Hey bro calm down that's called Montecarlo and it's a procedure literally everybody knows and my computer science teacher gave me this shit First day of second year High school
There have onther way to do so, and generate one point only use the random function once : ) 1: Take 1 as fixed radius, Hand choose a circle angle N ( any angle value with in (0, 180] ) Loop: (Reapet 2~3, get as many point as possible) 2: Use uniformly distributed random function generate number 0~1 and multiply it by N as random angle 3: Draw the point on polar coordinate system, format: (radius, angle) 4: Calculate the distance between two adjacent points and add them up to approximate Arc length ("adjacent " can be done easily by ordering the x, y value of coordinate) 5: Because: Arc length = N/180·Ïr, and r = 1; so: Ï â approximate Arc length * 180 / N More point we have, the smaller N we use, better the approximation we get, as this kind of prerequisite the only thing we can do is approximation.
It was actually really funny when he was 'acting' to be smart saying 'oh, you didn't know the formula to the area of a circle?' and then the next shot is a view of his computer screen looking up the area of the circle. It was clever how he did that transition ha.
To be honest, If someone applies for data scientist positions and doesn't even know the basic Monte Carlo, then he should definitely walk away. So many ML/DL junkies wanna be a data scientist these days, can't even pass a basic statistics test...
honestly, sometimes i find it really hard to tell whether joma is actually being serious to straight up trolling. Cuz him having a serious face can easily fool someone gullible like me.
Please do more of these 10 minute videos with interesting questions and their answers! This would be a great format to learners. As soon as you posed the question, I hopped on pycharm and coded it up! It was fun!
There is another way (more computing intensive though), you can transform the numbers into integer and check if they are coprime. The probability that two integers are coprime is equal to 6/Ï^2. And boom, you have a solution that is way longer to run
@@kolbstar the random generator is already ridiculously overboard, unnecessary and slow for pi calculation. So the root comment is just upping the game. And I am sure there are ways to calculate pi that are even slower. There HAS to be a competition for the slowest algorithm.
@@StrangerHappened I planted a tree in my back yard. I will wait for it to grow to maturity, cut it down, dry it, put it on a lathe to make a rod, then measure its circumference and diameter.
how bout - Using a presentation of a circle (center (0,0), r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2) - Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle - Calculate Pi from that area This way (if works) will require 0 call for random(), and gives you constant runtime + better accuracy.
My teacher inserted that in an exercise for the test... It was cool, but later I discovered that he copied everything in the tests from forums... So he knew only few things... Not as teacher at the University
@@jarolrivera640 I suppose you're from the US? Don't know how it is there, I know PhD students also teach but didn't know they're called professors too. I'm in the UK, and here only a fraction of PhD holders even have a chance of possibly becoming professor. It's the highest "rank" you can have in the academic ladder, and even that's not tenured. If you don't produce results in a certain time, you'll be asked to leave. Though by that time, you're probably quite crazy about the field you would probably produce results. With regard to lecturing, at research intensive universities would normally be working for at least 5-10 years or so to be a lecturer. Not sure if this is a good setup but that's how it is here.
Fun fact is that you don't need the random function to compute with this technique. Just using points on the grid would result the same. E.g. double loop for x and y. The finer the grid - the more precise is the result.
That's what I was thinking. Regular grids should even be slightly more accurate than random numbers. But be careful to exclude two of the four borders. Starting off with random numbers is probably just to confuse applicants, or test their ability to generalize and simplify.
yea but the point of the original question is to find pi with a bunch of seemingly ranodm numbers. if the question mentioned a grid, its a bit easier to come up with the solution.
I think itâs probably a habit from working on bigger projects, where a single letter variable isnât really a good idea if multiple people work on it
@@icaruscorp1529 but its a waste of time. You are passing n to the function and also counting it in side the loop, it adds extra computation for literally no reason
I look at coding interview questions everytime I wanna feel stupid Edit: I learned python, javascript,and c++ and algos still make me feel stupid đ.
tbh most usual interview questions are easy at companies outside of FANG and where the bosses understand anything about engineering. (Had some during the last year)
Dude... this is the second video of yours I've watched, and so far they're both hilarious. I genuinely thought you were calling me an idiot, and then the next cut shows the area of a circle Google results on screen hahaha.
