/sci/ - Science & Math » Thread #13927580

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Anonymous Sun 05 Dec 18:22:03 2021 No.13927580 View Reply Original Report

Quoted By: >>13927591 >>13928024 >>13928050 >>13928338 >>13928348 >>13929068 >>13929294 >>13929295 >>13930194 >>13933606 >>13933626

A square is dropped on a checkerboard, with any rotation and position. The square is as big as the checkerboard tiles. What's the probability it covers one of the corners of the checkerboard tiles?

Anonymous

Anonymous Sun 05 Dec 2021 18:24:13 No.13927591 Report

Quoted By: >>13927601

>>13927580
It is a 100%

Anonymous

Anonymous Sun 05 Dec 2021 18:26:51 No.13927601 Report

Quoted By: >>13927628

>>13927591
um no

Anonymous

Anonymous Sun 05 Dec 2021 18:33:44 No.13927628 Report

Quoted By: >>13927639

>>13927601
Do you mean that it covers any of the corners of the squares of the 8x8 board or that it covers one of the corners of the 8x8 board bitch

Anonymous

Anonymous Sun 05 Dec 2021 18:36:05 No.13927635 Report

Quoted By: >>13927703 >>13928021

50%, it either does or it doesn't

Anonymous

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Anonymous Sun 05 Dec 2021 18:37:07 No.13927639 Report

Quoted By: >>13928024

>>13927628
dude

Anonymous

Anonymous Sun 05 Dec 2021 18:57:40 No.13927703 Report

Quoted By: >>13927989 >>13929032

>>13927635
your confusing possibilities with probabilities retard

Anonymous

Anonymous Sun 05 Dec 2021 18:59:42 No.13927712 Report

Quoted By: >>13927810

With a question like this, how do you go about addressing precision?
Like, if a square lands almost perfectly on a space, but with a 0.000001 degree rotation, does it count?
I think to solve it you have to make assumptions like, it rotates in tenth degree increments, making 4 out of 3600 possible degree positions where it can land in at least the right orientation. But, in doing that, you are not ending up with just a close appreciation.

Anonymous

Anonymous Sun 05 Dec 2021 19:31:39 No.13927810 Report

Quoted By:

>>13927712
Just use calculus lol

Anonymous

Anonymous Sun 05 Dec 2021 19:43:19 No.13927843 Report

Quoted By:

May part of the dropped square go outside the borders of the entire checkerboard?

Anonymous

Anonymous Sun 05 Dec 2021 20:41:23 No.13927989 Report

Quoted By:

>>13927703
No, he is correct

Anonymous

Anonymous Sun 05 Dec 2021 20:54:50 No.13928021 Report

Quoted By: >>13928056

>>13927635
I guess there are infinitely many cases where it does and infinitely many cases where it doesn't, so 50% sounds right.

Anonymous

Anonymous Sun 05 Dec 2021 20:55:49 No.13928024 Report

Quoted By:

>>13927580
I guess the first order question is how likely it is for a square with random orientation and position to overlap a point on an infinite square grid. We can think about the probability distribution function as being continuous in angle and position space and consider only a periodic unit cell.

It's apparent that with zero rotation, the probability of intersecting a point is 100%. By drawing this sort of diagram >>13927639 it also becomes apparent that the position space without overlap is limited to the area remaining in the 'base' square when the 'dropped' square is cut out from it. This has some dependence on angle that I am too lazy to work out. If I did work it out, it would fully define the PDF, and integrating in both variables would give the probability.

I'm way, way too lazy to think about the boundaries tho.

Anonymous

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Anonymous Sun 05 Dec 2021 21:05:51 No.13928050 Report

Quoted By:

>>13927580
>What's the probability it covers one of the corners of the checkerboard tiles?
meaningless

Anonymous

Anonymous Sun 05 Dec 2021 21:07:42 No.13928056 Report

Quoted By:

>>13928021
Never mind. I tried it 30 times and it always landed on a corner. It's probably something like 99.9999%

Anonymous

Anonymous Sun 05 Dec 2021 22:22:23 No.13928338 Report

Quoted By: >>13928348 >>13933515

>>13927580
$\frac{6}{\pi}-1\approx91\%$

Anonymous

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Anonymous Sun 05 Dec 2021 22:25:12 No.13928348 Report

Quoted By: >>13928489 >>13929019

>>13927580
>>13928338
wtf
pic related is answer

Anonymous

Anonymous Sun 05 Dec 2021 22:54:15 No.13928489 Report

Quoted By: >>13932215

>>13928348
why

Anonymous

Anonymous Mon 06 Dec 2021 01:16:39 No.13929019 Report

Quoted By: >>13929076 >>13932215

>>13928348
Proof?

