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Using Constraint Programming to Identify Groups

From: andrew cooke <andrew@...>

Date: Mon, 29 Aug 2011 23:43:08 -0300

This is the text of my answer at
http://stackoverflow.com/questions/7076349/is-there-a-good-way-to-do-this-type-of-mining/7237972#7237972
- it would be better to read it there, but I wanted a local copy in case it
was deleted at some point.

Andrew

This is a little late, but this problem has been worrying me for some time.  I
was sure it could be solved with mixed integer / linear programming techniques
and asked for help in this question: http://stackoverflow.com/q/7137334/181772

least as I understand it, is so simple (when framed as a constraint program)
that you can solve it trivially with a simple program (which you already
knew).  In other words, constraint programming would be a cool way to solve
this, but, at least with the approach I found, would give you the same answer
as something much simpler.

I'll explain below my reasoning, how I would implement it with a constraint
solving package, and then give the final, trivial, algorithm.

Mixed integer programming solution
----------------------------------

The most important detail is the difference between horizontal and vertical
groups.  As far as i can see, anything that aligns vertically can be in the
same group.  But horizontal groups are different - components have to be close
together.

The hardest part of solving a problem with constraints seems to be finding a
way to describe the limits in a way that the solver can understand.  I won't
go into the details here, but solvers are frustratingly limited.  Luckily I
think there is a way to do this here, and it is to consider horizontal
neighbours:  if there are N points in a row then we have N-1 sets of
neighbours (for example, with 4 points A B C and D there are the three pairs
AB, BC, and CD).

For each pair, we can give a score, which is the number of spaces between them
(S_i) scaled by some factor K, and a flag (F_i) which is 0 or 1.  If the
pair are in the same horizontal group then we set the flag to 1, otherwise it
is zero.

It is critical to see that the set of flags for all the pairs *completely
defines a solution*.  For any row we can run along, placing pairs with a flag
of 1 in the same horizontal group, and starting a new horizontal group each
time the flag is 0.  Then, we can take all horizontal groups of size 1 and
convert them into vertical groups: any point that is not in a horizontal group
must be in a vertical group (even if it is a vertical group of just one).

So all we need now is a way to express an optimal solution in terms of the
flags.  I suggest that we want to minimise:

sum(1 - F_i) + sum(K * S_i * F_i)

This has two terms.  The first is the sum of "one minus the flag" for each
pair.  The flag is 1 when the points are in the same horizontal group and 0
otherwise.  So minimising this value is the same as saying that we want as
*few* horizontal groups as possible.  If this was the only constraint then we
could set it to zero by making all the F_i 1 - by making all pairs on a row
members of the same group.

But the second term stops us from choosing such an extreme solution.  It
penalises groups with gaps.  If a pair are in the same group, but are
separated by S_i spaces, then we have a "penalty" of K * S_i.

So we have a trade-off.  We want horizontal groups, but we don't want gaps.
The final solution will depend on K - if it is large then we won't include
any spaces in horizontal groups.  But as it is decreased we will start to do
so, until when it is very small (tends to zero) we place everything in a row
in a single group.

Analytic solution
-----------------

OK, cool.  At this point we have a way to express the problem that we can give
to a constraint engine.

But it's trivial to solve!  we don't need no stinkin' constraint engine to
solve this - we can just look at the expression:

sum(1 - F_i) + K * sum(S_i * F_i)

The two sums are over the same pairs, so we can move everything into the sum:

sum(1 - F_i + K * S_i * F_i)
sum(1 - F_i * (K * S_i - 1))

And then extract the constant (N here is the total number of pairs):

N - sum(F_i * (K * S_i - 1))

Now note that each term in the sum is independent (and additive).  so for each
term, we want the minimum value.  we have two options:

- if F_i is 0 then the entire term is 0.

- otherwise, F_i is 1 and the term is K * S_i - 1.

So the best choice depends on whether K * S_i is greater than 1.  if K *
S_i is greater than 1 then the smallest value of the term is 0, and F_i
should be 0.  Otherwise the second choice above is negative, and F_i should
be one.

Trivial algorithm
-----------------

What does this mean?  It means that for each pair we can simply look at the
number of spaces, S_i.  If that is greater than 1 / K then the two points
should be in separate groups.  otherwise they should be in the same group.

So all this fancy maths and optimisation and constraints and bullshitting
comes down to: how far apart are two points in neighbouring pairs?  If they
are closer than some cut-off, put them in the same horizontal group.
Otherwise, put them in separate groups.

So here, finally, is your algorithm:

choose some cut-off value, X
place each point in its own, singleton, horizontal group
for each row with more that one point:
for each neighbouring pair in the row:
if the space between the pair is less than X:
joint into a single horizontal group
for each column:
join any single singleton groups into a single vertical group

Conclusion
----------

- You can use constraint programming techniques to solve this problem.

- But such techniques are restricted to problems that can be described in
"the right" (typically, linear) way.

- The simplest such approach I can find is equivalent to a trivial, direct
algorithm that divides points in a row into horizontal groups depending on
the number of spaces between them.

- But this all depends on a whole pile of assumptions which may, of course,
be over-simplifications, or just plain wrong.