Feed Experiment

@rolfl This one could be because the username has "lol" in it. I don't know, but when I was experimenting it wasn't usual for normal search feeds to snowball like that. However, you get runaway feeds if you search for e.g. "stack" or anything else that's part of the url.

I tested this here a while ago:

chat.stackexchange.com/rooms/10121/test-room

rolfl

Hmm... could be... can be.. let's see.

12 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

another 18 minutes before it loops... so, I hope you have some patience.

Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

16 mins ago, by Jeremy Banks

You can, if you want.

in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

15 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

15 mins ago, by Stack Exchange

1 min ago, by Doorknob

Okay, let's see what can happen

15 mins ago, by Stack Exchange

2 mins ago, by Jeremy Banks

can-a-da

15 mins ago, by Stack Exchange

2 mins ago, by rolfl

can

15 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 3 mins ago, by RPi Awesomeness

I can't quite figure it out - it's one of the assignments for the chapter.

15 mins ago, by Doorknob

I can see it

15 mins ago, by Jeremy Banks

Ahh -- the messages are onboxed, so they don't directly contain the search string, so the search can't find them.

14 mins ago, by Jeremy Banks

@rolfl Can it? I'm probably being confused by caching, then, after not seeing them here.

in Ask Ubuntu General Room, 8 mins ago, by Nathan Osman

If you have g++-multilib installed, you can pass -m32 to the compiler to build for i386 on an amd64 machine.

in Ask Ubuntu General Room, 7 mins ago, by RPi Awesomeness

I can dream of unsigned long longs and of overloaded functions

in The h Bar, 5 mins ago, by Terry Bollinger

Passive aggression in the sense that if someone who has no understanding of a topic starts editing a piece on e.g. some obscure aspect of the standard model written by a genuine expert on that topic, the "winner" is always the one who can keep from blowing his or her stack -- which, surprise surprise, is usually the dufus who did not really understand or care in the first place, but knows that Wikipedia severely punishes losing your cool.

in Root Access, 4 mins ago, by Journeyman Geek

I'll need to work out what I need so I can keep myself fed. Roomie might be boarded, but its not confirmed eyt

3 mins ago, by rolfl

Hmm... could be... can be.. let's see.

3 mins ago, by rolfl

12 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

rolfl

2:38 AM

well, it's picking it's own messages up.

Logan M

Yes, it does seem that way.

rolfl

also, the the timing is odd... it was almost 30 minutes the last time.. this time only 15.

Logan M

It isn't consistent. I've seen everywhere from 6 minutes to hours between posts. I don't know what causes it to go faster or slower.

Hosch250

What's all this about?

Sorry if I interrupted the mod-fest.

rolfl

Hey Hosch - experimenting "officially" with the feed recursion.

Hosch250

2:41 AM

What feed recursion?

rolfl

in The 2nd Monitor, 1 hour ago, by Mat's Mug

5 mins ago, by Mat's Mug

it's for science

in The 2nd Monitor, 1 hour ago, by rolfl

OMG, there's an RSS feed for search results in the chat room....

Logan M

This is a bit unusual. I was fairly sure that when I tested, the search feeds went by the post's underlying text, not the displayed message. But clearly that isn't the case right now, so either I remembered that incorrectly or it's changed.

Hosch250

Lol.

rolfl

The real question is, do you spend your time waiting for feeds to post, or do you brush your teeth, etc.

Hosch250

Both?

rolfl

2:55 AM

18 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

Stack Exchange

in The 2nd Monitor, 25 mins ago, by Jamal

But a mod can convert an answer to a comment, regardless of the post owner's rep.

24 mins ago, by Stack Exchange

3 mins ago, by rolfl

12 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

24 mins ago, by Stack Exchange

3 mins ago, by rolfl

Hmm... could be... can be.. let's see.

24 mins ago, by Stack Exchange

in Root Access, 4 mins ago, by Journeyman Geek

I'll need to work out what I need so I can keep myself fed. Roomie might be boarded, but its not confirmed eyt

24 mins ago, by Stack Exchange

in The h Bar, 5 mins ago, by Terry Bollinger

Passive aggression in the sense that if someone who has no understanding of a topic starts editing a piece on e.g. some obscure aspect of the standard model written by a genuine expert on that topic, the "winner" is always the one who can keep from blowing his or her stack -- which, surprise surprise, is usually the dufus who did not really understand or care in the first place, but knows that Wikipedia severely punishes losing your cool.

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 7 mins ago, by RPi Awesomeness

I can dream of unsigned long longs and of overloaded functions

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 8 mins ago, by Nathan Osman

If you have g++-multilib installed, you can pass -m32 to the compiler to build for i386 on an amd64 machine.

24 mins ago, by Stack Exchange

14 mins ago, by Jeremy Banks

@rolfl Can it? I'm probably being confused by caching, then, after not seeing them here.

24 mins ago, by Stack Exchange

15 mins ago, by Jeremy Banks

Ahh -- the messages are onboxed, so they don't directly contain the search string, so the search can't find them.

24 mins ago, by Stack Exchange

15 mins ago, by Doorknob

I can see it

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 3 mins ago, by RPi Awesomeness

I can't quite figure it out - it's one of the assignments for the chapter.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by rolfl

can

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by Jeremy Banks

can-a-da

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Doorknob

Okay, let's see what can happen

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

in Mathematics, 18 mins ago, by Soham Chowdhury

Can I prove that or does it take high-powered tools I don't have?

in Mathematics, 17 mins ago, by anon

oh no you can prove it

in Mathematics, 7 mins ago, by Soham Chowdhury

Is it okay to still be in shock that such a simple fact (which I can understand) is so connected to so many other things?

