * ... * v_k. I want to apply an estimator on the data in
each cell.

So, basically, V = (v_1, v_2, ... , v_{k-1}, v_k) can be regarded as
an abstract, k-dimensional vector, right?

If I understand your revised problem statement correctly, what you
really want to do is build a graph of these vectors, where graph
adjacency is equivalent to adjacency in your sense. That is, imagine
you have V_1, V_2, ... , V_n all sitting out in front of you,
represented abstractly as simple points. Draw a line between V_i and
V_j if they are "adjacent" in your sense of the word. What you have
then is a graph structure where your version of adjacency exactly
corresponds to graph adjacency. Then, in your language, a stratum is
simply a path in this graph, and finding those is easy.

That is, unless I've got this all wrong.

 

[pball.benetech]

So, basically, V = (v_1, v_2, ... , v_{k-1}, v_k) can be regarded as
an abstract, k-dimensional vector, right?
Yes.


If I understand your revised problem statement correctly, what you
really want to do is build a graph of these vectors, where graph
adjacency is equivalent to adjacency in your sense.  That is, imagine
you have V_1, V_2, ... , V_n all sitting out in front of you,
represented abstractly as simple points.  Draw a line between V_i and
V_j if they are "adjacent" in your sense of the word.  What you have
then is a graph structure where your version of adjacency exactly
corresponds to graph adjacency.  Then, in your language, a stratum is
simply a path in this graph, and finding those is easy.

You're solving an earlier part of the problem which I call stratum
generation. I've never thought to use a graph representation for
stratum generation -- very interesting, and I'll pursue it. Would you
be willing to outline how you'd do it here?

I've had fun "breeding" strata using genetic algorithms, and a lot of
interesting ideas came out of that experiment, in particular the
utility of building randomly shaped strata to do indirect estimates
(the objective of the set arithmetic described here). I used GA's
because I think that the space of possible strata is too big to search
by brute force.

I'd still like to have the graph representation for the brute force
solution. Or, in another random algorithm, the graph could be a
sampling frame from which random paths could be pulled.

In a real dataset, some of the strata will contain adequate data to
make an estimate, and these are called valid strata. Other legal
strata (as shown by a path through the graph) may not have adequate
data to make an estimate (thus, legal but invalid).

We've done this problem by hand: observing that we can estimate a
stratum (1,2,3) and another (2,3) but not (1). So
E(1) = E(1,2,3) - E(2,3)
There are issues with the variance of E(1), but that's a very
different problem. Using the valid strata E(1,2,3) and E(2,3) we've
made an indirect estimate of E(1).

I'm looking for an algorithm which automates the search for
combinations like the one above. Given a set of valid strata[1], yield
combinations of valid strata which, when combined via union or
difference[2], evaluate to a stratum (adjacent but possibly not
valid). The combination of valid strata is called an indirect
estimate

[1] strata can be generated via many methods; they all end up in the
same database
[2] Note that the set of operations is shrinking

Thanks in advance for your thoughts -- PB.

 

[Arnaud Delobelle]

Ok, that's quite right -- when I just tried to define "any function,"
I found that all the solutions I came up with were combinations of the
set operations defined for immutable sets. Let me improve the spec as
the following:

There may be arbitrarily many set elements (denoted by integers
1,2,3,...), arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any of python's set operations
or combination of those operations.

OK then, if you only allow union, intersection, difference, symmetric
difference then I think it would be easy to prove (I haven't done it!)
that if there is a solution to you problem, then the method below
yields a solution:

*** warning: all this is untested AND written in a rush ***

let S be the set of all your sets except s0 (i.e. s1, s2, s3, ...)


# X is the "domain", C is the set of complements in X of elements of S

X = reduce(set.union, S)
C = set(X - s for s in S)


# Find the "fibers" at each element x of s0. A fiber at x is the set
# of all s in S or C that contain x

from collections import defaultdict
fibers = defaultdict(list)

for s in S | C:
for x in s & s0:
fibers[x].append(s)


# Find the "seeds" at each x in s0. A seed at x is the set of all
# elements contained in all s in the fiber at x.

# Intutively, a seed at x is the smallest set containing x that can be
# built from S

seeds = [reduce(set.intersection, F) for F in fibers.itervalues()]


# Now we know if there is a solution:

sol = reduce(set.union, seeds)

if sol != s0:
print "No solution"
else:
print "Solution: union(intersection(fibers[x]) for x in s0)"


I think this solution will be found in O(m*n**2) time, where:

* n is the size of X (i.e. the total number of elements)
* m is the size of S (i.e. the total number of sets)

and assuming that set operations are linear.

 

[Arnaud Delobelle]

Ok, that's quite right -- when I just tried to define "any function,"
I found that all the solutions I came up with were combinations of the
set operations defined for immutable sets. Let me improve the spec as
the following:

There may be arbitrarily many set elements (denoted by integers
1,2,3,...), arbitrarily many combinations of the elements composing
the sets s_i (s0, s1, ...). We can use any of python's set operations
or combination of those operations.

OK then, if you only allow union, intersection, difference, symmetric
difference then I think it would be easy to prove (I haven't done it!)
that if there is a solution to you problem, then the method below
yields a solution:

*** warning: all this is untested AND written in a rush ***

let S be the set of all your sets except s0 (i.e. s1, s2, s3, ...)


