The Nineteenth Byte

Eric Tressler

7:01 PM

i guess you could tie people down and force-feed them bread

Geobits

That does seem like the easiest screening method.

githubphagocyte

Wait a moment, both @Geobits and @Rainbolt used the word "icing". Do you not use "frosting"? Or is that only for cakes?

Geobits

@githubphagocyte It's pretty much interchangeable, but to my mind, frosting is softer, while icing is more of a glaze.

Eric Tressler

god DAMMIT

Geobits

I'm pretty sure poptarts themselves are labelled 'frosted', though.

Eric Tressler

7:02 PM

I was trying to make a joke and I left this page twice

Geobits

That seems like a terrible joke.

githubphagocyte

@Geobits - interesting. I had frosting/icing as one of my US/UK translations but sounds like it's another one to add to the long list of "it's not that simple"...

Eric Tressler

:|

it was a fine joke

githubphagocyte

Usually when a joke back fires it doesn't literally translocate you...

Geobits

Well, spit it out. This buildup is intense :D

Eric Tressler

7:04 PM

i actually forget, it was just something about the US v UK

all i remember is that when i press backspace, my browser goes back

Geobits

Oh. Yea, mine does that if I'm not focused in a text field.

user2179021

@Geobits hi

githubphagocyte

@EricTressler I get that problem a lot. I can't learn to avoid pressing backspace because most of the time it does what I expect, but every so often when timing is critical pressing tab prints a weird character instead of autocompleting the username, and then pressing backspace performs a browser back.

user2179021

hi @githubphagocyte

githubphagocyte

@user2179021 hi :)

Geobits

7:07 PM

Umm. hi?

Eric Tressler

@githubphagocyte yeah, I don't really know what the parameters are; but the chat here is pretty close to ideal

githubphagocyte

@Geobits do you not find numeric usernames memorable either?

Geobits

@githubphagocyte Nah, I know it's Lembik. He comes in and says hi often, generally to whoever's active at the moment.

Eric Tressler

pure curiosity: are you guys in school?

Geobits

Me? No. My son is.

githubphagocyte

7:09 PM

@Geobits my memory isn't much better even with recognisable names...

Nathan Merrill

I am

user2179021

@githubphagocyte I at last posted my question. Took me days to get it vaguely right

Eric Tressler

@Geobits neither am I; I was just wondering

githubphagocyte

@EricTressler when you say school do you mean mandatory school or further education (college/university)? No in either case

user2179021

please do tell me if I can improve it more .. codegolf.stackexchange.com/questions/37269/…

or just upvote it :)

Eric Tressler

7:10 PM

I mean any form of school; I've been out of school for 5 years

but i taught a whole lot of kids

I didn't go to class a lot during grad school

maybe that was a mistake

Geobits

I didn't go to class a lot during high school. Probably wasn't a mistake :p

Eric Tressler

maybe not. you can learn on your own, and probably better than from a mediocre teacher

githubphagocyte

I think most people who skip classes do it on purpose :)

Eric Tressler

I'm reading the Feynman lectures right now

Geobits

In my case, I got suspended for "hacking" (idiot teacher) a computer. Lost credits for the semester, so dropped out and got a GED two weeks later.

Eric Tressler

7:15 PM

you can overcome that

Geobits

Oh, I know. It hasn't ever held me back. That was 14, maybe 15 years ago :p

githubphagocyte

Bingo. I worked out what makes my backspace turn into a back button during a chat message - I am leaving the text box after all. If I mistype the first letter of a username and instead type a letter that doesn't bring up any possible username, then tab just shifts focus out of the text box, so that backspace becomes a back button. So it is my fault after all...

@user2179021 if you are going to use your own computer as the speed criterion then you might want to mention the specifications of your computer.

user2179021

@githubphagocyte thanks. Done

COTO

@PeterTaylor: As mentioned earlier, writing out script tags and event attributes wouldn't be allowed. Are there other ways of executing code by simply writing to the DOM? This is one of the main reasons I raised the idea here. I wanted some feedback as to which loopholes to eliminate to make it an interesting competition.

@PhiNotPi: You pinged me for something?

githubphagocyte

@user2179021 this might need clarifying: " You clearly aren't allowed to record the rotation from when you created the data and use that! "

COTO

7:22 PM

lol You guys discovered timecube.com, did you?

That site is epic.

Nathan Merrill

I got through about 1/3 of it

before I stopped reading

githubphagocyte

@NathanMerrill at the bottom it says "Next Page"...!!