5 mins ago, by rolfl

18 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

in The 2nd Monitor, 1 min ago, by QPaysTaxes

Aaagh I can't find an equivalent method to Hash#sort in the Ruby 2.2.0

in Game Development, 40 secs ago, by Mutch95

Can a method return 2 variables?

Anonymous

3:11 AM

@rolfl It can possibly be due to due to oddly overlapping cache intervals.

Anonymous

Are you at all tempted to add multiple similar feeds to the room? can

rolfl

that would compound it ...

Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Root Access, 4 mins ago, by Journeyman Geek

I'll need to work out what I need so I can keep myself fed. Roomie might be boarded, but its not confirmed eyt

12 mins ago, by Stack Exchange

in Game Development, 40 secs ago, by Mutch95

Can a method return 2 variables?

12 mins ago, by Stack Exchange

in The 2nd Monitor, 1 min ago, by QPaysTaxes

Aaagh I can't find an equivalent method to Hash#sort in the Ruby 2.2.0

12 mins ago, by Stack Exchange

5 mins ago, by rolfl

18 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

12 mins ago, by Stack Exchange

in Mathematics, 7 mins ago, by Soham Chowdhury

Is it okay to still be in shock that such a simple fact (which I can understand) is so connected to so many other things?

12 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by anon

oh no you can prove it

12 mins ago, by Stack Exchange

in Mathematics, 18 mins ago, by Soham Chowdhury

Can I prove that or does it take high-powered tools I don't have?

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Doorknob

Okay, let's see what can happen

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by Jeremy Banks

can-a-da

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by rolfl

can

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 3 mins ago, by RPi Awesomeness

I can't quite figure it out - it's one of the assignments for the chapter.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Doorknob

I can see it

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Jeremy Banks

Ahh -- the messages are onboxed, so they don't directly contain the search string, so the search can't find them.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

14 mins ago, by Jeremy Banks

@rolfl Can it? I'm probably being confused by caching, then, after not seeing them here.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 8 mins ago, by Nathan Osman

If you have g++-multilib installed, you can pass -m32 to the compiler to build for i386 on an amd64 machine.

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 7 mins ago, by RPi Awesomeness

I can dream of unsigned long longs and of overloaded functions

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in The h Bar, 5 mins ago, by Terry Bollinger

Passive aggression in the sense that if someone who has no understanding of a topic starts editing a piece on e.g. some obscure aspect of the standard model written by a genuine expert on that topic, the "winner" is always the one who can keep from blowing his or her stack -- which, surprise surprise, is usually the dufus who did not really understand or care in the first place, but knows that Wikipedia severely punishes losing your cool.

in Game Development, 10 mins ago, by MLM

Also see if you can refactor to avoid the situation

in Mathematics, 10 mins ago, by Soham Chowdhury

@anon, can the composition of two set-functions $\phi: G\rightarrow H, \varphi: H\rightarrow K$ be a homomorphism if neither of $\phi, \varphi$ are?

in Mathematics, 8 mins ago, by Alex Clark

@Anon In $\Bbb Z[x]$ all elements of the ideal $\langle p \rangle$ are of the form $rp$ for some $r\in\Bbb Z[x]$ since we are closed under ring multiplication and addition, and can always keep that $p$ factor out.

Now to prove that $\langle p \rangle$ is a prime ideal, I know that $ab\in\langle p\rangle\implies ab=rp$ for some $r\in\Bbb Z[x]$

in Mathematics, 4 mins ago, by Alex Clark

I can't get from $ab\in(p)\implies ab=rp$ to $a\in(p)$ or $b\in(p)$?

in Mathematics, 3 mins ago, by anon

if you really want to do nuts and bolts, you can write $a(x)b(x)=pc(x)$, compare leading coefficients and then induct on degree

2 mins ago, by Jeremy Banks

@rolfl It can possibly be due to due to oddly overlapping cache intervals.

1 min ago, by Jeremy Banks

Are you at all tempted to add multiple similar feeds to the room? can

rolfl

Yeah, like adding "Stack Exchange" as a search feed... these messages all show up ^^^

Also, it looks like it's started dropping off the RSS feed.

Stack Exchange

7 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Are you at all tempted to add multiple similar feeds to the room? can

7 mins ago, by Stack Exchange

2 mins ago, by Jeremy Banks

@rolfl It can possibly be due to due to oddly overlapping cache intervals.

7 mins ago, by Stack Exchange

in Mathematics, 3 mins ago, by anon

if you really want to do nuts and bolts, you can write $a(x)b(x)=pc(x)$, compare leading coefficients and then induct on degree

7 mins ago, by Stack Exchange

in Mathematics, 4 mins ago, by Alex Clark

I can't get from $ab\in(p)\implies ab=rp$ to $a\in(p)$ or $b\in(p)$?

7 mins ago, by Stack Exchange

in Mathematics, 8 mins ago, by Alex Clark

@Anon In $\Bbb Z[x]$ all elements of the ideal $\langle p \rangle$ are of the form $rp$ for some $r\in\Bbb Z[x]$ since we are closed under ring multiplication and addition, and can always keep that $p$ factor out.

Now to prove that $\langle p \rangle$ is a prime ideal, I know that $ab\in\langle p\rangle\implies ab=rp$ for some $r\in\Bbb Z[x]$

7 mins ago, by Stack Exchange

in Mathematics, 10 mins ago, by Soham Chowdhury

@anon, can the composition of two set-functions $\phi: G\rightarrow H, \varphi: H\rightarrow K$ be a homomorphism if neither of $\phi, \varphi$ are?