My first thought was to use the randomly generated points to do numerical integration. I had to first generate points, then sort them, then integrate numerically using trapezoidal sums (average of upper and lower sums). Since i used a simple sorting algorithm that would be in n^2 time it doesnât seem to be a very efficient solution. I like the way you did it without getting overly complicated. Iâve spent so much time doing heavy pure maths proofs so my brain must have been stuck up there and missed the easier solution. Good job!
Hey btw if anyone is trying to plot the points in the circle graph like I did, remember that the points range from 0-1. So only in the upper right quadrant. If you want to plot them in a full circle, you have to change the range from -1 to 1.
in python: use command help('modules') to see all modules installed, use command help(module) to see documentation, use command dir(module) to see available commands
This is one of the first cases of Monte Carlo you learn! Very simple and elegant solution to estimate Pi. The funny thing is that you can do the same calculation using a cardboard, a baby powder and a scale!
The fact that I'm doing my homework and its a very similar problem to this, so I'm using it for a reference. So glad I watched this earlier, great content man!
I like the content and appreciate how you guide through the act of creation in a way that shows quick and easy how to connect dots, where to look and how to improve, and all in a funny, entertaining and overall very polished looking way. I feel betrusted in my intellect but yet not overestimated :D First content I saw of yours was the apple-type blinds commercial on your other channel.
Better solution without the random function (could suggest both solutions, although they never said use the random function anyway) Calculate the area under the first quadrant of the unit circle curve using a riemann sum and multiply it by 4 x = np.linspace(0,1,1000) pi_approx = sum(np.sqrt(1-x**2)*x[1])*4 Also for the solution using the random function just create arrays of random numbers and perform the operations once instead of looping: x = random(n) y = random(n) pi_approx = (sum(x**2+y**2
When you're finished, you can show them another way and name-drop Euler for extra credit. double x = 0; for(int i = 1; i < some_limit; i++) { x += 1 / (i * i) } print(sqrt(x * 6)); // fixed by Eutu Salli The other, modern, fast formulas (Ramunajan and Chudnovsky) include polynomials and factorials. This would either require using libraries or writing your own functions.
how bout - Using a presentation of a circle (center (0,0) r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2) - Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle - Calculate Pi from that area This way (if works) will require 0 call for random(), and gives you constant runtime.
@@thaituanly8955 the integral of sqrt(r^2 - x^2)dx is 0.5 * (x * sqrt(r^2 - x^2) + r^2 * arctan(x / sqrt(r^2 - x^2))) Assuming you can pull this formula out of your head or derive it (I cannot), the arctan() function becomes a problem. If you can't code an arctan function, you'll need to rely on a math library. If you can access the math library, looking up pi will be far faster. Converting to the polar coordinate system creates a far easier integration and avoids arctan, but then you're integrating in radians which require the use of pi.
@@thaituanly8955 you can do a Reimann sum where you accumulated the areas of very thin rectangles from 0 to 1 whose heights are the sqrt(1 - x.position^2). The issue with that is the same issue with the Monte Carlo simulation in that you're calling a sqrt function n times which is slow.
Are you sure your formula there is correct? I implemented your solution and got results pretty far from pi (around 7,7). I'm pretty sure I triple checked the implementation to be correct.
@@etusax you are correct. It should be sqrt(x * 6). I posted a link to a working implementation on an online compiler/debugger, but CZcams deleted the post. I must be on a naughty list.
I really love learning from this channel, it makes learning complicated stuff fun, with a perfect blend of humour. Also, I really like that Patrick, let the artist know it's great work, whoever it might be.
