Firstly, how many cut sets do graphs have? Do they have more than one ? If so, is there a way of knowing how many of them there are? Also, what does it mean for a disconnecting set of edges to not be a cut set? Isn't disconnecting the graph the main definition of a cut set ?
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1Some graphs on $n$ vertices are already disconnected. You should try some examples, say with four vertices, to get a better understanding of what you are asking. – hardmath Aug 08 '17 at 18:39
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One easily finds a definition of cut set in Wikipedia, among other places on the Web. – hardmath Aug 08 '17 at 18:45
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the definitions contradict with the answers given to problems by my professor – april symmetry Aug 08 '17 at 19:04
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You are not giving me enough information to guess what's inconsistent or how to resolve your confusion. Perhaps the exercises involve the minimum size of a cut set for particular graphs? – hardmath Aug 08 '17 at 19:14
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Maybe consider block-diagonal examples for a simple-to-understand example. – mathreadler Aug 08 '17 at 19:32
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A cut set is the set of edges between the two parts of a bipartition of the nodes in a graph. So if a graph has $n$ nodes, there are $2^{n-1} - 1$ ways to partition those nodes into two nonempty subsets (bipartitions).
Unfortunately the number of cut sets is not quite in one-to-one correspondence with those bipartitions (which we easily counted). In particular, if the graph $G$ on $n$ nodes is disconnected, there can be various bipartitions that give us the same empty cut set.
So $2^{n-1} - 1$ is an upper bound on the number of cut sets, but it is not tight. Consider the examples of graphs on four nodes, and see if you can find a pattern for connected graphs versus disconnected graphs.
While every cut set is a "disconnecting set of edges" (removing them guarantees a disconnect graph will result), not every disconnecting set of edges can be realized as a cut set. This is true even if we assume graph $G$ is connected. Again you should be able to pick out an example of this using graphs on four nodes.
hardmath
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Edit: this hint may make most sense if you have studied the spectral clustering connection to graph cuts.
To build some intuition on this consider the masking matrix $${\bf M}= {\bf I}_n\otimes {\underset{m\times m \text{ block of 1:s}}{\underbrace{({\bf 1}_m{{\bf 1}_m}^T)}}}$$
Where $\otimes$ is Kronecker product. Introduce a random matrix ${\bf R}=uniform([0,1],m\times n)$. Then we reset ${\bf R = I+R+R}^T$ to ensure a symmetric matrix with large diagonal elements which is what we would expect from a candidate affinity matrix.
We can now consider $\bf A = M \circ R$ ($\circ$ Schur/Hadamard product) being a fake affinity matrix of a block-diagonal matrix from which we can construct a normalized Laplacian (${\bf D} = \text{diag}(\text{diag}({\bf A}))$):
$$\bf L = D^{-1/2}AD^{-1/2}-I$$
Now maybe you have some ideas of how to proceed?
EDIT Ok maybe my hint was a bit cryptic. The next step is to calculate and sort eigenvalues and check corresponding eigenvectors. As happens the $n$ largest eigenvalues (which will be all eigenvalues $>0$) will correspond to eigenvectors which are non-zero for the $m$ elements uniquely corresponding to a particular $m\times m$ block and zero everywhere else. If you are still unsure what it means, you can experiment with switching the order of the Kronecker factors where $\bf M$ is constructed and see what happens to those eigenvectors.
mathreadler
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Note that the OP is struggling with the definition of cut set, and with counting them. It seems doubtful that they have already "studied the spectral clustering connection to graph cuts." – hardmath Aug 08 '17 at 20:58
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@hardmath: Depends on which angle you approach graph cuts from. If you are an engineer learning Linear Algebra, calculus and loads of continous math first and hardly any set theory or discrete maths then spectral clustering will be an obvious introduction. – mathreadler Aug 08 '17 at 21:01