7 mins ago, by Stack Exchange

in Game Development, 10 mins ago, by MLM

Also see if you can refactor to avoid the situation

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in The h Bar, 5 mins ago, by Terry Bollinger

Passive aggression in the sense that if someone who has no understanding of a topic starts editing a piece on e.g. some obscure aspect of the standard model written by a genuine expert on that topic, the "winner" is always the one who can keep from blowing his or her stack -- which, surprise surprise, is usually the dufus who did not really understand or care in the first place, but knows that Wikipedia severely punishes losing your cool.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 7 mins ago, by RPi Awesomeness

I can dream of unsigned long longs and of overloaded functions

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 8 mins ago, by Nathan Osman

If you have g++-multilib installed, you can pass -m32 to the compiler to build for i386 on an amd64 machine.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

14 mins ago, by Jeremy Banks

@rolfl Can it? I'm probably being confused by caching, then, after not seeing them here.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Jeremy Banks

Ahh -- the messages are onboxed, so they don't directly contain the search string, so the search can't find them.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Doorknob

I can see it

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 3 mins ago, by RPi Awesomeness

I can't quite figure it out - it's one of the assignments for the chapter.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by rolfl

can

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

2 mins ago, by Jeremy Banks

can-a-da

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Doorknob

Okay, let's see what can happen

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

1 min ago, by Jeremy Banks

Whatever it is, it can't be good.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 18 mins ago, by Soham Chowdhury

Can I prove that or does it take high-powered tools I don't have?

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by anon

oh no you can prove it

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 7 mins ago, by Soham Chowdhury

Is it okay to still be in shock that such a simple fact (which I can understand) is so connected to so many other things?

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

5 mins ago, by rolfl

18 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The 2nd Monitor, 1 min ago, by QPaysTaxes

Aaagh I can't find an equivalent method to Hash#sort in the Ruby 2.2.0

in The Pod Bay, 7 mins ago, by Vedant Chandra

@duzzy I think that if this group of people cannot come up with a functioning Hyperloop, then for the foreseeable future no one can.

in Ask Ubuntu General Room, 6 mins ago, by Nathan Osman

I'll see if I can get a shot inside.

Stack Exchange

4:15 AM

53 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 6 mins ago, by Nathan Osman

I'll see if I can get a shot inside.

53 mins ago, by Stack Exchange

in The Pod Bay, 7 mins ago, by Vedant Chandra

@duzzy I think that if this group of people cannot come up with a functioning Hyperloop, then for the foreseeable future no one can.

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The 2nd Monitor, 1 min ago, by QPaysTaxes

Aaagh I can't find an equivalent method to Hash#sort in the Ruby 2.2.0

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

5 mins ago, by rolfl

18 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 7 mins ago, by Soham Chowdhury

Is it okay to still be in shock that such a simple fact (which I can understand) is so connected to so many other things?

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by anon

oh no you can prove it

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 18 mins ago, by Soham Chowdhury

Can I prove that or does it take high-powered tools I don't have?

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

in Mathematics, 51 mins ago, by Soham Chowdhury

he went off because he can't stand you.

in Ask Ubuntu General Room, 48 mins ago, by Nathan Osman

iTunes never ran on 98, the last supported Firefox version was 2.x, and any executable can write to C:\Windows\System - no file permissions.

in The Bridge, 46 mins ago, by Lazers 2.0

0

Q: Minecraft Bukkit Code: Only mine certain spawners?

so I have this code here from the plugin http://pastebin.com/ES5qaCcs and I want to make it so players can only mine Zombie and Skeleton spawners with the wrench tool. I believe it is something to do with EntityType. but I have no idea how to implement it into the code. Also being able to m...

minecraft

in Genealogy Conference, 43 mins ago, by Jan Murphy

0

Q: Are you interested in Canadian genealogy and Family History?

If you have Canadian ancestors and some unsolved questions regarding their genealogy and your family history here's your chance to try and clear them up! The latest Topic Challenge, which will run for about two weeks comes from @JanMurphy: One thing I have neglected so far is research in Can...

discussion featured challenge

in Genealogy Conference, 39 mins ago, by Jan Murphy

Sorensen's presentation gave a good overview of what census records were created, which ones survive and which were lost, and where records and indexes can be accessed online or on microfilm.

in The Bridge, 38 mins ago, by Newly Opened Proposals for Entertainment

1

Open Study

Proposed Q&A site for anyone engaging in learning through either a specific textbook or a specific course. The goal is to create a more structured, organized, and searchable forum that can be used by students for years to come. Creation inspired by / for MOOC students

Currently in definition.

in The Bridge, 38 mins ago, by Lazers 2.0

0

Q: Wasteland 2: Can I walk while crouching?

In Wasteland 2 I often crouch all my team members to get a better position before a battle begins. Then I want to move a character closer and I click to move him. Is there a way to move a character without leaving crouch mode?

wasteland-2

in Ask Ubuntu General Room, 38 mins ago, by TheX

I think it is a CX650 (I can see the box but not the print from here)

in Ask Ubuntu General Room, 36 mins ago, by Nathan Osman

(I can never remember the names of the connectors.)

in The Bridge, 35 mins ago, by GodEmperorDune

-1

Q: How To Ask Questions on Open Study and who this is for

Proposal: Open Study This site was inspired by how horrendously organized http://openstudy.com/ is. Please, only ask questions on this site if it can be connected to either a specific textbook or a specific lesson / section from a specific class. If you have general purpose questions please use...

life

in Minecraft Talk, 33 mins ago, by Tim Post

@Ieuan Working at something you can at least tolerate while you build up your open source presence isn't a horribad idea.

in VBA Rubberducking, 33 mins ago, by Duga

> `RubberduckParser` is already logging parse errors in debug builds (logging is disabled in release builds). What would be nice, is if we would report parse exception details in a docked window that we could bring up and populate as the parser blows up, a bit like Visual Studio compile errors/warnings.

Let's implement a docked toolwindow that lets user navigate exactly where the parse errors are, and that can be exported to a text file, or into the clipboard.

This will be great for gramma

in The Bridge, 24 mins ago, by Lazers 2.0

1

Q: Can I put my party in a line automatically?