I swear I laugh the most watching your videos. So many programming/coding/tech channels try to be funny and itâs just awkward. But joma always delivers
Determining whether random points fall in a circle seems more convoluted than it needs to be. A random point between 0 and 1 used as just one x or y component is sufficient since you can then derive the other component using pythagorean theorem and do 1 - the value squared and take the square root of that. Then you can use the property that triangles have pi radians and derive all the angles in radians from the values using simple trig. Sum them together, and you have pi.
Interesting question, thanks! This took me way longer to program than it should have (nearly 30min in C). I did learn that floats can get less precise as the numbers get more massive (which I somehow forget; it's obvious), and to use doubles in this case. My INT_MAX iterations got me to 3.141575.
Hey! I wanted to add a little side note for those that are interested in the maths behind it. Lets imagine that we are throwing arrows at a square of side length 1 (Lets call it "A"). Then the Probability of our arrow landing in the Part of the circle that intersects the square (Lets call it "C") is: Area(C) / Area(A) = pi/4 / 1 = pi/4. Now let X_k be a random variable, with X_k : [0,1)^2 -> {0 , 1} | (x, y) -> { 1, if |(x,y)| < 1 ; 0 in any other case We therefore have: P(X_k = 1) = pi/4, P(X_k = 0) = 1 - pi/4 => Expected Value: E(X_k) = P(X_k = 1) * 1 + P(X_k = 0) * 0 = pi/4 X_k represents our k-th throw. You get 1$ if your arrow lands in the circle and 0$ if it doesnt. We now use the "Weak law of large numbers" which states that 1/N * ( sum(X_k) from 1 to N ) -> E(X_k). Here we are just taking the mean of all our payments gained from throwing arrows. This converges (in probability) towards E(X_k) = pi/4. By using the markov-inequality we can even calculate the minimum number of throws we need to have a certain probality that we are closer than a>0 to pi, for any a>0.
You should use central limit theorem for more accuracy. You can do it while using n=100, and run it a few times let's say m times. The average of your m data would be a better estimator for pi
I actually understood this omg. Thanks Joma!!! thanks for the vid might also add that youre âweirdlyâ entertaining like you know how to do it somehow
you could also use the Central Limit Theorem and the fact that there's pi in the PDF of the Normal Distribution. actually there's a huge no. of places pi shows up in math
Here for area of circle you have to r/2 as a radius and r as a side of square, for randomizing the number from 0 to 1 , otherwise you have to randomize -1 to 1=> for clarity đ€
Why no one is talking about how funny at the same time educational his videos aređđđđđđđđđ !!!! You are doing a great job JOMA âšâšâšâšđ„đ„đ„đ„đ„
Very cool algorithm. Messing around with this a bit, and implemented it in C++ and C# as well. The results end up being very close to the actual value of Pi, just as shown here in the video. The cool this is that the C# version executes just about as fast as the C++ version does. I guess that's to be expected though, given that the heavy-lifting is done in the loop.
One day interview questions and the job at hand might become more in sync (Exp: 6 years at FANG as SoftwareDev and it mostly all been about managing complexity from code to large systems, not algorithm development)
Supposedly, they should be near impossible questions that aren't realisitically solvable within the confines of an interview, and the purpose is to show the approach and problem solving of the interviewee. A kobayashi maru, perhaps. Obviously, it doesn't quite work that way.
when he started typing "p" in google all the search results and auto texts were blurred and the titles are long and there are dashes and xs and stuffs.... đ just kidding. nice video by the way cheers!
I worked as a janitor at the college next to where I lived they had this on the blackboard unfinished and I solved it as well as all of the other equations left on the blackboard.
Check out my second channel: czcams.com/channels/WhwybsQg-EeBJpsqtmbscw.html for snackable educational content
Joma Tech what laptop or screen are you using to draw? Ipad? What might you recommend to do the same thing using linux/windows machines?