Anonymous

Anonymous Mon 06 Dec 2021 01:21:20 No.13929032 Report

Quoted By:

>>13927703
It’s bait anon

Anonymous

Anonymous Mon 06 Dec 2021 01:34:10 No.13929068 Report

Quoted By: >>13929185

>>13927580
What positions of the square count as it being on the checkerboard? Does the whole thing need to be on the board or is it on the board if any part of the square overlaps with the board?

Anonymous

Anonymous Mon 06 Dec 2021 01:36:16 No.13929076 Report

Quoted By:

>>13929019
>the proof is left as an exercise for the reader

Anonymous

Anonymous Mon 06 Dec 2021 02:07:00 No.13929185 Report

Quoted By:

>>13929068
It is an infinite checkerboard and the square falls anywhere on it

Anonymous

Anonymous Mon 06 Dec 2021 02:11:21 No.13929196 Report

Quoted By: >>13929216 >>13929229

Why are math and programmers like this? I know it’s this site, so everyone must be a dick
But honestly the sharing and teaching aspect is always hurrr figure it out yourself
Well I get it
You don’t want to constantly hold hands but is tossing a link to start the journey of learning to difficult? No

I knew pi was involved but probably from a misguided idea: there was a way to compute the number of pi by dropping sticks on a similarish board of squares
It’s a cool thing and I never knew, until someone told me

https://www.sciencefriday.com/articles/estimate-pi-by-dropping-sticks/

Anonymous

Anonymous Mon 06 Dec 2021 02:18:14 No.13929216 Report

Quoted By: >>13929239

>>13929196
Thanks anon i appreciate your share. OP here, this post was a problem on the recent putnam exam, i think i got a partial proof but wanted to see what you all think

Anonymous

Anonymous Mon 06 Dec 2021 02:21:07 No.13929229 Report

Quoted By:

>>13929196
If there wasn’t info on this site that sometimes isn’t very well broadcasted, that unfortunately shows up on /pol/ the maximum retard asshole board I’d probably stop coming here
But it’s like 1 in thousands of posts
I don’t have time for this shit honestly what am I doing here now?
Dodging out on doing shit, procrastinating and being lazy

For example I figured the euro vax qr pass was forgable because if you’re a comp sci moron you would guess so

But why wade through all the racist shit for someone to finally kick out the info
Not like it matters anyway because it’s just public/private key shit and you really want that private key without it being shared with every retard on tor and whatnot

The best way is to get inside somewhere as a legit employee and then you already have to take the tardjuice

Fuck it that’s right I’ll disrupt the thread
God give me strength to leave this timesink forever

The internet sucks now anyway the early 2000s are done and gone
Reeeeee

Anonymous

Anonymous Mon 06 Dec 2021 02:23:42 No.13929239 Report

Quoted By:

>>13929216
You’re welcome good luck with your studies

Anonymous

Anonymous Mon 06 Dec 2021 02:41:17 No.13929294 Report

Quoted By:

>>13927580
Dear someone smarter than me,

Can’t you just add like p different possible rotations and q different possible landing places relative the the center of any given square and find what the probability converges to as you increase q and p? Someone do it please

Anonymous

Anonymous Mon 06 Dec 2021 02:41:59 No.13929295 Report

Quoted By: >>13929327

>>13927580
This is super interesting! Here is my reasoning so far :

First of all, for the square to cover two corner it require it rotation to be of 0' and for it to cover all four corner it need a rotation of 0' and a specific (x,y) center. Since we are in the world of the infinite here we can discard these scenario since the probably of them happening is 0.

If the square lend exactly in the middle of a corner, it does not matter what the rotation is. There is a link between what % of the rotation allow for us to cover a corner and how far the center of our square is from the nearest corner.

Anonymous

Anonymous Mon 06 Dec 2021 02:51:59 No.13929327 Report

Quoted By: >>13929335

>>13929295
Not OP but if you are intrigued, check out Buffon's needle problem; this is reminiscent of that.