Before I enter a battle I often place my party members in a straight line. But it's a bit annoying doing it all by hand by individually selecting each character and putting them in a straight line one by one. Is there a way to put auto-align my party? Or put them in a pre set position with a hot...

wasteland-2

in RPG General Chat, 16 mins ago, by doppelgreener

@KorvinStarmast Editors of a post can be pinged in its comments. So, if it's truly awesome, use a comment, and then remember to delete it after a while.

in The Bridge, 16 mins ago, by Lazers 2.0

0

Q: How can I capture a zombie dwarf?

A dwarf died in my fortress and has turned into a zombie. I had placed a few cage traps around, but the zombie just walks pass it. I have a squad of hammer dwarves at the ready, but I would like to capture it instead of destroying it. How can I do that?

dwarf-fortress

in Ask Ubuntu General Room, 9 mins ago, by TheX

but anyone can buy a post-it note regardless of your operating system

in Mos Eisley, 8 mins ago, by Science Fiction & Fantasy

0

Q: In Battlestar Galactica, how does Earth 1 and Earth 2 make sense?

I'm a bit late - but if anyone can answer this I would be mighty thankful - Kara describes the planet Earth she found to Lee as Green and beautiful and blue, so how could it be the ruined Earth? Or am I remembering incorrectly? :) So is there an answer or did the writers mess up? I do not underst...

battlestar-galactica-2004

in Off the record, 6 mins ago, by Unihedron

I'm wondering if there can be an audio equivalent of reverse image search.

Anonymous

4:35 AM

room topic changed to Feed Experiment: [can]

Stack Exchange

22 mins ago, by Stack Exchange

in Off the record, 6 mins ago, by Unihedron

I'm wondering if there can be an audio equivalent of reverse image search.

22 mins ago, by Stack Exchange

in Mos Eisley, 8 mins ago, by Science Fiction & Fantasy

0

Q: In Battlestar Galactica, how does Earth 1 and Earth 2 make sense?

I'm a bit late - but if anyone can answer this I would be mighty thankful - Kara describes the planet Earth she found to Lee as Green and beautiful and blue, so how could it be the ruined Earth? Or am I remembering incorrectly? :) So is there an answer or did the writers mess up? I do not underst...

battlestar-galactica-2004

22 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 9 mins ago, by TheX

but anyone can buy a post-it note regardless of your operating system

22 mins ago, by Stack Exchange

in The Bridge, 16 mins ago, by Lazers 2.0

0

Q: How can I capture a zombie dwarf?

A dwarf died in my fortress and has turned into a zombie. I had placed a few cage traps around, but the zombie just walks pass it. I have a squad of hammer dwarves at the ready, but I would like to capture it instead of destroying it. How can I do that?

dwarf-fortress

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in RPG General Chat, 16 mins ago, by doppelgreener

@KorvinStarmast Editors of a post can be pinged in its comments. So, if it's truly awesome, use a comment, and then remember to delete it after a while.

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in The Bridge, 24 mins ago, by Lazers 2.0

1

Q: Can I put my party in a line automatically?

Before I enter a battle I often place my party members in a straight line. But it's a bit annoying doing it all by hand by individually selecting each character and putting them in a straight line one by one. Is there a way to put auto-align my party? Or put them in a pre set position with a hot...

wasteland-2

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in VBA Rubberducking, 33 mins ago, by Duga

> `RubberduckParser` is already logging parse errors in debug builds (logging is disabled in release builds). What would be nice, is if we would report parse exception details in a docked window that we could bring up and populate as the parser blows up, a bit like Visual Studio compile errors/warnings.

Let's implement a docked toolwindow that lets user navigate exactly where the parse errors are, and that can be exported to a text file, or into the clipboard.

This will be great for gramma

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in Minecraft Talk, 33 mins ago, by Tim Post

@Ieuan Working at something you can at least tolerate while you build up your open source presence isn't a horribad idea.

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in The Bridge, 35 mins ago, by GodEmperorDune

-1

Q: How To Ask Questions on Open Study and who this is for

Proposal: Open Study This site was inspired by how horrendously organized http://openstudy.com/ is. Please, only ask questions on this site if it can be connected to either a specific textbook or a specific lesson / section from a specific class. If you have general purpose questions please use...

life

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in Ask Ubuntu General Room, 36 mins ago, by Nathan Osman

(I can never remember the names of the connectors.)

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in Ask Ubuntu General Room, 38 mins ago, by TheX

I think it is a CX650 (I can see the box but not the print from here)

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in The Bridge, 38 mins ago, by Lazers 2.0

0

Q: Wasteland 2: Can I walk while crouching?

In Wasteland 2 I often crouch all my team members to get a better position before a battle begins. Then I want to move a character closer and I click to move him. Is there a way to move a character without leaving crouch mode?

wasteland-2

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in The Bridge, 38 mins ago, by Newly Opened Proposals for Entertainment

1

Open Study

Proposed Q&A site for anyone engaging in learning through either a specific textbook or a specific course. The goal is to create a more structured, organized, and searchable forum that can be used by students for years to come. Creation inspired by / for MOOC students

Currently in definition.

22 mins ago, by Stack Exchange

in Genealogy Conference, 39 mins ago, by Jan Murphy

Sorensen's presentation gave a good overview of what census records were created, which ones survive and which were lost, and where records and indexes can be accessed online or on microfilm.

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in Genealogy Conference, 43 mins ago, by Jan Murphy

0

Q: Are you interested in Canadian genealogy and Family History?

