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positive and negative relationships in .NET Embed Code 39 Full ASCII in .NET positive and negative relationships

positive and negative relationships using barcode drawer for .net control to generate, create code39 image in .net applications. barcode pdf417 X 1 X 5 label as X or Y Figure 5.10. If a signed gr USS Code 39 for .

NET aph contains a cycle with an odd number of negative edges, then it is not balanced. Indeed, if we pick one of the nodes and try to place it in X , then following the set of friend/enemy relations around the cycle produces a con ict by the time we get to the starting node..

by symmetry.) Then, because node 2 is friends with node 1, it too must belong to X. Node 3, an enemy of 2, must therefore belong to Y ; hence, node 4, a friend of 3, must belong to Y as well.

And node 5, an enemy of 4, must belong to X. The problem is that, if we continue this reasoning one step further, node 1, an enemy of 5, should belong to Y but we had already decided at the outset to put it into X. We had no freedom of choice during this process so this shows that there is no way to divide the nodes into sets X and Y to satisfy the mutual-friend/mutual-enemy conditions of structural balance.

Hence, the signed graph in Figure 5.10 is not balanced. The reasoning in the previous paragraph sounds elaborate, but in fact it follows a simple principle: we were walking around a cycle, and every time we crossed a negative edge we had to change the set into which we were putting nodes.

The dif culty was that getting back around to node 1 required crossing an odd number of negative edges, and so our original decision to put node 1 into X clashed with the eventual conclusion that node 1 ought to be in Y . This principle applies in general: if the graph contains a cycle with an odd number of negative edges, then this implies that the graph is not balanced. Indeed, if we start at any node A in the cycle and place it in one of the two sets, and then we walk around the cycle and place the other nodes where they must go, the identity of the set where we re placing nodes switches an odd number of times as we go around the cycle.

Thus, we end up with the wrong set by the time we make it back to A. A cycle with an odd number of negative edges is thus a very simple-to-understand reason why a graph is not balanced: You can show someone such a cycle and immediately convince him or her that the graph is not balanced. For example, the cycle in Figure 5.

8 consisting of nodes 2, 3, 6, 11, 13, 12, 9, and 4 contains ve negative edges, thus supplying a succinct reason why this graph is not balanced. But are there other, more complex reasons why a graph is not balanced In fact, though it may seem initially surprising, cycles with an odd number of negative edges are the only obstacles to balance. This is the crux of the following claim [97, 204]:.

Claim: A signed graph is ba VS .NET Code 39 Extended lanced if and only if it contains no cycle with an odd number of negative edges..

advanced material + + + + . Figure 5.11. To determine i .

net vs 2010 Code-39 f a signed graph is balanced, the rst step is to consider only the positive edges, to nd the connected components using just these edges, and to declare each of these components to be a supernode. In any balanced division of the graph into X and Y , all nodes in the same supernode must be placed in the same set..

We now show how to prove th is claim. The proof proceeds by designing a method that analyzes the graph and either nds a division into the desired sets X and Y or else nds a cycle with an odd number of negative edges. Proving the Characterization: Identifying Supernodes.

Let s recall what we re trying to do: Find a division of the nodes into sets X and Y so that all edges inside X and Y are positive, and all edges crossing between X and Y are negative. When we produce a partition into sets X and Y with these properties it is called a balanced division. We now describe a procedure that searches for a balanced division of the nodes into sets X and Y ; either it succeeds, or it stops with a cycle containing an odd number of negative edges.

Because these are the only two possible outcomes for the procedure, this will provide a proof of the claim. The procedure works in two main steps: the rst step is to convert the graph to a reduced form that only contains negative edges, and the second step is to solve the problem on this reduced graph. The rst step works as follows.

Notice that whenever two nodes are connected by a positive edge, they must belong to the same set, either X or Y , in a balanced division. So we begin by considering what the connected components of the graph would be if we were to consider only positive edges. These components can be viewed as a set of contiguous blobs in the overall graph, as shown in Figure 5.

11. We will refer to each of these blobs as a supernode: each supernode is connected internally via positive edges, and the only edges going between two different supernodes are negative. (If there were a positive edge linking two different supermodes, then we should have combined them together into a single supernode.

) Now, if any supernode contains a negative edge between some pair of nodes A and B, then we already have a cycle with an odd number of negative edges, as illustrated.
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