Anonymous

Anonymous Mon 06 Dec 2021 02:56:04 No.13929335 Report

Quoted By:

>>13929327
btw, Buffon's needle problem also has a quantity-over-pi kind of solution.

Anonymous

Anonymous Mon 06 Dec 2021 09:28:01 No.13930194 Report

Quoted By:

>>13927580
The problem can be reduced to
>What is the probability of dropping a stick on two corners of a square, given that every throw guarantees that the stick at least partially lands inside the square

This can further be reduced to
>What is the probability of dropping a stick of length L right between two parallel line of distance L

Anonymous

View Same Google iqdb SauceNAO square.png, 76KiB, 863x2271

Anonymous Mon 06 Dec 2021 09:30:13 No.13930203 Report

Quoted By:

The problem can be reduced to
>What is the probability of dropping a stick on two corners of a square, given that every throw guarantees that the stick at least partially lands inside the square

This can further be reduced to
>What is the probability of dropping a stick of length L right between two parallel line of distance L

Anonymous

Anonymous Mon 06 Dec 2021 21:04:07 No.13932215 Report

Quoted By: >>13932218 >>13932241 >>13933596

>>13928489
>>13929019
Let s be the length of the sides of the tiles. Wherever the dropped square lands, its center will lie in a tile. Every tile is the same, so we can restrict the possible positions of the dropped square to within a single tile -- [-s/2, s/2] x [-s/2, s/2] -- and generalize whatever answer we get for a single tile to the whole checkerboard.

We denote by Sq(c, ?) a square of side length s centered at point c and rotated counter-clockwise by ? radians. Given a point P and angle ?, the set of points Q such that Sq(Q, ?) contains P is Sq(P, ?). The dropped square will not cover a corner only if its center does not lie in Sq(P, ?) for corners P. So we need to find the area of the green figure (pic related). The green figure must be a square centered at (0, 0) due to symmetries.

To find the area of the green square, we first find the length of the red line segment. This line segment starts at (0, 0) and is angled from the right by ?, so it is perpendicular to the light blue line. The light blue line can be represented by cos(?)(x-s/2) + sin(?)(y-s/2) = -s/2. Or equivalently, cos(?)x + sin(?)y = (s/2)(cos(?) + sin(?) - 1). So the length of the red line segment is (s/2)(cos(?) + sin(?) - 1). The area of the green square is then s^2 (cos(?) + sin(?) - 1)^2 = 2s^2 (1 + cos(?)sin(?) - cos(?) - sin(?)). Divide by the area of a tile s^2 to get the probability that a square rotated ? radians will cover a corner. Then integrate d? from 0 to ?/4 and divide by ?/4 to get the probability a square of any orientation will cover a corner. Subtract that from one to get the answer, 6/?-1.

Anonymous

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Anonymous Mon 06 Dec 2021 21:05:09 No.13932218 Report

Quoted By: >>13933596

>>13932215
Shit, forgot pic related

Anonymous

Anonymous Mon 06 Dec 2021 21:09:34 No.13932241 Report

Quoted By:

>>13932215
>probability a square of any orientation will cover a corner
will not* cover a corner

Anonymous

Anonymous Tue 07 Dec 2021 05:35:33 No.13933515 Report

Quoted By:

>>13928338
Based Buffon-needle poster

Anonymous

Anonymous Tue 07 Dec 2021 06:11:21 No.13933596 Report

Quoted By: >>13934171 >>13934193

>>13932215
>>13932218
Impressive. I don't get how you computed the red length though.

Anonymous

Anonymous Tue 07 Dec 2021 06:16:04 No.13933606 Report

Quoted By:

>>13927580
Depends on how you drop it. You cannot put a number on such a variable unless it's defined- then we have the control and are able to make a sound calculation. Maybe I'm wrong but there's a lot more to the equation and I have questions

Anonymous

Anonymous Tue 07 Dec 2021 06:28:39 No.13933626 Report

Quoted By:

>>13927580
8/?=0

Anonymous

Anonymous Tue 07 Dec 2021 10:50:31 No.13934171 Report

Quoted By:

>>13933596
For a line ax + by = c, if a^2 + b^2 = 1 then c is the perpendicular distance between (0, 0) and the line.

Anonymous

Anonymous Tue 07 Dec 2021 10:56:09 No.13934193 Report

Quoted By:

>>13933596
For a line ax + by = c, if a^2 + b^2 = 1 then |c| is the perpendicular distance between (0, 0) and the line.

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