God damn, man I found your channel today, I loved it the way that you put some funny things in videos, congrats! I thxs 4 this tip
Hey bro calm down that's called Montecarlo and it's a procedure literally everybody knows and my computer science teacher gave me this shit First day of second year High school
There have onther way to do so, and generate one point only use the random function once : )
1: Take 1 as fixed radius, Hand choose a circle angle N ( any angle value with in (0, 180] )
Loop: (Reapet 2~3, get as many point as possible)
2: Use uniformly distributed random function generate number 0~1 and multiply it by N as random angle
3: Draw the point on polar coordinate system, format: (radius, angle)
4: Calculate the distance between two adjacent points and add them up to approximate Arc length ("adjacent " can be done easily by ordering the x, y value of coordinate)
5: Because: Arc length = N/180·Ïr, and r = 1; so: Ï â approximate Arc length * 180 / N
More point we have, the smaller N we use, better the approximation we get, as this kind of prerequisite the only thing we can do is approximation.
num_point_total will be n đ. You should try with 3 or more dimensions. Monte Carlo is made for bigger dimensions.
My favorite interview question is
*âWhatâs your name?â*
and i sometimes get it wrong
Thats a pretty hard question
Such a tough question.
Too bad. My name is harder to pronounce.
its my favorite as well, but it can get difficult at times
They asked me this at my interview for "assistant to the cashier" at McDonald's. I aced it.
Lmao the whole interview should be can you count and stand a lot?
Damn you're a cashier's assistant!? Jealous!
@@steelfalconx2000 assistant to the cashier
@@vancekangyishu Dwight.
He is kidding ,he actually knows it
I feel like he only made this video to flex on us, especially his Patrick drawing skills
He does that in every video
@@aaryashah2462 true lol
If you have been watching his videos you would know he is a major flexerđbut the good kind
Man, i am so dumb.
bruh. the boy is not humble and i love it.
It was actually really funny when he was 'acting' to be smart saying 'oh, you didn't know the formula to the area of a circle?' and then the next shot is a view of his computer screen looking up the area of the circle. It was clever how he did that transition ha.
I didn't even realise đđđ
It was so well done that I didn't even notice it
đđđ
What the- đđđ didn't realize that!
I was about to call you captain obvious but seems like many people didnÂŽt realize it so the comment is fine.
"Fuckin' idiot..." made me almost cry from laughing so hard. You bastard, gotta call me out like that xD
It was funny, he had the equation looked up in google on his screen
Through jokes its clear who's full of shit.
Yeah I felt offended too, but also I laughed like never before.
@Nathan Magro hahahahahaha wth
Long time I didnât laugh with yt video
3:04 I needed to pause the video because I was laughing too hard
Same, lol
Same here đđđđđ
I didn't even realize at first. That's amazing
@Haze The Space Commie Look at his monitor at 3:04 ...
Same đ€Ł
I felt like I was being humiliated this whole time
lol same here
When you realize he actually searched up for circle area formula. đđđ
Did you like it? "50 shades of Joma Tech"-like?
Bro đđ
I thought he wasn't serious when he said "calculate Ï"
3:04 googling the formula lmao, nice touch
lol
"Oh U didn't know? Fucking idiot"
Pointing out the obvious makes it unfunny.
He got the joke boys, pack it up.
@@l.1244 I didn't pay attention to that detail so his comment's still relevant
7:32 Actually 2.92 is closer to Pi than 3.6
thats called the BrAiN GaMeS
Oh you didn't know? *chuckles* idiot.
@@Luis-Torres "Google: What is more close to pi, 2.92 or 3.6????????????????"
I saw that too. Xd
damn brilliant
If someone asks me this kind of question, I would just walk away...
Facts bro đđđ
To be honest,
If someone applies for data scientist positions and doesn't even know the basic Monte Carlo, then he should definitely walk away. So many ML/DL junkies wanna be a data scientist these days, can't even pass a basic statistics test...
@@faizanali193 who the hell claims themselves DS guy if they don't know Monte Carlo? i mean, you could do it without any coding skill.
3:01 my math teacher every time I ask him something đđđ
Lol, I'm a math teacher. đđ
honestly, sometimes i find it really hard to tell whether joma is actually being serious to straight up trolling. Cuz him having a serious face can easily fool someone gullible like me.
