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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 ...


5

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

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

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

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

Counterexample to your conjecture:


2

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


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

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

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

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

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

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

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

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

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

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

Great question. Try this: R.T. Rockafellar, Network Flows and Monotropic Optimization, J. Wiley, (1984). Rockafellar is a reknowned expert in optimisation. His books and materials are excellent examples of clarity and build quickly from an overview to advanced topics in network theory. If you go to his website, he has kindly made the full text of this ...


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

Showing (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i) is sufficient. So, you can prove (i) $\Rightarrow$ (ii) by contraposition, NOT (ii) $\Rightarrow$ NOT (i). That is, suppose $T$ doesn't have the cut property. Then you can find $e' \notin T$ in a cut defined by some $e \in T$ such that $c(e') < c(e)$. But adding $e'$ to $T$ creates ...


1

A fundamental theorem of graph theory flow is the Max-Flow/Min-Cut theorem, which states that if you can find a cut whose capacity is equal to any valid flow, then the flow is a maximum and the cut is a minimum. A cut is a partition of the vertexes of the graph into 2 sets, where the sink is in one set and the source is in the other, and both sets are ...


1

Put it each column the corresponding variable, last column is used for constants. So rewrite each equation in the form $...=0$. Then, for $A$ we get $x_1-x_2+x_4=0$, $B$: $x_2-x_3-100=0$, $C$: $x_3-x_4+80=0$. Put that in a matrix: $ \begin{pmatrix} 1 & -1 & 0 & 1 & 0\\ 0 & 1 & -1 & 0 & -100\\ 0 & 0 & 1 & -1 & ...


1

Is it possible this is a trick question? The entering basic variable is a nonbasic variable that you increase from $0$, while keeping the other nonbasics constant. Doing this will turn it into a basic variable for the next iteration. The leaving basic variable is determined by the minimum ratio test. This is determined by which basic variable drops to $0$ ...


1

There seems to be some small confusion here. Your primal is the path-flow formulation of the multicommodity flow problem, which orders separate path-flow to every path in some given path set $P$ (which in turn can be of exponential size), while your dual is in fact the dual linear program corresponding to the arc-flow formulation (and hence has polynomial ...


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 ...



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