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12:00 AM
@Semiclassical honestly, if one posts a potentially inflammatory joke and it isn't directly personally at someone... I don't really care whether it is race themed or what-have-you. Puns are puns and jokes are jokes. They're spontaneous and if someone takes offense to them... well they can be the guys producing diamonds.
anyway can we please move onto a nicer subject?
 
We could talk about the puzzle? :P
 
i don't really have anything to say about it? :p
 
yeah, I can't say I do either. Other than I'm sure this is a problem in percolation theory.
 
probability and combinatorics are my weak points. They kind of... bore me.
 
There's some nice physics to the question, though.
I know, because one of my profs was a person who did a lot of work re: percolation theory in solid state physics.
 
12:02 AM
if I could relate the boolean values of existence to whether or not there is a path, then I could do a blatant sum to find the percentage of successes
 
To give a sense of why one might find it interesting: Suppose you think of the n(n+1) cell as a block of material, and you want to know whether or not it conducts.
 
then maybe I could relate it to the probability of failure to connect. But honestly, problems that are not practical bore me.
@Semiclassical fair point, but I am studying math and computer science.
 
Yeah, that's fair.
 
there isn't a particular usage for that concept
i mean... unless you talk about communication systems
but tbh that is pathfinding and finding existence through brute force.
 
Some applications for it are here: www1.coe.neu.edu/~emelas/apps.htm
 
12:04 AM
or in the case of DTN's... just routing using some algorithm that has a decent chance of success.
but let's change the subject
DTN is what I'm doing for that research thingy and I probably shouldn't talk about it here
 
delay-tolerant-networks
 
mmkay
 
basically, networks where your phone act as routers as well as any devices connected
they move around a lot and have connections intermittently, so the general issue is "how to route efficiently"
I'm looking at different protocols
aside from that I should probably stop
 
Mostly it's interesting in physics because there's such a sharp phase transition based on what the probability of having a link/not having a link (which physically would be the result of some kind of doping)
 
12:06 AM
mmm
ok
 
So that below, you're very unlikely to have conduction, but above that threshold you're very likely to have it work.
 
fair enough
 
So you have a transition between a configuration whose communication, so to speak, isn't very affected by disorder, and one where it very much is.
 
conductivity in cells sounds like Dr. Frankenstein to me though
or getting electrocuted
 
heh.
I think with this guy's work it was more like electrons hopping from site to site.
It was low-temperature physics stuff.
(like, really low)
 
12:08 AM
I didn't read the article
i was just referring to your thing about conductivity
 
That's fair.
 
Really?
 
so does low conductivity mean resistance to lightning?
 
idk if it's percolation.
 
Low conductivity means that you put a certain amount of voltage across and you don't get much current through it.
High conductivity means you do.
 
12:10 AM
oh I see it.
 
yeah, but what does that mean for if lightning struck the cell?
less damage? more damage?
 
I dunno. That's not what it's talking about.
 
im asking theoretically
 
see I thought the puzzle was a lattice theory problem
but it seems that percolation theory and lattice theory overlap
 
like... does electrical resistance raise the chance of a high voltage causing damage?
or does it lessen it by mediating the flow?
 
12:11 AM
Problem is, when you talk about conductivity to a physicist, you're almost inevitably talking about 'linear response theory.' Which basically means "how does the system respond to relatively small changes in applied inputs."
A lightning strike is very much not linear.
 
im just thinking in terms of
 
So I really have no idea what the correlation with resistance would be.
 
"lighting strikes two people. Which one has more cellular damage?"
but fair enough
(and I do mean on average of course)
 
I could see two directions, and that's why I'm not sure.
One is that higher conductivity means more current flows through you, so more charge every second
Which obviously bodes ill.
On the other hand, high resistance means that what current flow there is will require a lot of energy transfer. That leads to the resistor having to dissipate a lot of heat.
Which practically means that if you force a lot of current through an insulator then it'll get really hot.
 
@Semiclassical the reason i ask is because in a few things I've read over the years, there are stories about giants or whatever. Greek mythology. One gets struck by lighting and just staggers back. all it does it leave a burn mark. However I always thought "it shouldn't work that way. If it burned it like that, it should've electrocuted it."
 
12:14 AM
Which is obviously bad.
 
so I was just thinking of whether or not there was such a thing as "resistance to lightning"
if that makes sense biologically speaking
your statement made that pop into my head
 
Of course, what's also tricky is that a lightning strike is very different in terms of hazard than, say, touching a live wire in your house.
 
well it is 8.21 gigawatts for starters
enough to travel through time
 
If that live wire represents an AC current, then the main hazards are 1) the current keeps flowing, so your hand can clamp onto the wire just due to how nerves work
and 2) the frequency of that electric current can mess with your heart's electrical activity.
 
and then you catch on fire
 
12:16 AM
yeah.
it's bad all around.
 
or worse:
 
With lightning, there's obviously a huge amount of charge that goes through you if you're struck by it.
 
the power trips out in the middle of your desktop pc updating and you're like "NOOOOOOOOOOOOOO!"
 
