Bellman-Ford – is number of interations greater than diameter?

Diameter of a connected, undirected graph is the smallest natural number d, so that between any two vertices of the graph exist path of length at most d.

Prove or disprove: in Bellman-Ford is the number of iterations always equal or lower than d.

I’m trying to solve this issue. What I tried was sketching a lot of graphs, however I have failed to find a single graph where the number of iterations would be higher than the diameter.

The only graph where the number of iterations wouldn’t be <= than diameter would be a graph with negative edges, however I found out that in undirected graph there can’t be any negative edges, otherwhere there would be a negative cycle.

So, AFAIK the statement is correct. However, how would I prove such a statement? I don’t even know how to start. Thanks for any help

Calculation of the diameter of a connected network in the CONGEST model

I need to show that in the CONGEST model, the diameter of a (connected) network can be calculated in $ O(m) $ rounds, with $ m $ indicating the number of edges of the network.

My idea would be the following: First a leader is determined by using the radius-growth algorithm which has a runtime of $ O (n) $ rounds. Because $ n \leq m – 1$ , that works.

After the leader was determined, the leader starts a BFS, which needs $ O(D)$ rounds (with $ D $ being the diameter of the network, which is at that point unknown, but of course $ D \leq m $ ). Then we can determine the node with the highest distance from the leader in $ O(D) $ rounds, let’s call this node $ a $ . Then we start another BFS from the node $ a $ and, again, determine the node with the highest distance from $ a $ , let’s call this node $ b $ . Then $ dist(a, b)$ should be the diameter $ D $ of the network.

The runtime seems to be okay, so it would be possible in $ O (m) $ rounds with this approach. The only problem I have is that I am not sure if this approach is correct, at least I am not able to prove it formally. I need to prove, if the $ dist(a, b) = D$ , then the two BFS find the nodes $ a $ and $ b$ . My idea would be that if the leader, which is determined by using the radius-growth algorithm, is either the node $ a $ or $ b $ , then trivially it would be correct. If it is neither $ a $ nor $ b $ (nor any other node at the ‘start’ or ‘end’ of a path with length $ D $ ), then during the first BFS we would at some point reach the path which represents the diameter. If without loss of generality the node $ b $ has a higher distance to the leader, then the leader would find $ b$ after the first BFS, because if it would find another node (let’s call this node $ x $ ), then $ dist(a, b) \neq D $ , because $ dist(a, x) $ would be the diameter and not $ dist(a, b)$ which would be a contradiction.

Is that right?

If a circle with a diameter of 12 contains a chord of length 6 sqrt2, what is the length of the minor arc intercepted by the chord?

I’ve been stuck on this question for a while now… the main problem is that I don’t know the way I should draw it.

Should I draw it so that the diameter is perpendicular to the chord (so it can bisect the chord & its arc), or should I create an inscribed angle with the chord and diameter?

I tried both ways and I still have no idea how to even get an answer. Any help is much appreciated!

Minimum diameter spanning tree problem

Minimum diameter spanning tree (MDST) problem is defined as following: given the connected weighted graph $ G(V, E)$ , weight function $ w: E \rightarrow R, w(e) > 0\ \forall e \in E$ , find the spanning tree $ T(V, E’)$ of $ G$ such as maximal distance between vertices in $ T$ is minimal.

I’ve done my research and haven’t found much literature on this particular problem (I’ve seen Euclidean MDST and bounded MDST, but not MDST itself) and wondered if there’s any good explanation of polynomial solution to that problem. Also, I saw a theorem stating that $ \Theta(n^3)$ solution exists here, but no full solution is anywhere.

I believe this community would benefit if solution is described here.

Diameter for permutations of bounded support

Let $ S\subset \textrm{Sym}(n)$ be a set of permutations each of which is of bounded support, that is, each $ \sigma\in S$ moves $ O(1)$ elements of $ \{1,2,\dotsc,n\}$ . Let $ \Gamma$ be the graph whose set of vertices is the set of ordered pairs $ (i,j)$ , $ 1\leq i,j\leq n$ , $ i\ne j$ , and whose edges are $ ((i,j), (i,j)^\sigma)$ , $ \sigma\in S$ . (In other words, $ \Gamma$ is a Schreier graph.) Assume $ \Gamma$ is connected, i.e., $ S$ generates a $ 2$ -transitive group.

Must it be the case that $ \textrm{diam}(\Gamma) = O(n)$ ? Could it be the case that $ \textrm{diam}(\Gamma)\gg n^{1+\delta}$ , $ \delta>0$ , or even $ \delta=1$ ?

What are the answers to these questions if we assume that $ \langle S\rangle$ is all of $ \textrm{Alt}(n)$ or $ \textrm{Sym(n)}$ ?

Where can I find lens mounts for small 2 mm diameter lenses?

I need a lens mount for one of my projects. I have a CCD chip, a single lens (2 mm diameter), and I know the mechanical position and tolerances. I would like to find a lens mount for that lens or see whether I can take advantage of some existing small lens mounts, like the Raspbery Pi camera module lens mount. Here are my questions:

What type of lens mount is used on the Raspbery Pi camera module? It seems that the Infenion Tof camera uses the same lens mount.

Which company produces these lens mounts and from which vendor can one buy these small or similar lens mounts (like for mobile phones or other compact devices) that attach to a PCB?

The smallest lens mounts I found was an S Mount M12x0.5 type. For test purposes I bought a couple of female S type lens holder for webcams and then screwed in some of my small S Type lenses. Mechanical play here was too big. I had expected a very tight fit since these are opto-mechanical components, but the lens could still move even when almost fully screwed in.

If $(N,\|.\|)$ is a non trivial vector space with norm, then the diameter of the unit sphere is $\ge2$

It is easy to see that $ \delta(B_1(0))\le 2$ because $ $ \|x-y\|\le\|x\|+\|-y\|=\|x\|+|-1|\|y\|=\|x\|+\|y\|<1+1=2$ $ so $ \sup\{\|x-y\|:x,y\in B_1(0)\}\le 2$

But to show that $ \delta(B_1(0))\ge 2$ we chose $ a\in B_1(0)\backslash\{0\}$ and $ s\in[0,1)$ and observe that $ \|s\frac{a}{\|a\|}\|=s$ $ \implies s\frac{a}{\|a\|},-s\frac{a}{\|a\|}\in B_1(0)$ and $ \|s\frac{a}{\|a\|}-(-s\frac{a}{\|a\|})\|=2s$

But why does that imply that $ \delta(B_1(0))\ge2s$ ?

This is how I think I understand it: the diameter of $ B_1(0)$ is greater than that of $ B_s(0)$ for all $ s\in[0,1)$ because we it contains it. And the diameter of $ B_s(0)$ is $ 2s$ for each $ s$ , which is true because there exists a sequence $ (\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})$ of pairs of points, with $ 0\le\epsilon_n<s~\forall n$ and $ \lim\limits_{n\rightarrow\infty}\epsilon_n=s$ such that it is always contained in $ B_s(0)$ and $ \lim\limits_{n\rightarrow\infty}d(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})=2s$

I’m not sure if I got it right…

What is the relationship between lens diameter and exposure?

All of the online calculators for exposure, I have seen, factor in time, film speed, and appature (which is a function of the real appature diameter and the focal length of the lens)

However I would have thought a lens with a larger diameter (with the same focal length and appature) would collect more light and therefore be faster!

Am I just looking at rubbish exposure calculators, or am I missing something?