If you have Canadian ancestors and some unsolved questions regarding their genealogy and your family history here's your chance to try and clear them up! The latest Topic Challenge, which will run for about two weeks comes from @JanMurphy: One thing I have neglected so far is research in Can...

discussion featured challenge

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in The Bridge, 46 mins ago, by Lazers 2.0

0

Q: Minecraft Bukkit Code: Only mine certain spawners?

so I have this code here from the plugin http://pastebin.com/ES5qaCcs and I want to make it so players can only mine Zombie and Skeleton spawners with the wrench tool. I believe it is something to do with EntityType. but I have no idea how to implement it into the code. Also being able to m...

minecraft

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in Ask Ubuntu General Room, 48 mins ago, by Nathan Osman

iTunes never ran on 98, the last supported Firefox version was 2.x, and any executable can write to C:\Windows\System - no file permissions.

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in Mathematics, 51 mins ago, by Soham Chowdhury

he went off because he can't stand you.

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

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in Ask Ubuntu General Room, 16 mins ago, by Nathan Osman

You can just create both functions and the compiler will usually know which one to call based on what you pass to it.

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

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53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

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24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

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in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

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in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

in The 2nd Monitor, 11 mins ago, by Captain Obvious

0

Q: Crawling 2D Matrix

I'm trying to create a class that's able to 'crawl' through a generic 2D matrix. Crawling through a Bitmap (which can be viewed as a 2D Matrix of Colors) should yield the same result of a flood fill. So the first thing I did was create a IMatrix (previous name was I2DMatrix. I renamed it since i...

c#

in The 2nd Monitor, 11 mins ago, by Captain Obvious

-1

Q: Merge sort in Java

I tried to write a simple code in Java for Merge sort. (I am new in Java), but the code does not sort arrays. Can anyone help me what the problem is? Any improvements and comments appreciated. import java.util.Arrays; public class Methods { //* --- Merger Sort Recursive Helper ---- * p...

java

in Mathematics, 7 mins ago, by Soham Chowdhury

where can i learn a bit more?

in Mathematics, 6 mins ago, by anon

you can learn what coproducts and free products are on wikipedia

Stack Exchange

5:05 AM

27 mins ago, by Stack Exchange

in Mathematics, 6 mins ago, by anon

you can learn what coproducts and free products are on wikipedia

27 mins ago, by Stack Exchange

in Mathematics, 7 mins ago, by Soham Chowdhury

where can i learn a bit more?

27 mins ago, by Stack Exchange

in The 2nd Monitor, 11 mins ago, by Captain Obvious

-1

Q: Merge sort in Java

I tried to write a simple code in Java for Merge sort. (I am new in Java), but the code does not sort arrays. Can anyone help me what the problem is? Any improvements and comments appreciated. import java.util.Arrays; public class Methods { //* --- Merger Sort Recursive Helper ---- * p...

java

27 mins ago, by Stack Exchange

in The 2nd Monitor, 11 mins ago, by Captain Obvious

0

Q: Crawling 2D Matrix

I'm trying to create a class that's able to 'crawl' through a generic 2D matrix. Crawling through a Bitmap (which can be viewed as a 2D Matrix of Colors) should yield the same result of a flood fill. So the first thing I did was create a IMatrix (previous name was I2DMatrix. I renamed it since i...

c#

27 mins ago, by Stack Exchange

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

27 mins ago, by Stack Exchange

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

27 mins ago, by Stack Exchange

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

in The Overlook Hotel, 22 mins ago, by Stack Exchange

0

Q: What is the correct formatting for actions taken before dialogue?

What would be the correct way to format the following exchange in narrative prose? Is it: Alice sits on the bench, silent. Eventually, I speak. "What are we going to do?" After a long pause, she responds. "I don't think we can do anything." Or is it: Alice sits on the bench, silent. Eventuall...

formatting dialogue

in Mathematics, 20 mins ago, by anon

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

in TeX, LaTeX and Friends, 18 mins ago, by Joseph Wright

@SeanAllred The copyright holder can license code however they like. For example, one sees GPL code that is also available in proprietary versions. Provided all contributors accept this when they add stuff that is fine.

in Mathematics, 17 mins ago, by Soham Chowdhury

@anon i'd prefer the comments section were. one can dream.

in Mathematics, 17 mins ago, by Soham Chowdhury

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

in Mathematics, 15 mins ago, by Paul Plummer

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

in Mathematics, 15 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

in Mathematics, 14 mins ago, by Soham Chowdhury

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

in The 2nd Monitor, 14 mins ago, by QPaysTaxes

I think it can be simplified to just a[1] <=> b[1]

in The 2nd Monitor, 12 mins ago, by QPaysTaxes

Oh looks like it can

in Mathematics, 9 mins ago, by Soham Chowdhury

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

in Mathematics, 5 mins ago, by Paul Plummer

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

in Discussion between AjmalW and Brian Tung, 2 mins ago, by AjmalW

@BrianTung Thanks for your comments. 1) If $\mu(x,x_{0})>10$ it requires that $n$ should be greater than 10. Since $n=11$ is the minimum value in this case, so there must be $\mu(x,x_{0})=11$. 2) Rewriting it, we get $1/\mu(x,z)\leq \max\left \{ 1/\mu(x,y),1/\mu(y,z) \right \}$ which implies $\mu(x,z)\geq \max\left \{ \mu(x,y),\mu(y,z) \right \}$. So I have to show that $\mu(x,z)\geq \mu(x,y)$ or $\mu(x,z)\geq \mu(y,z)$. I think I can show it if $x\leq y\leq z$ is assumed. Am I right?

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

As regards Problem $2$, rewrite the expression in terms of $\mu(\cdot, \cdot)$ and see how far you can get.