On the other hand, it's all over and done with in an instant.
 
fair point
but people don't really survive lightning do they?
and does the size of the person at all change its effect?
 
12:17 AM
Well, people do survive it. I'm not sure how well, though.
 
@Semiclassical their marbles certainly don't
 
Per this NOAA site: "According to the NWS Storm Data, over the last 30 years (1987-2016) the U.S. has averaged 47 reported lightning fatalities per year. Only about 10% of people who are struck by lightning are killed, leaving 90% with various degrees of disability. More recently, in the last 10 years (2007-2016), the U.S. has averaged 30 lightning fatalities."
 
wait really
fair point
 
but... nobody has just walked away from lightning, right?
i mean like "I just casually got struck by lighting in the middle of eating my hamburger out in the woods"
 
12:20 AM
I wouldn't be surprised if there's a range of response, to be honest. But the phrase 'various degrees of disability' doesn't bode will.
 
I was thinking of whether someone could remain standing up and not collapse
 
or for that matter react no worse than getting hit in the face with mud
 
two different ones, in fact
(and really quite different, in fact)
 
fair enough
 
12:22 AM
i wonder how astyx solved the puzzle.
 
@Dodsy It might be like that infection problem from before. Perhaps it is a truly trivial solution.
 
Not entirely surprising that it would have an effect, if one thinks about how electric-shock therapy has been used clinically for treatment of mental health.
 
true
 
(That's one of those things that has a lot of bad stereotypes which are held over from earlier decades.)
 
well Ty, it needs to start with a horizontal edge, and end with a horizontal edge.
 
12:23 AM
Just to be clear, are we talking an edge collection like this:
 
@Semiclassical I'm trying to think about how a 100-200 foot tall giant could survive a lightning strike and shrug it off like a bee sting when a human gets knocked on their ass at bare minimum.
 
Well, we are talking mythology.
 
@Dodsy true, but perhaps it's just the probability of one column having at least one edge in it.
 
Not exactly starting off at a very high bar for scientific respectability.
`-+-+-+-`
` | | | | `
 
@Semiclassical true, but I'm thinking about the scientific aspect.
because it would be ironic for me to use that character and lighting actually work
because of the fact that the size of the target has no effect on the reponse
 
12:25 AM
Well that's what i was saying, for $n=1$ there is a 75% chance that there is a horizontal edge at the beginning and a 75% chance that there is a horizontal edge at the end, and a 50% chance that there is a vertical edge in the middle.
 
okay I have no idea how to make ascii graphics here :/
 
cause an electric surge isn't lessened by volume, right?
 
You're honestly in a realm of electromagnetism I know little about.
 
And for that case we found that the chance of two horizontal edges being connected was ~46% meaning that it all hinges on the vertical edge.
 
Argle bargle
 
12:26 AM
i know electricity is a wave, so I would presume that it has the same impact and damage to scale on anything regardless of size.
so for instance: a planet sized turtle gets the same damage as a human
even though that sounds counter intuitive
 
my internet is terrible
 
@Dodsy no. Your internet is spotty.
 
unless you are connecting to 4chan in which it is EVIL!
 
this is $n=1$
D:
I don't go on 4chan
 
12:29 AM
So, to be clear, we're interested in the probability of going from either corner on one side to either corner on the other
 
left to right
using edges
 
Right.
 
they can be vertical or horizontal edges.
 
@Dodsy fair enough. I'm just clarifying that terrible means that the content is terrible. Granted, you could say that your internet service provider is terrible to which I would answer: "they all are"
 
From what (little) I remember of percolation theory, my immediate question is: Given a particular graph of edges which conducts, what does the dual graph looks like?
 
12:30 AM
but, we don't even need to look at the far right or far left edges.
 
true, you only care about 5 edges
 
True.
 
By that I mean: Given one particular graph, I can immediately draw a second one where all of the missing edges are present and vice versa.
 
true.
 
and cutting away that extraneous information now would be equivalent to some fractional cancellation later after counting all the possibilities.
 
12:31 AM
I seem to remember that being a good thing to do.
 
Well it's an interesting development
 
i just want to find the probability of any random grid of that size being connected
 
@Typhon Yeah, that's a typical question
 
@Semiclassical well the riddle says that each one has a 50/50 chance of connection
 
I would call this an NP-Hard problem though, wouldn't you?
 
12:33 AM
no
 
Correction
 
it is harder to verify than it is to solve
 
the riddle says that each edge has a 50% chance of being there.
 
yeah
so each edge has a 50/50 chance of connection
 
Another version of that would be: Suppose that I take the probability of having an edge to be small. Then I doubt you'll have full connectivity. However, you'd expect that even in this scenarios you'd still get connected portions/clusters.
 
12:33 AM
but we do know that the answer is 50%
 
as in a connection existing at that location
 
So you can ask how the cluster size varies with the probability.
 
@Dodsy not the final answer. I'm just saying the same thing you did in different words.
 