That's cuz ur stupid it's ok
@@barryrims9098 its ok i know u aint got friends
@@choiceindot its ok i know your a low energy human being trying to find the meaning of your existence ill just give it to you
@@choiceindot your actually an idiot lol
@@choiceindot you implicitly admitted you`re an idiot lol nice job idiot
This dude is flexed so hard that youtube had no other choice but to recommend this to everyone
Please do more of these 10 minute videos with interesting questions and their answers! This would be a great format to learners. As soon as you posed the question, I hopped on pycharm and coded it up! It was fun!
There is another way (more computing intensive though), you can transform the numbers into integer and check if they are coprime. The probability that two integers are coprime is equal to 6/Ï^2. And boom, you have a solution that is way longer to run
That's using a lot more than just a uniform random variable generator. You're using multiplication, square root, and the definition of coprime.
@@kolbstar the random generator is already ridiculously overboard, unnecessary and slow for pi calculation. So the root comment is just upping the game. And I am sure there are ways to calculate pi that are even slower. There HAS to be a competition for the slowest algorithm.
@@StrangerHappened I planted a tree in my back yard. I will wait for it to grow to maturity, cut it down, dry it, put it on a lathe to make a rod, then measure its circumference and diameter.
how bout
- Using a presentation of a circle (center (0,0), r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2)
- Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle
- Calculate Pi from that area
This way (if works) will require 0 call for random(), and gives you constant runtime + better accuracy.
@@StrangerHappened thatâs not true the random generator is a pretty damn good empirical way of calculating pi.
That's like the very first example any professor would show when teaching about Monte Carlo.
My teacher inserted that in an exercise for the test... It was cool, but later I discovered that he copied everything in the tests from forums... So he knew only few things... Not as teacher at the University
@@TrioLOLGamers at least here in Spain you have to be a doctor to be a professor
@@CamaradaArdi really?
@@CamaradaArdi Isn't that how it is everywhere?
@@jarolrivera640 I suppose you're from the US? Don't know how it is there, I know PhD students also teach but didn't know they're called professors too. I'm in the UK, and here only a fraction of PhD holders even have a chance of possibly becoming professor. It's the highest "rank" you can have in the academic ladder, and even that's not tenured. If you don't produce results in a certain time, you'll be asked to leave. Though by that time, you're probably quite crazy about the field you would probably produce results. With regard to lecturing, at research intensive universities would normally be working for at least 5-10 years or so to be a lecturer. Not sure if this is a good setup but that's how it is here.
5:32 that googling google to go to Google killed me đ€Łđ€Łđ€Łđ€Ł
Fun fact is that you don't need the random function to compute with this technique.
Just using points on the grid would result the same. E.g. double loop for x and y.
The finer the grid - the more precise is the result.
That's what I was thinking. Regular grids should even be slightly more accurate than random numbers. But be careful to exclude two of the four borders. Starting off with random numbers is probably just to confuse applicants, or test their ability to generalize and simplify.
yea but the point of the original question is to find pi with a bunch of seemingly ranodm numbers. if the question mentioned a grid, its a bit easier to come up with the solution.
you would have to deal with edge cases like what if the point was right on the line, but that's probably a trivial solution to it
1:40 to 1:43 can cure depression, end wars and achieve world peace
don't believe me? click it back to back
@@AashishMishra_Nygma shure i do
Joma: *gives function total number of points as n*
Joma: *proceeds to calculate total number of points as num_point_total*
@Light Myst :v Then why won't he name the func param like that. It's simply because of the order of his thoughts.
I think itâs probably a habit from working on bigger projects, where a single letter variable isnât really a good idea if multiple people work on it
Legit happens to me
@@icaruscorp1529 but its a waste of time. You are passing n to the function and also counting it in side the loop, it adds extra computation for literally no reason
you got me b*tch!
2:58 The greatest sense of humour I have ever seen!
This was literally one of the exam question in a programming and microcontroller course I took this semester.
More of these pls I love them, also the coding explanations really help beginners. Love u joma
Hey Joma! I really loved this content, gonna share with my friends to blow their minds.