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

Do you understand intuitively what $\mu(x, y)$ represents? It's the minimum $n$ such that the $n$th element of $x$ differs from the $n$th element of $y$. In other words, if $\mu(x, y) = k$, then $x$ and $y$ agree through the first $k-1$ bits (assuming $\mathbb{N}$ starts with $1$), and then differ for the first time at the $k$th bit. So if $x_0$ is an infinitely long sequence of zeros, and $\mu(x, x_0) > 10$, what can you say about the first ten bits of $x$?

in Mathematics, 1 min ago, by Soham Chowdhury

because you can't reduce the word $gh$ any further.

in Mathematics, 58 secs ago, by anon

you mean $g$ and $h$ can't have a nontrivial relation in $K$? or in $G*H$?

in Mathematics, 52 secs ago, by Soham Chowdhury

i don't think they can in the latter

in Mathematics, 25 secs ago, by Paul Plummer

If you can't reduce a word any further then it is not trivial...

Stack Exchange

5:21 AM

15 mins ago, by Stack Exchange

in Mathematics, 25 secs ago, by Paul Plummer

If you can't reduce a word any further then it is not trivial...

15 mins ago, by Stack Exchange

in Mathematics, 52 secs ago, by Soham Chowdhury

i don't think they can in the latter

15 mins ago, by Stack Exchange

in Mathematics, 58 secs ago, by anon

you mean $g$ and $h$ can't have a nontrivial relation in $K$? or in $G*H$?

15 mins ago, by Stack Exchange

in Mathematics, 1 min ago, by Soham Chowdhury

because you can't reduce the word $gh$ any further.

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

Do you understand intuitively what $\mu(x, y)$ represents? It's the minimum $n$ such that the $n$th element of $x$ differs from the $n$th element of $y$. In other words, if $\mu(x, y) = k$, then $x$ and $y$ agree through the first $k-1$ bits (assuming $\mathbb{N}$ starts with $1$), and then differ for the first time at the $k$th bit. So if $x_0$ is an infinitely long sequence of zeros, and $\mu(x, x_0) > 10$, what can you say about the first ten bits of $x$?

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

As regards Problem $2$, rewrite the expression in terms of $\mu(\cdot, \cdot)$ and see how far you can get.

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by AjmalW

@BrianTung Thanks for your comments. 1) If $\mu(x,x_{0})>10$ it requires that $n$ should be greater than 10. Since $n=11$ is the minimum value in this case, so there must be $\mu(x,x_{0})=11$. 2) Rewriting it, we get $1/\mu(x,z)\leq \max\left \{ 1/\mu(x,y),1/\mu(y,z) \right \}$ which implies $\mu(x,z)\geq \max\left \{ \mu(x,y),\mu(y,z) \right \}$. So I have to show that $\mu(x,z)\geq \mu(x,y)$ or $\mu(x,z)\geq \mu(y,z)$. I think I can show it if $x\leq y\leq z$ is assumed. Am I right?

15 mins ago, by Stack Exchange

in Mathematics, 5 mins ago, by Paul Plummer

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

15 mins ago, by Stack Exchange

in Mathematics, 9 mins ago, by Soham Chowdhury

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

15 mins ago, by Stack Exchange

in The 2nd Monitor, 12 mins ago, by QPaysTaxes

Oh looks like it can

15 mins ago, by Stack Exchange

in The 2nd Monitor, 14 mins ago, by QPaysTaxes

I think it can be simplified to just a[1] <=> b[1]

15 mins ago, by Stack Exchange

in Mathematics, 14 mins ago, by Soham Chowdhury

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Paul Plummer

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

@anon i'd prefer the comments section were. one can dream.

15 mins ago, by Stack Exchange

in TeX, LaTeX and Friends, 18 mins ago, by Joseph Wright

@SeanAllred The copyright holder can license code however they like. For example, one sees GPL code that is also available in proprietary versions. Provided all contributors accept this when they add stuff that is fine.

15 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by anon

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

15 mins ago, by Stack Exchange

in The Overlook Hotel, 22 mins ago, by Stack Exchange

0

Q: What is the correct formatting for actions taken before dialogue?

What would be the correct way to format the following exchange in narrative prose? Is it: Alice sits on the bench, silent. Eventually, I speak. "What are we going to do?" After a long pause, she responds. "I don't think we can do anything." Or is it: Alice sits on the bench, silent. Eventuall...

formatting dialogue

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

in Mathematics, 15 mins ago, by Soham Chowdhury

@anon can $g$ and $h$ have some nontrivial relation in $G*H$?

in Mathematics, 13 mins ago, by Soham Chowdhury

17 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

in Mathematics, 8 mins ago, by Soham Chowdhury

can I show that the free product satisfies a universal property?

in The Pod Bay, 7 mins ago, by TildalWave

@VedantChandra No idea, that's not up to me and there's sites with a lot better stats that are still in beta. Tho OK, we can't be expected to have like 32 questions per day and we have less of the drive-by traffic from the trilogy sites. Our stats seem stable and steadily increasing tho. And the community is growing stronger too, both in numbers and members with higher site privileges that actually use them (like reviewing et al.).

Stack Exchange

5:35 AM

14 mins ago, by Stack Exchange

in The Pod Bay, 7 mins ago, by TildalWave

@VedantChandra No idea, that's not up to me and there's sites with a lot better stats that are still in beta. Tho OK, we can't be expected to have like 32 questions per day and we have less of the drive-by traffic from the trilogy sites. Our stats seem stable and steadily increasing tho. And the community is growing stronger too, both in numbers and members with higher site privileges that actually use them (like reviewing et al.).

14 mins ago, by Stack Exchange

in Mathematics, 8 mins ago, by Soham Chowdhury

can I show that the free product satisfies a universal property?

14 mins ago, by Stack Exchange

in Mathematics, 13 mins ago, by Soham Chowdhury

17 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

14 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Soham Chowdhury

@anon can $g$ and $h$ have some nontrivial relation in $G*H$?

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

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7 mins ago, by Stack Exchange

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16 mins ago, by Doorknob

Can we go home now?

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15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The Overlook Hotel, 22 mins ago, by Stack Exchange

0

Q: What is the correct formatting for actions taken before dialogue?

