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Database Forum / General DB Topics / DB Theory / June 2008

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Can anybody help me to find a potential application for this     mathematical problem

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simon - 13 Jun 2008 08:19 GMT
Suppose there are N parties. Each party has R records, each of which
has D dimensions.

For example, N=3, parties are P1, P2, P3.
D=4, dimensions are named as (A,B,C,D).
R=2, P1 has records T1=(a1,b1,c1,d1),T2=(a2,b1,c2,d1);
    P2 has records T3=(a1,b1,c1,d1),T4=(a2,b1,c2,d3);
    P3 has records T5=(a1,b1,c1,d1),T6=(a1,b1,c2,d4);

Then, the 3-party union on dimension A ={T1.A, T2.A, ..., T6.A}={4*a1,
2*a2};
union on B={T1.B, T2.B, ..., T6.B}={6*b1};
union on C={3*c1,3*c2};
union on D={4*d1,d3,d4};

The mathematical problem is: given the parties and their records, the
parties want to find whether each record T has the property that the
value on T.A appears at least k1 times in the 3-party union on
dimension A, the value on T.B appears at least k2 times in the union
on B, the value on T.C appears at least k3 times in the union on C,
and the value on T.D appears at least k4 times in the union on D.

For example, when k1=k2=k3=k4=2, T2 has this property, because a2, b1,
c2, d1 appear at least 2 times in the 3-party union on dimension
A, B, C, D respectively. T1, T3, T5 have also the property, but T4 and
T6 haven't.

Is there any application for this problem? I mean, given the parties
and their records, what are the motivations to find such a property? I
find some applications in customer management in chain stores, and
intrusion alert in distributed network monitoring, but
feel that they are still not very appropriate.

Can anybody help me? Thanks a lot!

Simon
-CELKO- - 16 Jun 2008 17:42 GMT
>> Is there any application for this problem? <<

This sounds like a version of the "golfer problem" -- given a bunch of
people, break them into groups of (n) such that no two people are in a
grouping more than (k) times.  The idea is to keep you playing with
some new blood each round.  Look at the recreational math newsgroup;
there ought to be something there.
 
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