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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 ...


6

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).


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 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

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

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

Counterexample to your conjecture:


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

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 ...


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}, ...


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

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 ...


2

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 ...


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

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 ...


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

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

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. ...


2

A maximum flow is a flow that attains the highest flow value possible for the given network. A maximal flow is a flow whose value cannot be increased without decreasing the flow along some arc. All maximum flows are maximal flows. Not all maximal flows are maximum flows. Figure 3.9 in the Bang-Jensen and Gutin textbook, first edition illustrates an ...


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

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


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

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 ...


2

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 ...


1

I found the answer to this in the book "Probability on Trees and Networks" by Russel Lyons and Yuval Peres in Chapter 2 in the Section 2.4 on energy. The proof does indeed come "easily" after a bit of overhead with some very useful definitions and results made in this book.


1

I hesitate to write this up as an answer, but I guess I don't have the 'comment' privilege yet. These types of issues are important to understand, as they force you to understand the problem in more depth and come up with a model that accurately describes it without too much fluff. Anytime you encounter a case that breaks your model, you can try to add ...


1

You misunderstood the result. Whether a flow is possible or not is not relevant. In your first case, where the graph is $ s \to a \, \ t $ the minimal cut (=partition of the nodes into two sets, one containing s and the other t) is ({s,a}, {t}) which has capacity 0 because there are no edges from one set to the other. The other cut ({s},{a,t}) has ...


1

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 ...



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