Nathan Merrill

I know. I also looked at each of the links on the second page

my favorite link is this one: thewisesthuman.com

it looks sooo horrid

COTO

It hasn't been updated in quite a while. Mr. Gene Ray, the presumptive author, looks pretty old in the photo posted, at that's now at least 6-7 years old. He may have since died.

githubphagocyte

That lovely backdrop only reaches almost half way down.

COTO

7:30 PM

We can all be grateful to him for teaching us the unvarnished truth: "Educators are lying bastards. -1 x -1= +1 is WRONG, it is academic stupidity and is evil."

Nathan Merrill

...what is the proof behind that?

like, academically, why do we know that two negatives multiplied is a positive?

COTO

That's actually a question on Math.SE

Or more specifically: why is -1/-1 = 1

There are a lot of good answers.

Geobits

> This convention has been adopted for the simple reason that any other convention would cause something to break.

> For example, if we adopted the convention that (-1)(-1) = -1, the distributive property of multiplication wouldn't work for negative numbers:

(-1)(1 + -1) = (-1)(1) + (-1)(-1)

(-1)(0) = -1 + -1

0 = -2

COTO

There are a lot of things wrong with it. That isn't the half of it.

githubphagocyte

Start with -1 apples. Call that a bunch. Now gather together -1 bunches and count how many apples in it. 1, right?

Rainbolt

7:34 PM

Whatever happened to @ProgrammerDan? I miss that guy.

COTO

People tend to understand it best in terms of debts and credits with money.

Nathan Merrill

Sure, but if I were to prove it, you can't do intuition

COTO

Of course, in a field with characteristic 2, -1 = 1, hence -1 x -1 = -1, hence maybe Mr. Ray was onto something. ;)

Nathan Merrill

unless it's an axiom

COTO

You can explain it in terms of more fundamental axioms, but these aren't public-friendly.

Essentially -1 and 1 form a group of order 2, and we extend the natural numbers through a Cartesian product with this group.

githubphagocyte

7:38 PM

One intuitive route is to see multiplication by -1 as negation.

Rainbolt

@COTO Do you have a link to that question? I want to read the answers

COTO

I'll see if I can find it again.

Geobits

40

Q: Why negative times negative = positive?

Someone recently asked me why a negative * a negative is positive, and why a negative * a positive is negative, etc. I went ahead and gave them a proof by contradiction like so: Assume $(-x) * (-y) = -xy$ Then divide both sides by $(-x)$ and you get $(-y) = y$ Since we have a contradiction, t...

Nathan Merrill

6

A: Why negative times negative = positive?

Simple Answer: $$ (-a)b + ab = (-a)b + ab $$ $$(-a)b + ab = b(a-a) $$ $$(-a)b + ab = b(0) $$ $$(-a)b + ab = 0 $$ $$(-a)b = -ab $$ $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$ $$(-a)(-b) + (-ab) = (-a)(b-b) $$ $$(-a)(-b) + (-ab) = (-a)(0) $$ $$(-a)(-b) + (-ab) = 0 $$ $$*(-a)(-b) = ab $$ Hope this h...

Geobits

Boom!

Nathan Merrill

7:40 PM

heh, ninja'ed

that answer I posted is the one I was looking for

githubphagocyte

I tried googling for it but I didn't put quotes around the minus signs so I just got this

Nathan Merrill

hehe, that's funny

Rainbolt

Could you just expand multiplication into a bunch of additions and then ask an easier question about addition?

(-3) * (-4) = - (-3) - (-3) - (-3) - (-3)

COTO

You can't expand a multiplication into a negative number of terms. That's one of the reasons some people have trouble with the concept.

Here's one of my favourite false proofs:

Let a = b <> 0.
Thus ab = bb.
Thus ab - aa = bb - aa.
Thus a(b - a) = (b + a)(b - a).

Thus a = b + a.

Thus a = 2a.

Thus 1 = 2.

Geobits

Too many zeroes ><

Nathan Merrill

7:59 PM

I actually don't know how to disprove that @Geobits

oh, now I do

its basis is (-1)^2 = (1)^2 and sqrt both sides

COTO

@Geobits: I've seen that one too.

Everybody in Canada is the same age.

How do I know this?

By induction.

Suppose we have n people.

We want to show that if any n-1 of these people have the same age, then all n have the same age.

That's simple enough. First, choose a group of n-1 people. This will exclude one person, who we'll call A. There will also be a member in the group of n-1 people we'll call C.

Next we create another group of n-1 people that includes A. This group will necessarily include C.

Hence C's age is the age of the first group of n-1 people, and C's age is the age of the second group of n-1 people, and the union of those two groups is the full n people.