Omg, i love this guy!!
I'm a data science major and I am so happy I stumbled across your channel. I'm gonna sign up with your services.
8:40 you could also calculate root mean square of M estimates of size N, and get better precision while maintaining O(n + m) complexity
This was a outstanding video. Thanks for making it. I'm just starting to learn Python so appreciate the coding segment in this video as well.
I was going to comment: I could learn this all day if you would teach it.
And then joma academia.
Nice thanks
This is the classic Monte Carlo introduction problem. Ptsd to my thesis days.
âNow calculate the mass of the sunâ đđ I died
I look at coding interview questions everytime I wanna feel stupid
Edit: I learned python, javascript,and c++ and algos still make me feel stupid đ.
tbh most usual interview questions are easy at companies outside of FANG and where the bosses understand anything about engineering. (Had some during the last year)
Lmaoo
Lol
@@DrRAZI99 glad that made you giggle đ
Its ok. Noone can know everything.
Dude... this is the second video of yours I've watched, and so far they're both hilarious. I genuinely thought you were calling me an idiot, and then the next cut shows the area of a circle Google results on screen hahaha.
My first thought was to use the randomly generated points to do numerical integration. I had to first generate points, then sort them, then integrate numerically using trapezoidal sums (average of upper and lower sums). Since i used a simple sorting algorithm that would be in n^2 time it doesnât seem to be a very efficient solution. I like the way you did it without getting overly complicated. Iâve spent so much time doing heavy pure maths proofs so my brain must have been stuck up there and missed the easier solution. Good job!
how the fuck is this your first thought
Hey btw if anyone is trying to plot the points in the circle graph like I did, remember that the points range from 0-1. So only in the upper right quadrant. If you want to plot them in a full circle, you have to change the range from -1 to 1.
in python:
use command help('modules') to see all modules installed,
use command help(module) to see documentation,
use command dir(module) to see available commands
or just use google....
This is one of the first cases of Monte Carlo you learn! Very simple and elegant solution to estimate Pi. The funny thing is that you can do the same calculation using a cardboard, a baby powder and a scale!
The fact that I'm doing my homework and its a very similar problem to this, so I'm using it for a reference. So glad I watched this earlier, great content man!
I like the content and appreciate how you guide through the act of creation in a way that shows quick and easy how to connect dots, where to look and how to improve, and all in a funny, entertaining and overall very polished looking way. I feel betrusted in my intellect but yet not overestimated :D
First content I saw of yours was the apple-type blinds commercial on your other channel.
Better solution without the random function (could suggest both solutions, although they never said use the random function anyway)
Calculate the area under the first quadrant of the unit circle curve using a riemann sum and multiply it by 4
x = np.linspace(0,1,1000)
pi_approx = sum(np.sqrt(1-x**2)*x[1])*4
Also for the solution using the random function just create arrays of random numbers and perform the operations once instead of looping:
x = random(n)
y = random(n)
pi_approx = (sum(x**2+y**2
This is called the Montecarlo simulation
Wow, I remember that. I had it once in my high school classes. Good to know it can be an interview question
When you're finished, you can show them another way and name-drop Euler for extra credit.
double x = 0;
for(int i = 1; i < some_limit; i++) {
x += 1 / (i * i)
}
print(sqrt(x * 6)); // fixed by Eutu Salli
The other, modern, fast formulas (Ramunajan and Chudnovsky) include polynomials and factorials. This would either require using libraries or writing your own functions.
how bout
- Using a presentation of a circle (center (0,0) r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2)
- Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle
- Calculate Pi from that area
This way (if works) will require 0 call for random(), and gives you constant runtime.
@@thaituanly8955 the integral of sqrt(r^2 - x^2)dx is
0.5 * (x * sqrt(r^2 - x^2) + r^2 * arctan(x / sqrt(r^2 - x^2)))
Assuming you can pull this formula out of your head or derive it (I cannot), the arctan() function becomes a problem. If you can't code an arctan function, you'll need to rely on a math library. If you can access the math library, looking up pi will be far faster.
