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8

From what I understand there are broadly two distinct applications of mathematics to neuroscience. One uses mathematics to study the biological/chemical/physical aspects of the mechanisms in the brain, such as action potentials and the interactions between neurons. The type of math used here is differential equations/dynamic systems. Relevant wikipedia ...


7

Imagine a complex pipeline with a common source and common sink. You start to pump the water up, but you can't exceed some maximum flow. Why is that? Because there is some kind of bottleneck, i.e. a subset of pipes that transfer the fluid at their maximum capacity--you can't push more through. This bottleneck will be precisely the minimum cut, i.e. the set ...


4

These notes contain a proof. The digraph used in this proof is a little more complicated than the one that you have in mind: each point of the partial order corresponds to two vertices of the digraph. Let $\langle P,\preceq\rangle$ be the partially ordered set; I’ll write $p\prec q$ to indicate that $p\preceq q$ and $p\ne q$. For each $p\in P$ the digraph ...


4

Well spotted. In a case like this, it's a good idea to check the article's history (using the "View history" link at the top). In the present case, the error was introduced only two days ago by an anonymous user in this edit (which I just reverted).


4

Yes, you should increase the capacity of reverse edge by flow sent. Each time sending some flow by edge you should update its reverse edge too, so that flow passes only in one direction and the reverse edge has capacity = initial capacity + flow. http://stackoverflow.com/questions/7687732/maximum-flow-ford-fulkerson-undirected-graph ...


4

There are many ways to associate a group to a graph, some interesting, some artificial. I'm not sure what the motivation is behind the question. Given a graph $G$, the automorphisms of $G$ form a group called $\mbox{Aut}(G)$. Given a connected graph $G$, think of it as a topological space, pick a base point, and consider the fundamental group ...


3

First of all, the algorithm terminates because at least one edge is saturated in each step and there is a finite number of edges. Now let $f$ be the resulting flow, and assume $f'$ is some greater flow through the tree. For $f'$ to be a greater flow than $f$, we must have $f'(e)\gt f(e)$ for at least one edge $e$ leading away from the source. Now ...


3

In the same vein as Didier's answer providing bounds on the extinction probability $q$, we can also obtain bounds on the transmission probability $p$ required for the probability of $N$ signals dying out to go to $0$ as $N\to\infty$, which is the probability required for the extinction probability $q$ not to be $1$. Didier's equation for $q$ can be ...


3

This does not compute an exact value but some rigorous upper bounds of the probability $r$ that every site in $\mathbb Z^2$ is reached, eventually. Consider the Galton-Watson branching process where the progeny of every individual has the distribution of the number of signals transmitted at time $n+1$ by any site reached at time $n$. Thus the progeny of ...


3

Commonly used models in mathematical physics for collections of neurons are so-called neural networks. Here is some general explanation of the models. Some popular models are the Hopfield model, Ashkin-Teller neural networks, Blume-Emery-Griffiths neural networks,... Techniques used to solve them involve statistical mechanics which result in partition ...


3

Probably you'll find this one very useful. Chapter 5 ($M/M/C$ queue) corresponds to your model, where it is assumed that there are $c$ servers, the service time is exponentially distributed, and so are the interarrival times (of course, with different means). Anyway, the key is $M/M/C$ (or $M/M/N$, etc.) queue.


3

What you're looking for is described on pages 247-249 of Robert Vanderbei's Linear Programming: Foundations and Extensions (2nd edition), although Vanderbei has some of his sign conventions opposite of yours. Take a close look in particular at Figure 14.4 on page 249 ("Adding a new node to accommodate an arc $(i,j)$ having an upper bound $u_{ij}$ on its ...


3

A cut set is not necessarily an edge cut. Think about it: If $X$ in the definition of cut set is not itself connected, then you need to restore more than one edge to reconnect $G$. Also, cut sets appear to be defined even for a graph that is not connected to begin with.


3

There are no graphs for which the Tutte polynomial is $0$. One thing that would go wrong if there were such a graph: The chromatic polynomial is contained within the Tutte polynomial; if the Tutte polynomial were $0$, then the graph would not be $k$-colourable for all $k \geq 0$. But this is impossible since e.g. we can colour each vertex a different ...


3

Counterexample to your conjecture:


3

A side remark: it is called a skew-symmetry, because of the minus sign (source). Also, your third rule should be $$\sum_{v \in V} f(u, v) = 0, \quad \text{for all $u \in V \setminus \{s,t\}$},$$ because the flow between each vertex and start/target one should also be considered in the sum (see my explanation of the conservation of flow a bit lower), but it ...