What would be the correct way to format the following exchange in narrative prose? Is it: Alice sits on the bench, silent. Eventually, I speak. "What are we going to do?" After a long pause, she responds. "I don't think we can do anything." Or is it: Alice sits on the bench, silent. Eventuall...

formatting dialogue

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by anon

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in TeX, LaTeX and Friends, 18 mins ago, by Joseph Wright

@SeanAllred The copyright holder can license code however they like. For example, one sees GPL code that is also available in proprietary versions. Provided all contributors accept this when they add stuff that is fine.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

@anon i'd prefer the comments section were. one can dream.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Paul Plummer

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 14 mins ago, by Soham Chowdhury

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 14 mins ago, by QPaysTaxes

I think it can be simplified to just a[1] <=> b[1]

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 12 mins ago, by QPaysTaxes

Oh looks like it can

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 9 mins ago, by Soham Chowdhury

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 5 mins ago, by Paul Plummer

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by AjmalW

@BrianTung Thanks for your comments. 1) If $\mu(x,x_{0})>10$ it requires that $n$ should be greater than 10. Since $n=11$ is the minimum value in this case, so there must be $\mu(x,x_{0})=11$. 2) Rewriting it, we get $1/\mu(x,z)\leq \max\left \{ 1/\mu(x,y),1/\mu(y,z) \right \}$ which implies $\mu(x,z)\geq \max\left \{ \mu(x,y),\mu(y,z) \right \}$. So I have to show that $\mu(x,z)\geq \mu(x,y)$ or $\mu(x,z)\geq \mu(y,z)$. I think I can show it if $x\leq y\leq z$ is assumed. Am I right?

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

As regards Problem $2$, rewrite the expression in terms of $\mu(\cdot, \cdot)$ and see how far you can get.

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

Do you understand intuitively what $\mu(x, y)$ represents? It's the minimum $n$ such that the $n$th element of $x$ differs from the $n$th element of $y$. In other words, if $\mu(x, y) = k$, then $x$ and $y$ agree through the first $k-1$ bits (assuming $\mathbb{N}$ starts with $1$), and then differ for the first time at the $k$th bit. So if $x_0$ is an infinitely long sequence of zeros, and $\mu(x, x_0) > 10$, what can you say about the first ten bits of $x$?

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 1 min ago, by Soham Chowdhury

because you can't reduce the word $gh$ any further.

in The Nineteenth Byte, 12 mins ago, by Sp3000

Hmm I think I can do sub 200

in Board and Card Games, 5 mins ago, by doppelgreener

definitely i can see them avoiding this being very deliberate

in Pin 13, 52 secs ago, by Nick Alexeev

@AnnonomusPenguin Went looking for a framework that would facilitate the creation of code for external I2C-linked peripherals. My initial thinking: “I can write a parent class from which all drivers for I2C-linked peripherals would inherit. Wait, somebody probably have done that already… Then why reinvent the wheel?”

Stack Exchange

5:50 AM

15 mins ago, by Stack Exchange

in Pin 13, 52 secs ago, by Nick Alexeev

@AnnonomusPenguin Went looking for a framework that would facilitate the creation of code for external I2C-linked peripherals. My initial thinking: “I can write a parent class from which all drivers for I2C-linked peripherals would inherit. Wait, somebody probably have done that already… Then why reinvent the wheel?”

15 mins ago, by Stack Exchange

in Board and Card Games, 5 mins ago, by doppelgreener

definitely i can see them avoiding this being very deliberate

15 mins ago, by Stack Exchange

in The Nineteenth Byte, 12 mins ago, by Sp3000

Hmm I think I can do sub 200

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 1 min ago, by Soham Chowdhury

because you can't reduce the word $gh$ any further.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

Do you understand intuitively what $\mu(x, y)$ represents? It's the minimum $n$ such that the $n$th element of $x$ differs from the $n$th element of $y$. In other words, if $\mu(x, y) = k$, then $x$ and $y$ agree through the first $k-1$ bits (assuming $\mathbb{N}$ starts with $1$), and then differ for the first time at the $k$th bit. So if $x_0$ is an infinitely long sequence of zeros, and $\mu(x, x_0) > 10$, what can you say about the first ten bits of $x$?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

As regards Problem $2$, rewrite the expression in terms of $\mu(\cdot, \cdot)$ and see how far you can get.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by AjmalW

@BrianTung Thanks for your comments. 1) If $\mu(x,x_{0})>10$ it requires that $n$ should be greater than 10. Since $n=11$ is the minimum value in this case, so there must be $\mu(x,x_{0})=11$. 2) Rewriting it, we get $1/\mu(x,z)\leq \max\left \{ 1/\mu(x,y),1/\mu(y,z) \right \}$ which implies $\mu(x,z)\geq \max\left \{ \mu(x,y),\mu(y,z) \right \}$. So I have to show that $\mu(x,z)\geq \mu(x,y)$ or $\mu(x,z)\geq \mu(y,z)$. I think I can show it if $x\leq y\leq z$ is assumed. Am I right?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 5 mins ago, by Paul Plummer

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 9 mins ago, by Soham Chowdhury

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 12 mins ago, by QPaysTaxes

Oh looks like it can

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 14 mins ago, by QPaysTaxes

I think it can be simplified to just a[1] <=> b[1]

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 14 mins ago, by Soham Chowdhury

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Paul Plummer

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

@anon i'd prefer the comments section were. one can dream.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in TeX, LaTeX and Friends, 18 mins ago, by Joseph Wright

@SeanAllred The copyright holder can license code however they like. For example, one sees GPL code that is also available in proprietary versions. Provided all contributors accept this when they add stuff that is fine.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by anon

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The Overlook Hotel, 22 mins ago, by Stack Exchange

0

Q: What is the correct formatting for actions taken before dialogue?