# X is the "domain", C is the set of complements in X of elements of S

X = reduce(set.union, S)
C = set(X - s for s in S)


# Find the "fibers" at each element x of s0. A fiber at x is the set
# of all s in S or C that contain x

from collections import defaultdict
fibers = defaultdict(list)

for s in S | C:
for x in s & s0:
fibers[x].append(s)


# Find the "seeds" at each x in s0. A seed at x is the set of all
# elements contained in all s in the fiber at x.

# Intutively, a seed at x is the smallest set containing x that can be
# built from S

seeds = [reduce(set.intersection, F) for F in fibers.itervalues()]


# Now we know if there is a solution:

sol = reduce(set.union, seeds)

if sol != s0:
print "No solution"
else:
print "Solution: union(intersection(fibers[x]) for x in s0)"


I think this solution will be found in O(m*n**2) time, where:

* n is the size of X (i.e. the total number of elements)
* m is the size of S (i.e. the total number of sets)

and assuming that set operations are linear.

I forgot to add, using reduce() is probably not a great idea but it
allowed me to write down my idea quickly!

 

[pball.benetech]

I used Arnaud's suggestion with a few minor tweaks, as follows:

#-----
fset = frozenset
s0 = fset([1])
s1 = fset([1,2])
s2 = fset([2])

# let S be the set of all your sets except s0 (i.e. s1, s2, s3, ...)
S = set([s1,s2])

# X is the "domain", C is the set of complements in X of elements of
S
X = set()
for s in S:
X |= s
print "s0 is ", s0
print "S is ", S
print "X is ", X

C = set(fset(X - s) for s in S if fset(X-s))
print "C is ", C

# Find the "fibers" at each element x of s0.
# A fiber at x is the set of all s in S or C that contain x
from collections import defaultdict
fibers = defaultdict(list)
for s in S | C:
for x in s & s0:
fibers[x].append(s)
print "fibers are ", fibers

# Find the "seeds" at each x in s0.
# A seed at x is the set of all elements contained in all s in the
fiber at x.
# Intutively, a seed at x is the smallest set containing x that can
be
# built from S
seeds = set()
for F in fibers.itervalues():
seeds &= set(F)
print 'seeds are ', seeds

# Now we know if there is a solution:
sol = set()
for s in seeds:
sol |= s
if sol != s0:
print "No solution"
else:
print "Solution: union(intersection(fibers[x]) for x in s0)"
#-----

yields this:

s0 is frozenset([1])
S is set([frozenset([1, 2]), frozenset([2])])
X is set([1, 2])
C is set([frozenset([1])])
fibers are defaultdict(<type 'list'>, {1: [frozenset([1, 2]),
frozenset([1])]})
seeds are set([])
No solution


I follow your logic up to the point of the "seeds," where I get a bit
confused. Thanks much! for the idea, I'm going to keep thinking about
it. Any other ideas are *most* welcome. -- PB.

 

[pball.benetech]

Partial solution: the complements idea sparked a slightly new
direction. S is defined as above (all sets except s0, the target).

fset = frozenset
s0 = fset([1])
s1 = fset([1,2])
s2 = fset([2])
s3 = fset([2,3,4])
s4 = fset([3])
s5 = fset([1,2,3,4,5])
s6 = fset([5])
s7 = fset([2,3,4,5])
S = set([s1,s2,s3,s4,s5,s6,s7])

# solutions for s0:
# s1 - s2
# s5 - s7
# s5 - s3 - s6

def find_complement(S, s0):
for s in set(s for s in S if s & s0):
symm_diff = s ^ s0
if symm_diff in S:
yield s0, s, symm_diff

for c in find_complement(S, s0):
print c

(frozenset([1]), frozenset([1, 2]), frozenset([2]))
(frozenset([1]), frozenset([1, 2, 3, 4, 5]), frozenset([2, 3, 4, 5]))

So this algorithm can find partitions with 2 parts. It's helpless for
more complicated relationships -- in this case the combination s0 = s5-
s3-s6. Some kludgy ideas occur to me, and I'll work on them now. All
ideas welcome -- PB.

 

[Arnaud Delobelle]

I used Arnaud's suggestion with a few minor tweaks, as follows:

#-----
fset = frozenset
s0 = fset([1])
s1 = fset([1,2])
s2 = fset([2])

# let S be the set of all your sets except s0 (i.e. s1, s2, s3, ...)
S = set([s1,s2])

# X is the "domain", C is the set of complements in X of elements of
S
X = set()
for s in S:
X |= s
print "s0 is ", s0
print "S is ", S
print "X is ", X

C = set(fset(X - s) for s in S if fset(X-s))
print "C is ", C

# Find the "fibers" at each element x of s0.
# A fiber at x is the set of all s in S or C that contain x
from collections import defaultdict
fibers = defaultdict(list)
for s in S | C:
for x in s & s0:
fibers[x].append(s)
print "fibers are ", fibers

# Find the "seeds" at each x in s0.
# A seed at x is the set of all elements contained in all s in the
fiber at x.
# Intutively, a seed at x is the smallest set containing x that can
be
# built from S
seeds = set()
for F in fibers.itervalues():
seeds &= set(F)

Change this to:

seeds = []
for F in fibers.itervalues():
seed = X
for f in F:
seed &= f
seeds.append(seed)

And the whole thing should work

HTH

print 'seeds are ', seeds

# Now we know if there is a solution:
sol = set()
for s in seeds:
sol |= s
if sol != s0:
print "No solution"
else:
print "Solution: union(intersection(fibers[x]) for x in s0)"
#-----

yields this:

s0 is frozenset([1])
S is set([frozenset([1, 2]), frozenset([2])])
X is set([1, 2])
C is set([frozenset([1])])
fibers are defaultdict(<type 'list'>, {1: [frozenset([1, 2]),
frozenset([1])]})
seeds are set([])
No solution


I follow your logic up to the point of the "seeds," where I get a bit
confused. Thanks much! for the idea, I'm going to keep thinking about
it. Any other ideas are *most* welcome. -- PB.