If that probability goes up, then the cluster size should go up.
 
@Typhon the final answer is also 50%.
 
12:34 AM
0.0
ok let me think
 
I figured that out by thinking that Astyx meant only horizontal edges
 
@Dodsy that fact is part of why I think that the flip transformation is interesting.
 
and then doing badmath
 
there are 4 possible good paths out of the 32 middle segments for n=1
4/32 = 1/8
 
You're doing the same miscalculation I did.
 
12:36 AM
wait no
that is the minimum possibilities
 
We're not counting good paths, but simply arrangements which contain the good paths.
 
@BalarkaSen yeah I know
 
@Semiclassical right, because if $n=1$ then any opposite graph of an incorrect permutation would show a correct permutation.
At least, I believe so, without going through all of the permutations.
 
there are 2*8 paths along the horizontals and 2*4 paths along the twists step shaped paths. The number of cases including them is 2^n where n is the number of lines not in the path
that gives me 3*8 combinations
 
@BalarkaSen We know that the path must start on a horizontal and end on a horizonal, I was thinking for $n=1$ there is a 75% chance that there will be a horizontal to begin the path and a 75% chance that there will be a horizontal to complete the path. There is a 50% chance that there will be a vertical (only 1) meaning that there is a 50% chance that a path will be made for $n=1$
 
12:39 AM
24/32 = 3/4 probability for a 1 by 2 grid
 
Same remiscalculation I did :P
 
@BalarkaSen then explain my mistake please? I hate probability
 
What do you think of that?
 
You're counting two arrangements twice when you compute 24
i.e. overcounting is going on in that argument
@Dodsy I don't really have the energy to think about this problem anymore. Sorry :(
 
how so?
i compute the combinations with the top row filled, and then the bottom row filled...?
etc
 
12:43 AM
In the 2*8 arrangements which contain the top horizontal path, at least one contains the twist shaped path
Do you see this?
It looks like
 
oh geez
 
_ _
b|_
 
so is it 12/32?
 
Just to show why I think Mathematica is a good approach for this, here's the pic which GridGraph[{5,6}] gives
 
I think you end up with a lot of miscounts. It should be 16/32
 
12:44 AM
ok...
 
I gave up computing this way when I realized this was happening
 
so that's the prob of a connection given random grid
 
but what's the probability of a connection when an edge has a 50% chance of disappearing?
 
Hi
Ooh, picture
Has progress happened?
 
12:46 AM
no
 
Moreover, once you've created that graph you can pull up its edge list. What I"m trying to figure out right now is how to make it keep/drop each edge from that list with probability 1/2.
 
@Akiva neh
 
(ideally I want to allow arbitrarily probability)
 
Now fill half of them with \\ shapes and the other half with // shapes...
(joking)
 
go away /s
 
12:47 AM
:(
 
@Semiclassical I could do that with my 3D renderer by grabbing a random number... but it wouldn't be useful
 
hm
I don't think I'm good at probability at all
though it shouldn't be this hard.
 
me neither
wanna talk about geometry or something else?
like trying to create matrices with indices at 1/2 intervals?
still cannot rejog that thought
 
for instance for $n=1$ if i try to compute it algebraically I get a 94% chance of it happening...
which, is obviously wrong.
 
or is it
 
12:52 AM
Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers?
 
So P is not given a prime.
Wait, 1<x? So x and y can't be 1?
So P is not... twice a prime?
 
1 is less than the value of x, x is less than the value of y.
 
hold on
 
P is not a semiprime (product of two primes) as well
 
S and P are the names of the logicians :)
 
12:53 AM
The number you told P.
 
P cannot determine the two numbers so it is not a semiprime. However, we know that either both are odd or both are even.
 
Why? x+y<100, not x+y=100.
 
D:
@AkivaWeinberger darnit
 
In any case, S knows that P isn't given a semiprime
 
ok hold on
 
12:54 AM
So it's not the sum of two primes
 
S could conclude it was a semiprime
so it isn't 4
 
Zee
Primes are sooo boring
 
S could be given 11 perhaps
 
and it could be given any other even number
@Zee ok...?
 
This is a hard question too..
 
Zee
12:56 AM
Ya, idk why he asked me that
 
Wait hold on
 
I think one of S's numbers is prime.
 
How is this unique...?
Does 2 and 9 work?
S is given 11, P is given 18.
P has no idea.
 
Well P could easily find the factors of 18.
 
S knows that P would have no idea, since 11 is not the sum of two primes.
 
12:57 AM
and work out that is 9 and 2.
so he'd know it right away.
 
@Dodsy Could be 6 and 3.
 
or 3 and 6
 
@AkivaWeinberger the fact that P can solve it from knowing that S knew that P knew it was a semiprime is what makes it unique.
 
So he'd at least have an idea.
 
Oh wait
 
Zee
12:58 AM
IDs are issued by the DMV
 
wait
maybe you're right
but I think the further banter
maybe changes that
I think you're on the right track
we're looking for numbers that have a sum of less than 100
and who have more than one factor.
 

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