Thus we've proved our lemma that if any n-1 people have the same age, all n people have the same age.

Let's take this down to the first inductive step, n = 1

Well, with n = 1, there's only one person in the group, hence obviously everybody in the group has the same age: the age of the only individual.

Sparr

@COTO your second group of n-1 that includes A does not necessarily include C. He could be the one.

Geobits

8:14 PM

I'm not sure how you've proven your assumption concerning n-1.

Sparr

and that will certainly be the case when n=2

Geobits

That^

COTO

Yeah. The inductive step fails at n=2. It is, however, true for all other steps.

;)

Sparr

no

one of your statements is false

COTO

Sure.

Sparr

8:18 PM

such that it always fails at n=2, and sometimes fails at n>2

COTO

Never fails at n>2.

Sparr

"Next we create another group of n-1 people that includes A. This group will necessarily include C." <-- false

at n=3 there are three possible groups of n-1. one of the groups that includes A includes C, the other does not.

COTO

It should be "excludes A"

I'll fix that.

Sparr

A is the person you excluded in the first n-1

excluding him from both n-1s makes those the same group

Geobits

Much easier to prove that a healthy dog has 12 legs.

COTO

8:20 PM

At n>2, any n-1 group that does not contain A must contain C.

Geobits

Why? there are n possible groups, one of which excludes C

COTO

In fact, there is only one n-1 group that excludes A.

Geobits

If you exclude A both times, it's the same group and you're not comparing anything.

Sparr

^^

COTO

I'm defining A specifically as somebody who was in the first group that didn't include C.

Geobits

8:22 PM

> This will exclude one person, who we'll call A.

COTO

Let me be more rigorous and formal about it.

Let the group consist of n people, and assume that the proof holds up to n-1.

Select any random group of n-1 people. This will contain at least two individuals: A and B.

Because A and B are both in the group of n-1 people, A and B have the same age.

Geobits

Doesn't that assume they're all the same age to start with?

COTO

Yes. That's the whole point of the induction.

Assume truth for n-1 and then prove for n.

Let the person who isn't in the group of n-1 people be called C. C isn't either A or B, obviously.

Now form another group of n-1 people that includes C. This group of n-1 people must either include A or B or both.

And since it's a group of n-1 people, all of their ages are the same, hence C's age is the same as A's or B's, which are equal, and thus has the same age as everyone else in the original group.

The union of the two groups is the full n people, hence all n people have the same age.

As you pointed out, the induction fails at n=2.

Geobits

What I got out of that is "If you have n-1 people of the same age and add another person of the same age, you get a larger group of people with the same age.

COTO

Right.

Geobits

8:28 PM

Your assumption that a different group of n-1 people are all the same age only holds if C is already the same age, so it doesn't hold for any n.

githubphagocyte

While we're sharing wisdom, those little spiky things for eating sweetcorn have nothing on two sets of plastic tongs. I've just eaten 2 cobs of sweetcorn and I have zero holes in my fingers.

COTO

Well, if the above proof isn't clear, I'll leave you to ponder it. ;)

Sparr

it's clearly wrong

githubphagocyte

Something slips in that treats "there exists a set of n-1 people" as "every set of n-1 people".

user2179021

@githubphagocyte thanks

hi @COTO

COTO

8:30 PM

The induction fails at n=2, yes. But if I was able to show for a specific group of people that if any subgroup of 5 of them had the same age, then all 6 would have the same age. That's the essence of the proof.

Hi, long alphanumeric string. :)

githubphagocyte

@COTO yes, as such the proof is valid, just easy to overapply...

user2179021

let me advertise codegolf.stackexchange.com/questions/37269/… one last time :)

githubphagocyte

@user2179021 is your name being different in chat a deliberate choice or just the way it happened?

user2179021

@githubphagocyte it's completely random. I think it's because I share this computer and stackexchange aggresively remembers cookies

@githubphagocyte but I dno't really know

@githubphagocyte I am hoping someone will upvote one of my questions soon :)

githubphagocyte

@user2179021 there are a few mysteries. My email-based identicon only shows up on sites other than Area 51. On Area 51 I have a completely different one for some reason, even though it signs me in automatically based on being signed in to another site.

user2179021

8:36 PM

I also don't understand how, even if I delete all the cookies, I have to actually quit firefox and restart it for stackexchange to forget my identity

it's quite annoying

githubphagocyte

@user2179021 is that even if you close all stack exchange tabs?