3

This is not a good definition: Two graphs are equivalent if they have the same set of edges (ex. (A,B),(A,C)). It should be: Two graphs are equal if they have the same vertex set and the same set of edges. E.g. these two graphs are equal: (although they are drawn differently) and no two of these three graphs are equal: Equivalence ...


3

In your example the cut is $[\{s\},\{a,b,t\}]$; its capacity is $1$, since the only edge that is cut is $sa$. There are two min cuts in your example, $[\{s\},\{a,b,t\}]$ and $[\{s,a,b\},\{t\}]$, each with capacity $1$. The only other cut is $[\{s,a\},\{b,t\}]$, which has infinite capacity. Your ‘no edge from $S\cap T$ to $T\cap Y$’ should read ‘no edge ...


3

Judging from your question, I think you may be misinterpreting the sums. From left to right, you're summing over: The arcs that have "s" as their tail The arcs that have "s" as their head The arcs that have "t" as their head The arcs that have "t" as their tail. In particular, the sums in the lemma do not involve any arcs that are not incident to s or t. ...


3

Other ways to associate a group (but typically infinite ones) are the following: Graph of Groups: See e.g. Serre, "Trees" or the corresponding wiki-article. These are groups arising from actions on graphs (trees). See also its generalization, called Complexes of groups (see e.g. Bridson, Haefliger, "Metric spaces of non-positive curvature" RAAGs: Right ...


2

Your first balanced flow equation is incorrect, it should be: $$x_1 - x_2 = 400$$


2

The network in (b) shows the situation after deletion of the $(s^*, t^*)$ arc. The $(s^*, t^*)$ arc needs to be removed in order to have a one-to-one correspondence between $s$-$t$ cuts in the primal and paths from $s^*$ to $t^*$ in the dual. No, it doesn't belong to the dual network. Dual arcs must be between faces in the primal network. I'm not sure ...


2

Each of the basic flows you're using should conserve flow at nodes other than s and t. For convenience, write e.g [acba] as a flow of 1 unit on the cycle $a \to b \to c \to a$, and [sat] as 1 unit on $s \to a \to t$. So you could end up with something like $\frac{3}{2} [sat] + 2 [acba] + [abta]$ (just for illustration; that's not the answer here)


2

The rows represent the supply and demand nodes. The columns represent the decision variables $x_{ij}$; i.e., how much you send from supply node $i$ to demand node $j$. That's why there are $m+n$ rows and $mn$ columns. The matrix itself is the coefficient matrix for the constraints in the (balanced) transportation problem; in other words, the left-hand ...


2

I will try to show you why this is intuitive using counter example. Let us call the set reachable by unsaturated paths A, and the other set B. so s is in A, and t is in B. We know that by definition, each edge from A to B is in the minimum cut. Now suppose A were some other set such that it had a node v such that v was not reachable by an unsaturated path. ...


2

Everything regarding the question should be clear from the Wikipedia entry you referred to. Anyway, flow networks are directed graphs by definiton. For 'skew symmetry' Cormen's Introduction to Algorithms says: Skew symmetry is a notational convenience that says that the flow from a vertex $u$ to a vertex $v$ is the negative of the flow in the reverse ...


2

Make a vertex that is the source of the two inputs, and connect this vertex with the input vertices in your problem. Then apply a suited algorithm.


2

The trick would be to separate a vertex $v$ into two new vertices $v^-$ and a $v^+$ in. In a directed graph you then connect all incoming edges to $v$, denoted $\delta^-(v)$, to $v^-$ and all outgoing edges, denoted $\delta^+(v)$ to $v^+$. Finally, you add an edge from $v^-$ to $v^+$ with capacity equal to the capacity the vertex $v$ originally had. This way ...


2

Vertex coloring is useful in grouping. Suppose you have hazardous chemicals that cannot be stored together. What is the minimum number of storage units you need to store all your chemicals? Similarly, suppose you have a set of volunteers and certain volunteers who won't work with each other. How do you group them to minimize the number of groups? Dominating ...


2

Hint: Create $G'$ as follows: For any $w \in V$ create two vertices $w_\text{in}$ and $w_\text{out}$. Connect them with an edge of zero cost. For any $e \in E$ create two vertices $e_\text{in}$ and $e_\text{out}$. Connect them with an edge of zero cost. For any $e = (w \to w') \in E$ add edges $\{w_\text{out},e_\text{in}\}$ and $\{e_\text{out}, ...



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