Converting to the polar coordinate system creates a far easier integration and avoids arctan, but then you're integrating in radians which require the use of pi.
@@thaituanly8955 you can do a Reimann sum where you accumulated the areas of very thin rectangles from 0 to 1 whose heights are the sqrt(1 - x.position^2).
The issue with that is the same issue with the Monte Carlo simulation in that you're calling a sqrt function n times which is slow.
Are you sure your formula there is correct? I implemented your solution and got results pretty far from pi (around 7,7). I'm pretty sure I triple checked the implementation to be correct.
@@etusax you are correct. It should be sqrt(x * 6). I posted a link to a working implementation on an online compiler/debugger, but CZcams deleted the post. I must be on a naughty list.
I really love learning from this channel, it makes learning complicated stuff fun, with a perfect blend of humour. Also, I really like that Patrick, let the artist know it's great work, whoever it might be.
Instead of using another variable incrementing for total points, you can directly use "n" which came as input argument.
lmao
This is super smart. Never would have thought of this.
Easily the funnest programming tutorial I've ever seen, great information too
5:27 - You went to google.com to google google and clicked on the first google link to take you to google. BIG BRAIN!
Teaching recursion for free. đ
I genuinely died when he searched up the area of the circle đđđ
Yes
why are you still alive then
@@thelastilluminati6959 because he was revived by the funny
I've done this on my TI graphing calculator back when I was in high school and it was a ton of fun learning about the Monte Carlo method.
first number: 3.6 (0,46 away)
second number: 2.92 (0,22 away)
Joma: "that is not more accurate"
What kind of program do you use to draw like that? Thank you you're awesome
I swear I laugh the most watching your videos. So many programming/coding/tech channels try to be funny and itâs just awkward. But joma always delivers
That logic of thinking with the graph is where the main clue is... Awesome idea
Determining whether random points fall in a circle seems more convoluted than it needs to be.
A random point between 0 and 1 used as just one x or y component is sufficient since you can then derive the other component using pythagorean theorem and do 1 - the value squared and take the square root of that.
Then you can use the property that triangles have pi radians and derive all the angles in radians from the values using simple trig.
Sum them together, and you have pi.
I don't even like math, but can't stop watching videos of this dude
Interesting question, thanks!
This took me way longer to program than it should have (nearly 30min in C). I did learn that floats can get less precise as the numbers get more massive (which I somehow forget; it's obvious), and to use doubles in this case. My INT_MAX iterations got me to 3.141575.
OMG! This was so useful and also very fun, thanks!
You could do the same thing with a smaller circle, checking the distance to 0.5, 0.5 instead of the origin and including points within 0.5 of it.
num_point_total is really useful( ânâ is useless)
Dude this man is my favourite...except he shows up on my interview day lamo..i'm dead...
Hey! I wanted to add a little side note for those that are interested in the maths behind it.
Lets imagine that we are throwing arrows at a square of side length 1 (Lets call it "A").
Then the Probability of our arrow landing in the Part of the circle that intersects the square (Lets call it "C") is: Area(C) / Area(A) = pi/4 / 1 = pi/4.
Now let X_k be a random variable, with X_k : [0,1)^2 -> {0 , 1} | (x, y) -> { 1, if |(x,y)| < 1 ; 0 in any other case
We therefore have: P(X_k = 1) = pi/4, P(X_k = 0) = 1 - pi/4 => Expected Value: E(X_k) = P(X_k = 1) * 1 + P(X_k = 0) * 0 = pi/4
X_k represents our k-th throw. You get 1$ if your arrow lands in the circle and 0$ if it doesnt.
We now use the "Weak law of large numbers" which states that 1/N * ( sum(X_k) from 1 to N ) -> E(X_k). Here we are just taking the mean of all our payments gained from throwing arrows. This converges (in probability) towards E(X_k) = pi/4.
By using the markov-inequality we can even calculate the minimum number of throws we need to have a certain probality that we are closer than a>0 to pi, for any a>0.
Your sense of humor is great!
