将节点分配给图形的算法设计

Question

I have a graph-theoretic (which is also related to combinatorics) problem that is illustrated below, and wonder what is the best approach to design an algorithm to solve it.

Given 4 different graphs of 6 nodes (by different, I mean different structures, eg STAR, LINE, COMPLETE, etc), and 24 unique objects, design an algorithm to assign these objects to these 4 graphs 4 times , so that the number of repeating neighbors on the graphs over the 4 assignments is minimized . For example, if object A and B are neighbors on 1 of the 4 graphs in one assignment, then in the best case , A and B will not be neighbors again in the other 3 assignments.

Obviously, the degree to which such minimization can go is dependent on the specific graph structures given. But I am more interested in a general solution here so that given any 4 graph structures, such minimization is guaranteed as the result of the algorithm.

Any suggestion/idea of solving this problem is welcome, and some pseudo-code may well be sufficient to illustrate the design. Thank you.

Answer 1

Representation:

You have 24 elements, I will name this elements from A to X (24 first letters). Each of these elements will have a place in one of the 4 graphs. I will assign a number to the 24 nodes of the 4 graphs from 1 to 24.

I will identify the position of A by a 24-uple =(xA1,xA2...,xA24), and if I want to assign A to the node number 8 for exemple, I will write (xa1,Xa2..xa24) = (0,0,0,0,0,0,0,1,0,0...0), where 1 is on position 8.

We can say that A =(xa1,...xa24)

e1...e24 are the unit vectors (1,0...0) to (0,0...1)

note about the operator '.':

• A.e1=xa1
• ...
• X.e24=Xx24

There are some constraints on A,...X with these notations:

Xii is in {0,1} and

Sum(Xai)=1... Sum(Xxi)=1

Sum(Xa1,xb1,...Xx1)=1... Sum(Xa24,Xb24,... Xx24)=1

Since one element can be assign to only one node.

I will define a graph by defining the neighbors relation of each node, lets say node 8 has neighbors node 7 and node 10

to check that A and B are neighbors on node 8 for exemple I nedd:

A.e8=1 and B.e7 or B.e10 =1 then I just need A.e8*(B.e7+B.e10)==1

in the function isNeighborInGraphs(A,B) I test that for every nodes and I get one or zero depending on the neighborhood.

Notations:

• 4 graphs of 6 nodes, the position of each element is defined by an integer from 1 to 24. (1 to 6 for first graph, etc...)
• e1... e24 are the unit vectors (1,0,0...0) to (0,0...1)
• Let A, B...X be the N elements.

A=(0,0...,1,...,0)=(xa1,xa2...xa24)

B=...

...

X=(0,0...,1,...,0)

• Graph descriptions:

IsNeigborInGraphs(A,B)=A.e1*B.e2+... //if 1 and 2 are neigbors in one graph for exemple

• State of the system:

L(A)=[B,B,C,E,G...] // list of neigbors of A (can repeat)

actualise(L(A)):
for element in [B,X]
if IsNeigbotInGraphs(A,Element)
L(A).append(Element)
endIf
endfor

• Objective functions

N(A)=len(L(A))+Sum(IsneigborInGraph(A,i),i in L(A))

...

N(X)=...

Description of the algorithm

1. start with an initial position A=e1... X=e24
2. Actualize L(A),L(B)... L(X)
3. Solve this (with a solveur, ampl for exemple will work I guess since it's a nonlinear optimization problem):

Objective function

min(Sum(N(Z),Z=A to X)

Constraints:

Sum(Xai)=1... Sum(Xxi)=1

Sum(Xa1,xb1,...Xx1)=1... Sum(Xa24,Xb24,... Xx24)=1

You get the best solution

4.Repeat step 2 and 3, 3 more times.

Answer 2

If all four graphs are K_6 , then the best you can do is choose 4 set partitions of your 24 objects into 4 sets each of cardinality 6 so that the pairwise intersection of any two sets has cardinality at most 2. You can do this by choosing set partitions that are maximally far apart in the Hasse diagram of set partitions with partial order given by refinement. The general case is much harder, but perhaps you can still begin with this crude approximation of a solution and then be clever with which vertex is assigned which object in the four assignments.

Answer 3

Assuming you don't want to cycle all combinations and calculate the sum every time and choose the lowest, you can implement a minimum problem (solved depending on your constraints using either a linear programming solver ie symplex algorithm engines or a non-linear solver, much harder talking in terms of time) with constraints on your variables (24) depending on the shape of your path. You can also use free software like LINGO/LINDO to create rapidly a decision theory model and test its correctness (you need decision theory notions though)

Answer 4

If this has anything to do with the real world, then it's unlikely that you absolutely must have a solution that is the true minimum. Close to the minimum should be good enough, right? If so, you could repeatedly randomly make the 4 assignments and check the results until you either run out of time or have a good-enough solution or appear to have stopped improving your best solution.