What would be the correct way to format the following exchange in narrative prose? Is it: Alice sits on the bench, silent. Eventually, I speak. "What are we going to do?" After a long pause, she responds. "I don't think we can do anything." Or is it: Alice sits on the bench, silent. Eventuall...

formatting dialogue

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Soham Chowdhury

@anon can $g$ and $h$ have some nontrivial relation in $G*H$?

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 13 mins ago, by Soham Chowdhury

17 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 8 mins ago, by Soham Chowdhury

can I show that the free product satisfies a universal property?

in The Nineteenth Byte, 5 mins ago, by randomra

but I'm hoping some other approach can beat that

Stack Exchange

6:15 AM

24 mins ago, by Stack Exchange

in The Nineteenth Byte, 5 mins ago, by randomra

but I'm hoping some other approach can beat that

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 8 mins ago, by Soham Chowdhury

can I show that the free product satisfies a universal property?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 13 mins ago, by Soham Chowdhury

17 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Soham Chowdhury

@anon can $g$ and $h$ have some nontrivial relation in $G*H$?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in 13th Age, 11 mins ago, by Brian Ballsun-Stanton

fractional reserve PCs. Can I get a wizard mutual fund?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in The 2nd Monitor, 15 mins ago, by Duga

"iterator not dereferencable" - is that a compiler error message? If so provide the full text from the compiler and highlight the place in the code involved. If not, what exactly is your specific problem? This site is not for code reviews (there is such a site in the StackExchange family, but I can't remember which). — Tony D 19 secs ago

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

in Maid Café (メイド喫茶), 17 mins ago, by Logan M

@Mysticial Can you screencap it? It was just deleted.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by pjs36

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in Ask Ubuntu General Room, 23 mins ago, by Nathan Osman

I've done that with Stack Overflow more times than I can count.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

in The h Bar, 24 mins ago, by Terry Bollinger

The SE model is better because it does not reward passive aggression per se. You still need a lot of experts to get good results, and groups can still go south, but in general it lets experts truly and mutually reinforce good answers.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Doorknob

Can we go home now?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

27 mins ago, by Stack Exchange

22 mins ago, by Stack Exchange

53 mins ago, by Stack Exchange

7 mins ago, by Stack Exchange

12 mins ago, by Stack Exchange

24 mins ago, by Stack Exchange

16 mins ago, by Jeremy Banks

You can, if you want.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The Overlook Hotel, 22 mins ago, by Stack Exchange

0

Q: What is the correct formatting for actions taken before dialogue?

What would be the correct way to format the following exchange in narrative prose? Is it: Alice sits on the bench, silent. Eventually, I speak. "What are we going to do?" After a long pause, she responds. "I don't think we can do anything." Or is it: Alice sits on the bench, silent. Eventuall...

formatting dialogue

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 20 mins ago, by anon

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in TeX, LaTeX and Friends, 18 mins ago, by Joseph Wright

@SeanAllred The copyright holder can license code however they like. For example, one sees GPL code that is also available in proprietary versions. Provided all contributors accept this when they add stuff that is fine.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

@anon i'd prefer the comments section were. one can dream.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 17 mins ago, by Soham Chowdhury

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by Paul Plummer

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 15 mins ago, by anon

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 14 mins ago, by Soham Chowdhury

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 14 mins ago, by QPaysTaxes

I think it can be simplified to just a[1] <=> b[1]

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in The 2nd Monitor, 12 mins ago, by QPaysTaxes

Oh looks like it can

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 9 mins ago, by Soham Chowdhury

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Mathematics, 5 mins ago, by Paul Plummer

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by AjmalW

@BrianTung Thanks for your comments. 1) If $\mu(x,x_{0})>10$ it requires that $n$ should be greater than 10. Since $n=11$ is the minimum value in this case, so there must be $\mu(x,x_{0})=11$. 2) Rewriting it, we get $1/\mu(x,z)\leq \max\left \{ 1/\mu(x,y),1/\mu(y,z) \right \}$ which implies $\mu(x,z)\geq \max\left \{ \mu(x,y),\mu(y,z) \right \}$. So I have to show that $\mu(x,z)\geq \mu(x,y)$ or $\mu(x,z)\geq \mu(y,z)$. I think I can show it if $x\leq y\leq z$ is assumed. Am I right?

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

As regards Problem $2$, rewrite the expression in terms of $\mu(\cdot, \cdot)$ and see how far you can get.

24 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

14 mins ago, by Stack Exchange

15 mins ago, by Stack Exchange

in Discussion between AjmalW and Brian Tung, 2 mins ago, by Brian Tung

Do you understand intuitively what $\mu(x, y)$ represents? It's the minimum $n$ such that the $n$th element of $x$ differs from the $n$th element of $y$. In other words, if $\mu(x, y) = k$, then $x$ and $y$ agree through the first $k-1$ bits (assuming $\mathbb{N}$ starts with $1$), and then differ for the first time at the $k$th bit. So if $x_0$ is an infinitely long sequence of zeros, and $\mu(x, x_0) > 10$, what can you say about the first ten bits of $x$?

in Mathematics, 15 mins ago, by Paul Plummer

@SohamChowdhury Seems like you guys can not catch a break, earthquakes, hot weather. But doesn't normally get hot, during some parts of the year, was it it causing so many problems now?

in Mathematics, 9 mins ago, by Soham Chowdhury

I'd only done a few chapters. Been a long time, can't remember.

in Mathematics, 5 mins ago, by PerplexedGuest

Which can be interpreted as the number of self-avoiding walks on an $m$ by $n$ lattice that connect $(0,0)$ to $(m,n)$ which take exactly 1 left turn.

in Mathematics, 3 mins ago, by Paul Plummer

Maybe one of you combinatorialist can help with this problem here..

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