The question doesn't even say you need to use the random function to solve it.
You can actually just divide by n
Yea oops
How? Could you please explain?
nb_loop = 10000000
nb_point = 0
x = 0
y = 0
for _ in range(nb_loop):
y = x
x = random.uniform(0,1)
if x**2 +y**2
Super fast way:
from numpy import sum, random as r
N = 10000 (your choice)
sum(r.rand(N)**2 + r.rand(N)**2 < 1) * 4.0 / N
It doesn't matter, I'm not going to write any of these calcs for my applications anyways
Your videos are awesome man!
This was very satisfying thanks.
6:08 when he typed in x^2 first before x**2 đ gets me every time
2:52 don't mind me, just wanted the golden moment of the video to replay
You should use central limit theorem for more accuracy. You can do it while using n=100, and run it a few times let's say m times. The average of your m data would be a better estimator for pi
I just love how many subtle things like
2:58 ,my brain after exams
3:04 I see what you did there I nearly died laughing LMFAO
They taught us how to do this at a programming boot camp I was at before starting uni. (the boot camp was organized by the university)
I actually understood this omg. Thanks Joma!!! thanks for the vid
might also add that youre âweirdlyâ entertaining like you know how to do it somehow
I was able to solve after you drew the graph
All the knowledge inside this piece of humor, I had to stop and laugh for every joke. Really well done đđŒ
you could also use the Central Limit Theorem and the fact that there's pi in the PDF of the Normal Distribution. actually there's a huge no. of places pi shows up in math
Here for area of circle you have to r/2 as a radius and r as a side of square, for randomizing the number from 0 to 1 , otherwise you have to randomize -1 to 1=> for clarity đ€
3:04 i lit laughed so hard... after all the bragging....lol
Why no one is talking about how funny at the same time educational his videos aređđđđđđđđđ !!!! You are doing a great job JOMA âšâšâšâšđ„đ„đ„đ„đ„
Watching your video, and following along makes me feel smarter lol
Thanks for the content
I wish all interview questions would be like this...
"YOUR NAME" background in monitor.
Jesus, you got me when searching in google for google lmao.. i spit in my cereal
Great video! This is a nice example of a Monte Carlo method, pretty cool :)
This is actually very comprehending for ultra smol brain like mine, much of a mentally striking question than a complicated one
I can see this as being a useful way to test RNG data sources, the closer they get to pi, the more random they are.
Very cool algorithm. Messing around with this a bit, and implemented it in C++ and C# as well. The results end up being very close to the actual value of Pi, just as shown here in the video. The cool this is that the C# version executes just about as fast as the C++ version does. I guess that's to be expected though, given that the heavy-lifting is done in the loop.
Based off of just drawing the quadrant with all the dots: you find the proportion of points where x^2 + y^2
This was great thanks!
Joma you are amazing, I laughed hard at 5:30 đđ
Amazing content too
agree with u đ
đđ
that's like genius. Especially if the interviewers are staring at you ........judging you
One day interview questions and the job at hand might become more in sync (Exp: 6 years at FANG as SoftwareDev and it mostly all been about managing complexity from code to large systems, not algorithm development)
You are really amazing.â€
Everything you do is just amazing.
These questions are good for boosting the ego of the interviewer, nothing more.
Supposedly, they should be near impossible questions that aren't realisitically solvable within the confines of an interview, and the purpose is to show the approach and problem solving of the interviewee. A kobayashi maru, perhaps. Obviously, it doesn't quite work that way.
It's a classical problem. making an approximation of pi using proba. here we used the monte carlo method.
@@jeremyshaw1 i solved it in 5 mins.
@@ignaciosavi7739 I solved it in 2
@@daniele4839 great job!
when he started typing "p" in google all the search results and auto texts were blurred and the titles are long and there are dashes and xs and stuffs.... đ just kidding. nice video by the way cheers!
Pork đđđđ
I worked as a janitor at the college next to where I lived they had this on the blackboard unfinished and I solved it as well as all of the other equations left on the blackboard.
I really love this kind of videos