VisualMelon

@user2179021 I'm afraid I'm not good at reading, might you be able to throw together a simple visual aid (maybe related to the example) so I can better understand the question?

user2179021

@VisualMelon I am not good at drawing but I can explain anything isn't clear. Is my example timesX = [2, 4, 98] and timesY =[ 1,7,9,99] not helpful?

@githubphagocyte it seems to be!

@VisualMelon please ask

@VisualMelon maybe if I explain it well enough you could draw me a picture I could add?

VisualMelon

gladly, but I'm not all that good at drawing myself ;)
I'll re-read it a few times more, I think I'm missing the whole point of the distance

user2179021

@VisualMelon Let me try to explain it again

you have two circles of points named X and Y

we are interesting in the distance of the points in X to the points in Y

it's not symmetric

so for each point in X, just see how far it is from the closest point in Y

and add those numbers up

does that make sense?

COTO

8:43 PM

On the problem involving the optimizer, the problem is that there isn't really any structure to exploit in the problem. It's not convex; the objective function is riddled with local minima, hence steepest descent and Nelder-Mead simplex, etc. aren't going to be of any use. The objective function isn't LMS, hence Levenberg-Marquardt, etc. are out.

VisualMelon

@user2179021 "Here the distance is just the sum of the absolute distance from each point in X to its closest point in Y" - I'd not understood what "closest" means, I think I've got it now, thanks ;)

user2179021

@COTO right.. that's why it is fun :)

COTO

The only way I see of optimizing it is to just write a grid search over a single parameter.

user2179021

@VisualMelon cool. Closest is really just how far you would have to go along the circle to reach a point in Y

@VisualMelon of course you can go forwards or backwards

@COTO well there might be a sweep line algorithm.. not wanting to give any ideas away :)

@COTO is the problem at least clear now?

COTO

If by "sweep line", you mean just incrementing theta by tiny amounts and ultimately picking the best value by the time you've gone full-circle, then agreed. That's your dry toast grid search.

As for the clarity of the problem: I understood it from the original description. ;)

g2g

user2179021

8:47 PM

@COTO great! Please upvote :)

VisualMelon

@user2179021 yeah, I think the text is perfectly clear, I'm just not very good at reading

COTO

lol If it means that much to you, why not?

user2179021

@COTO I was thinking something maybe a little better than just increminging

incrementing

@COTO I spent days polishing the question :_)

@COTO consider a small rotation of X as it is aligned with Y

COTO

Then you obviously have a broader mind about the subject than I do, because I don't see iterative refinement as being at all useful in the large-n cases, and basically unnecessary in the small-n cases.

user2179021

@COTO until new points are aligned, you know exactly what the distance will be

@COTO so it seems, at least to me, that don't need to do lots of small increments

@COTO you can work it out mathematically until some point in X hits a point in Y when you need to start agani

at least that is my idea :)

so maybe you can it in jumps

maybe :)

in any case, the benchmark I have given is terrible :)

so it can't be hard to beat that

@COTO thank you

VisualMelon

8:56 PM

@user2179021 reading back, I don't think it's quite clear how the wrapping around the circumference works (but you can work it out from the example), and I don't think it is explicit whether it applies to the distance calculation (but it appears from your code sample that it does not)

Rainbolt

9:39 PM

~DayIDie

githubphagocyte

9:54 PM

@Rainbolt that took me a minute... :)

VisualMelon

@user2179021 threw a program together to render the problem, http://visualmelon.onl/spindist.png (red lines are Y values, blue dots are X values)
can you clarify how the wrapping works - because your example solution seems to presume that you can't wrap the distance (i.e. 99 is closer to 80 then 1)

(for T = 100, that is)

Rainbolt

10:09 PM

@githubphagocyte Hehe. That's a Rainbolt original. I was proud of it when I created it.

githubphagocyte

@Rainbolt there's a firm called Swinton and I half expect someone to paint a tilda in front of their sign overnight

Rainbolt

tildeswinton. This is like MadGab (if you've ever played that), and I would have failed on this one.

Don't worry. It'll come to me.

Martin Büttner

11:20 PM

Wow, for a second there I freaked out when I saw the title of this challenge: codegolf.stackexchange.com/q/37275/8478

@overactor you totally could have if I had been around

@overactor That is really weirdly worded. You're right with Alltagsabläufe. It basically translates to everyday-life processes (in the sense of rigid processes). Automatenprogramm might refer to something like the phases a washing machine goes through or how a lift operates but I'm not entirely sure what it's asking for.

Btw, I think CJ knows a bit of German, if you're stuck again. ;)

@githubphagocyte Yeah, might be a good idea.

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