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    <p><strong>Representation:</strong></p> <p>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.</p> <p>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.</p> <p>We can say that A =(xa1,...xa24)</p> <p>e1...e24 are the unit vectors (1,0...0) to (0,0...1)</p> <p>note about the operator '.':</p> <ul> <li>A.e1=xa1</li> <li>...</li> <li>X.e24=Xx24</li> </ul> <p>There are some constraints on A,...X with these notations :</p> <p>Xii is in {0,1} and</p> <p>Sum(Xai)=1 ... Sum(Xxi)=1</p> <p>Sum(Xa1,xb1,...Xx1)=1 ... Sum(Xa24,Xb24,... Xx24)=1</p> <p>Since one element can be assign to only one node.</p> <p>I will define a graph by defining the neighbors relation of each node, lets say node 8 has neighbors node 7 and node 10</p> <p>to check that A and B are neighbors on node 8 for exemple I nedd:</p> <p>A.e8=1 and B.e7 or B.e10 =1 then I just need A.e8*(B.e7+B.e10)==1</p> <p>in the function isNeighborInGraphs(A,B) I test that for every nodes and I get one or zero depending on the neighborhood.</p> <p><strong>Notations:</strong></p> <ul> <li>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...)</li> <li>e1... e24 are the unit vectors (1,0,0...0) to (0,0...1)</li> <li>Let A, B ...X be the N elements.</li> </ul> <blockquote> <p>A=(0,0...,1,...,0)=(xa1,xa2...xa24) </p> <p>B=... </p> <p>... </p> <p>X=(0,0...,1,...,0)</p> </blockquote> <ul> <li>Graph descriptions:</li> </ul> <blockquote> <p>IsNeigborInGraphs(A,B)=A.e1*B.e2+... //if 1 and 2 are neigbors in one graph for exemple</p> </blockquote> <ul> <li>State of the system:</li> </ul> <blockquote> <p>L(A)=[B,B,C,E,G...] // list of neigbors of A (can repeat)</p> </blockquote> <pre><code>actualise(L(A)): for element in [B,X] if IsNeigbotInGraphs(A,Element) L(A).append(Element) endIf endfor </code></pre> <ul> <li>Objective functions</li> </ul> <blockquote> <p>N(A)=len(L(A))+Sum(IsneigborInGraph(A,i),i in L(A))</p> <p>...</p> <p>N(X)= ...</p> </blockquote> <p><strong>Description of the algorithm</strong></p> <ol> <li>start with an initial position A=e1... X=e24</li> <li>Actualize L(A),L(B)... L(X)</li> <li>Solve this (with a solveur, <a href="http://www.ampl.com/" rel="nofollow">ampl</a> for exemple will work I guess since it's a nonlinear optimization problem):</li> </ol> <p><em>Objective function</em></p> <blockquote> <p>min(Sum(N(Z),Z=A to X)</p> </blockquote> <p><em>Constraints:</em></p> <blockquote> <p>Sum(Xai)=1 ... Sum(Xxi)=1</p> <p>Sum(Xa1,xb1,...Xx1)=1 ... Sum(Xa24,Xb24,... Xx24)=1</p> </blockquote> <p>You get the best solution</p> <p>4.Repeat step 2 and 3, 3 more times.</p>
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    1. COthanks for the proposal. But can you explain it in a bit detail that what you mean by e1, e2, etc, and what do you mean by N elements A, B, ..., X and their representations (xa1,xa2...xa24) ?
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    2. CO@Skyork, I updated the post, I added the paragraph **representation**
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