## reference request – Counting unions of unlabelled connected graphs

My question can be stated as follows: let $$X$$ be a hereditary family of unlabelled graphs closed under disjoint unions. Suppose we know, for each $$n$$, the number $$c_n$$ of connected graphs in X on $$n$$ vertices. How do we get from that the total number $$x_n$$ of graphs in $$X$$ on $$n$$ vertices?

I am vaguely aware that there are methods involving generating functions that answer (some version of) this question, but the only source I found for that online is a transcription of a set of lecture notes with no references whatsoever. I am not particularly well versed in the theory of generating functions, so any source which goes from the basics to the relevant derivation would be greatly appreciated.

However, if it is at all possible, I would ideally like to find an explicit formula for the coefficients $$x_n$$ in terms of the $$c_i$$, or at least a (tight-ish) lower bound for $$x_n$$ in terms of the $$c_i$$.

I can see the answer in some very simple cases: if, say, $$c_i = 1$$ for all $$i$$ (e.g., $$X$$ is the class of unions of cliques), then $$x_n$$ is just the partition function $$p(n)$$ from number theory (compare this to the labelled case, where we get the Bell numbers). However, I do not see how to generalise this. It seems to me very likely that this question would have been studied in the language of multisets, but once again, I am not particularly familiar with that area.

## c# – How to get ipv4 address exactly of another device connected same Wifi network that the client ID was known

How do I get ipv4 of another device connected same wifi network?
I used below code (C# – Unity):

``````string ipv4 = "";
string hostName = "anp.local";
var host = Dns.GetHostEntry(hostName);
{
Debug.Log("IP: " + ip.ToString());
{
ipv4 = ip.ToString();
}
}
``````

I got a wrong result: 125.235.4.59 on Window and Android.
Expect result should be 192.168.1.100 (work on Ipad)
My goal is able to connect another device in same wifi network on Window/Android/IOS platforms and I planed using ipv4 for it.

## solution verification – Any finite connected graph with every vertex has degree \$ge 2\$ has a circuit

Is my proof for the following statement correct?

Any finite connected graph with every vertex has degree $$ge 2$$ has a
circuit.

My attempt: Let $$G$$ be a finite connected graph. Let $$|G|=n$$. Suppose that degree of any vertex $$ge 2$$. Now, pick any vertex $$v_1$$. By hypothesis $$v_1$$ must have at least two distinct edges incident on it; pick one, call it $$e_1$$. Call the end vertex of $$e_1$$ different from $$v_1$$ as $$v_2$$. Now pick an edge $$e_2$$ different from $$e_1$$ incident on $$v_2$$. Let $$v_3$$ be the end vertex of $$e_2$$ different from $$v_2$$. If there is an edge joining $$v_3$$ and $$v_1$$, we are done. If not, pick any edge $$e_3$$ different from $$e_2$$ incident on $$v_3$$ and repeat the previous argument. Now, we proceed by induction and find vertices $${v_1 , v_2, ldots, v_n , v_{n+1} }$$ such that there is an edge between $$v_i$$ and $$v_{i+1}$$. Since $$G$$ is connected, the component of $$G$$. containing $$v_1$$ which is a superset of $${ v_1, v_2, ldots , v_{n+1} }$$ is equal to $$G$$. So, $$v_{n+1}=v_1$$ and hence we are done.

Is this proof correct? Is there an easier way to do this?

## For an infinite n-dimensional discrete grid. Each cell has probaility \$p\$ of being on. No infinite large connected piece?

A piece is called connected if it’s made up of on cells, and let `C` be the set of on cells in the piece, and $$forall c in C exists d in C$$ such that $$c$$ and $$d$$ are directly connected, that is their coordinates are the same, except for one value which differs by one.

The fact that there cannot be a connected piece of infinite size seems apparent to me. But it was debated. We agreed that the 1D case is clearly true i.e. you can’t have a piece of infinite size. And the “curse of dimensionality” kinda rings a bell here since the higher the dimension, the more sparse your grid and hence even less likely to have a connected piece.

So I am seeking proof or a theorem from somewhere that can settle this once and for all.

## MacBook Pro battery drains when connected to AC power

I’ve been experiencing a very annoying problem lately. Often, I’ve got my MacBook Pro connected to power when I’m working from the office. At home, I often leave it unplugged until the battery is too low to continue at which point I plug it in.

The problem occurs when my device is connected to a power supply. My battery indicator shows 100% charge which is fine. But sometimes the battery starts draining and the menu bar icon says “Battery not charging,” even when it’s connected to a wall socket, and it keeps draining until absolutely dry. It doesn’t even warn me the battery is running low because it shows “Power source: Power adapter” which I guess disables the Operating system low-battery warning. This has happened on multiple occasions.

I’ve already tried several things, I’ve swapped chargers, swapped wall sockets, started with as few tools/apps running as possible. But the problem comes back. I also can’t really tie this to a specific app or behavior, AFAIK.

Does anyone know what might be causing this behavior and what to possibly do about it?

## macbook pro – How to disable the “Your trackpad is wirelessly connected to this Mac” notification?

So I have a 2019 Macbook Pro connected to a Caldigit TS3 Plus dock. Connected to that I have a Magic Trackpad 2 via USB cable (to charge it). It is a Bluetooth device. Every time I connected my MBP to this docking station I get this stupid notification:

And I can’t for the life of me figure out how to disable this. There’s no Bluetooth setting for the device. Clicking on “Settings” takes you to the Bluetooth System Preferences. There’s nothing I can see under Notifications Systems Preferences that pertains to this.

Now I should also mention that I connect TWO different MBPs to this same docking station and it seems like only when I change laptops does this come up but I’m not 100% sure about that (as in, I’m not 100% certain it doesn’t show up when I power the same laptop on or wake it up).

I’m actually shocked that a Google search for “Your trackpad is wirelessly connected to this Mac” (including quotes) has exactly 0 matches.

## My Bose QuietComfort 35 II headphones are connected to Windows 7 and an iPad: how can I prevent Windows 7 from muting the audio coming from the iPad?

My Bose QuietComfort 35 II headphones are connected to Windows 7 and some iPad (iPad Pro MM172LL/A): how can I prevent Windows 7 from muting the audio coming from the iPad? Even when no audio is played on Windows 7, I have to disconnect the Bose QuietComfort 35 II headphones from Windows 7 to be able to hear the iPad.

## bitcoind – Bitcoin V0.1 Not Connected

You cannot.

There have been both changes to the P2P protocol that prevent this, and changes to the way that nodes are discovered.

The method which 0.1.0 discovered new nodes was by joining the #bitcoin channel on freenode. Your IP address would be encoded in a certain way and that would be the IRC nick for your node. However this has long since been removed from the Bitcoin software and any node that does connect to the #bitcoin channel will be kicked from it (the encoding is unique and identifiable).

Since 0.1.0 can’t connect to the IRC channel to discover new nodes, it is unable to learn about nodes that it can connect to.

Furthermore, the P2P protocol has changed since 0.1.0 was released and thus 0.1.0 is unable to communicate with modern nodes. This change was to add a checksum to the end of the P2P message header. Messages without this checksum will be rejected by modern nodes.

Lastly, 0.1.0 is too old so modern nodes won’t connect to it. They have a minimum supported protocol version number, and 0.1.0’s protocol version number is far too low. So even if you could get it to connect, the connection would be dropped.

The way to deal with the node discovery issue is to manually modify the peers.dat file to have the IP address and port of nodes that you want to connect to.

The only way to deal with the P2P protocol changes (checksum and version number) is to have a shim that sits in between all of your connections and modifies the P2P messages to work with modern node software. Or you can connect to a modified node which can deal with 0.1.0 traffic.

Even if you are able to connect to the Bitcoin network with 0.1.0, you won’t be able to mine anything as the CPU mining is no longer viable, and the CPU miner implementation in 0.1.0 is not efficient. You also wouldn’t be able to even sync as it syncs too slowly. Blocks will be found faster than it is able to process them. It will run into additional issues when it receives a new block before it is able to sync the blockchain that precedes that block.

## networking – How to configure a optical fiber router modem to allow port forwarding with other connected router/or from main router itself?

I recently got a new fibre connection, the ISP provided me with a wifi modem, although I have a router. Anyway back to the point, I was able to port forward with my old non-wifi modem and router set up
but now I can’t port forward. I have searched the internet fully and I found almost the same situation like this one – How to configure cable modem router to allow port forwarding to other connected router?
But this also did not help me, I hope anyone would help me in this topic relatively soon

## networking – Can I connect a VOIP phone to the internet from a laptop that is connected to the internet wirelessly?

Can I connect a VOIP phone, that does not have wireless capability, to the internet by using an ethernet cable to the phone from a laptop that is connected to the internet wirelessly? My home office is not hard-wired, so both my laptop and printer work off of wifi. There is a more expensive version (Polycom VVX250) of my VOIP phone that does have a USB port allowing a wifi card to be inserted, but my phone (Polycom VVX150) does not have that port. I really don’t want to return the phone I have and spend more money for that functionality if I don’t have to because we will be moving into an office in a few months that will be hard-wired so this won’t be an issue. Thoughts anyone? Thanks in advance.